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1. Stats help
A biology student wishes to assess the amount of calories in a certain brand of chocolate energy bars. A sample of bars is taken and the calories are 254, 305, 206, 550, 234, 451, 327, 234, 342, and 400 calories. Find a 90% confidence interval for the mean calorie content of this brand of energy bar, assuming an approximate normal distribution. Calculate the width of our interval and discuss whether or not our knowledge about is accurate?
Attempt n=10 x bar=330.3 s= 109.69 a=0.1 d.f. 9 ttable 1.89331
so Xbar +/- t*s/SQRT(n)
= 330.3-109.698
is it right?
Construct an upper 99% bound for the mean calories in the certain brand of chocolate energy bars for above given question . Explain what the bound means
2. Re: Stats help
[QUOTE=pankti_ptl;166436]A biology student wishes to assess the amount of calories in a certain brand of chocolate energy bars. A sample of bars is taken and the calories are 254, 305, 206, 550, 234, 451, 327, 234, 342, and 400 calories. Find a 90% confidence interval for the mean calorie content of this brand of energy bar, assuming an approximate normal distribution. Calculate the width of our interval and discuss whether or not our knowledge about is accurate?
Attempt n=10 x bar=330.3 s= 109.69 a=0.1 d.f. 9 ttable 1.89331
so Xbar +/- t*s/SQRT(n)
= 330.3-109.698
is it right?
/QUOTE]
If you meant 330.3 +/- one sigma for 90%, then no, that's incorrect since that only
covers 68% of the area (68% confidence level, not 90%). However, +/- one
sigma might be what the question means by "width of interval". It's unclear
to me.
You are asked if our knowledge is accurate. I was struck by the 550 data item
which sticks out like a sore thumb. So I did two plots of the PDF, one for the
data as given (white curve) and one with the 550 changed to 400 (blue curve).
The effect of the "sore thumb" on the variance and thus the on width of the bell
curve is enormous, as you can see. I think far more data needs to be collected
before we could claim "sufficient knowledge" (to create confidence levels we could
have any confidence in).
BTW, the SUMs are of all the PDF values from 0 to 1200. Notice that the white
curve left tail doesn't reach zero at X = 0 (left origin). This is reflected in a
small effect on the SUM for that plot (SUMs should equal 1). That sort of thing
can be handled but it's good to be aware of it.
Art
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# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2011 | June | Q#9
Hits: 917 Question a. Calculate the sum of all the even numbers from 2 to 100 inclusive, 2 + 4 + 6 + …… + 100 b. In the arithmetic series k + 2k + 3k + …… + 100 k is a positive integer and k is a factor of 100. i. Find, in terms of k, […]
# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2011 | June | Q#5
Hits: 34 Question A sequence is defined by , Where is a positive integer. a) Write down an expression for in terms of k. b) Show that c) i. Find in terms of k, I its simplest form. ii. Show that is divisible by 6. Solution a) We are given that sequence is defined by We are […]
# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2011 | January | Q#6
Hits: 104 Question An arithmetic sequence has first term a and common difference d. The sum of the first 10 terms of the sequence is 162. a. Show that 10a + 45d =162 Given also that the sixth term of the sequence is 17, b. write down a second equation in a and d, c. find the value of a and […]
# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2011 | January | Q#4
Hits: 10 Question A sequence is defined by Where is a constant. a) Find an expression for in terms of c. Given that , b) Find the value of c. Solution a) We are given that sequence is defined by We are required to find . We can utilize the given expression for general terms beyond first term […]
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Statistics > QUESTIONS & ANSWERS > HLT 362 Applied Statistics for Health Care Professionals Quiz 2. Grand Canyon University (All)
# HLT 362 Applied Statistics for Health Care Professionals Quiz 2. Grand Canyon University
### Document Content and Description Below
HLT-362V-Applied Statistics for Health Care Professionals Quiz 2 1. To obtain a sample of 20 patients in ICU, clinician goes to the ICU and selects the current patients. This is an example of a: A... . Judgement sampling B. Snowball sampling C. *Convenience sampling D. Simple random sampling 2. A Type I error is committed when _____. A. We do not reject a null hypothesis that is true. B. We do not reject a null hypothesis that is false. C. *We reject a null hypothesis that is true. D. We reject a null hypothesis that is false. 3. Which of the following would be an appropriate null hypothesis? A. The mean of a population is greater than 65. B. *The mean of a population is equal to 65. C. The mean of the sample is greater than 65. D. The mean of a sample is equal to 65. 4. Quantitative research strives for quality and the ability to apply the analysis to a broader population. This is referred to as _____. A. Normality B. Reliability C. *Generalization D. Validity 5. Scenario Based Question: If you were conducting a study of blood pressure readings in a hospital unit, compared AM and PM readings, and assumed the data were normally distributed and variances were equal, what type of statistical test would be conducted? A. F-test B. Paired t-test C. Pooled variance t-test D. Separate variance t-test 6. Which of the following can be reduced by proper interviewer training? A. Neither sampling error nor measurement error B. Sampling error C. *Both sampling error and measurement error D. Measurement error 7. A Type II error is committed when _____. A. *We reject a null hypothesis that is true. B. We do not reject a null hypothesis that is false. C. We do not reject a null hypothesis that is true. D. We reject a null hypothesis that is false. [Show More]
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## Sunday, January 10, 2010
### The Number 24
24 in its own right is a fascinating number.
Firstly it represents all the permutations of 4 (containing 4 elements) that can be made from 4 which is 4 * 3 * 2 * 1.
However there is another very interesting property that if we add up the squares of the consecutive numbers from 1 to 24 (inclusive) that the result 4900 will be the square of an exact whole number i.e (70).
This is the only case known where the sum of squares of successive natural numbers is equal to the square of another whole number!
Interestingly the sum of 1 + 2 + 3 +....+ 24 = 300, while the sum of the prime numbers between 1 and 24 = 100!
24 - as we shall see - plays a key role in Ramanujan functions, which in turn provides a direct link to the number of dimensions in one of the earlier string theories.
24 also plays a crucial role in the search for the Monster Group (the largest known symmetrical object) which again provides a direct connection with string theory.
As we know, if the proper divisors of a perfect number are summed, the total is the same number. So for example 1, 2 and 3 are the proper divisors of 6 and the sume of 1 + 2 + 3 = 6. So the ratio here of proper divisors to the number = 1
Now 24 is not a perfect number. However interestingly the ratio of its proper divisors to the number = 1.5.
Another highly interesting relationship to 24 that equally results in 1.5 is worth commenting on in greater length.
I mentioned in an earlier contribution that I had found a mathematical justification as to why the 1, 2, 4 and 8 dimensions play an especially important role in an overall integral approach.
Once again the holistic qualitative notion of dimension is directly linked to the quantitative structure of its corresponding root (of unity).
So 1 qualitatively is related directly to the 1st root (of unity) which is identical. Thus in effect, no clear distinction exists as between qualitative and quantitative interpretation with respect to the 1st dimension.
To obtain the nth root of any number (which in qualitative terms corresponds with its nth dimension) we simply obtain cos (360/n) + i sin (360/n)
The value of both cos and sin will range in absolute terms as between 0 and 1.
Then when we add both terms and square the result, the absolute value will range between 1 and 2.
Now when the root and dimension is 1 the value is 1 + i (0) = 1.
So here we have a maximum in terms of the real part representing the absolute dominance of rational understanding according to one dimensional i.e. linear interpretation. Equally we have a minimum in terms of the square of sum of sin and cos values = 1.
When root and dimension is 2 the value is cos 180 + i sin 180 = - 1 + i (0) = - 1.
Now we have a maximum in terms of the negative real part. This reflects in turn the dominance of negative linear (i.e. intuitive) understanding.
Equally again the square of the absolute sum of sin and cos values = 1 (which again is a minimum in terms of what can be achieved).
When root and dimension is 4 the value is cos 90 + i sin 90 = 0 + i (1) = i.
Now we have a maximum in terms of the positive imaginary part. This reflects in turn the dominance of imaginary linear (i.e. the indirect rational expression of intuitive) understanding.
Equally again the square of the absolute sum of cos and sin values = 1 (which again is a minimum in terms of what can be achieved).
Finally when root and dimension is 8, the value is cos 45 + i sin 45 = 1/(sq. root of 2) + i/(sq. root of 2).
Now we have a balanced equality in terms of real and imaginary parts. This reflects in turn the harmonisation of rational (conscious) and intuitive (unconscious) understanding.
Equally again the square of the absolute sum of cos and sin values = 2 (which now is a maximum in terms of what can be achieved, representing the limit in terms of pure contemplative integration of experience).
So what we see is that 1, 2, 4 and 8 dimensional interpretation represent specialised extremes in terms of rational (real), intuitive, rational (imaginary) and empty understanding (as the equality of conscious and unconscious) respectively.
One could perhaps suggest therefore that the ideal radial balance (in terms of overall understanding) should come at the midpoint between the two extreme values i.e. 1 and 2 (for the square of absolute sum of cos and sin values).
Now this in fact happens when n = 24.
Therefore from this perspective, 24-dimensional understanding represents the ideal in terms of overall balanced understanding.
Coming back to the previous post on Personality Types, this would suggest that for truly balanced radial understanding of reality, one would need to successfully combine attributes of all 24 Personality Types.
So when we look at development from a radial perspective , the key goal is to successfully differentiate (to a degree) traits associated with each Personality Type before then integrating all in a simultaneous manner.
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# Math Sequence
posted by .
the nth term of a series is 9n-7. what is the result of subtracting the kth term from the (k+1)th term?
(with solution pls, thnx)
• Math Sequence -
looks like we can just substitute and evaluate
term(n ) = 9n-7
term(k+1) = 9(k+1) - 7 = 9k + 2
term(k) = 9k - 7
so term(k+1) - term(k)
= 9k+2 - (9k-7) = 9
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Q: I am working on homework4 and came across a minor discrepency in the description. According to the contour map in section 3 it appears that the start point is at ~ (2.0,0.6,0.5) and the finsih is at ~ (2.4,0.7,0.5) (which is what I based my raw data plot on). However, later in the homework description in section 4.4.2 it references the location of food and start points oppositely.
Just wanted to know which data to use or if it matters.
A: The BOBposition data has the start and end points contained in it. The first point is the start, the last is the end, go ahead and base your analysis on those points.
Q: I think I understand what gradient descent does: if y=x^2, G.D. will take you to the point (0,0) which is the local (and global) minimum of the function, is this right?
Now, I think it makes sense to assume that in HW 4 G.D. does not take you to the bottom of the terrain, rather, it's minimizing something else.
Let's say it minimizes the energy, that would be the 'z' components of the terrain matrix.
But, if it's minimizing time, then, considering that we have no information about the time, nor anything like 't(x,y)', how can we minimize it? At first I thought that minimizing the time would be to take the straight line between the 2 points, but since the terrain is rather complex, the shortest path might be a path other than the straight line. So, how does G.D. minimize time?
And,
BOB's movement toward the food has a definite path: BOBposition matrix.Therefore there's nothing to be minimized to it. So my guess is that we're trying to compare BOB's path to the minimum energy (or time) path that we get through G.D. The one that has a smaller difference, is the one that BOB is optimizing (energy or time). Is this correct?
But then again if we can't minimize time, how can we compare it to BOB's path?
A: I think I can clear up your confusion by helping you see how we are applying gradient descent in this case. Think of gradient descent as similar to least squares - in this case we are using it for fitting a function to our data. We choose an equation with unknown parameters, then use gradient descent to fit that function to our data. So we might have y = ax + b, with a and b unknown, and we have data for x and y. We can use gradient descent to solve that problem quickly. Also, gradient descent has advantages when dealing with many many parameters. The minimum gradient descent tries to find is the minimum error between the fit and the data.
The equation we choose to attempt to fit is up to us. ie, I gave you the suggestions of the fits. You try to fit them (if you follow the steps you will do so), then whichever one fits better shows you which is more likely to be optimized - the equation for energy, or the equation for time.
For time, you DO have that information, if you read the description is states that time starts at 0, and increments at each data point by 10 seconds. You have the right instinct, if BOB moves straight towards the food (straight in an x,y sense, since he would have to climb the mountain to go straight toward the food), he's optimizing time, since it takes less steps to get there than to go around at a constant z.
The item you compare is just error between the equation you predict and the data for each equation (the linear, and the nonlinear one). No numerical value of time comparisons is needed, since it is not asked for. Remember also that you are only using G.D. on the linear case, for simplicity, and you'll use fminsearch for the nonlinear case. Just read the description carefully on that one, it's meant to tell you almost exactly each step to make in matlab.
Q: For problem 4.4.2, you said "beginning a new m-file such as BOBModel.m". What's the prupose of doing that? Is it served as a function to calculate Ax=b? Also, what should be inside the m-file function? Is it the gradient descend code only?
I tried to run the whole statement u provided , but get some thing like "Error: "res" was previously used as a variable, conflicting with its use here as the name of a function or command."
A: I was suggesting you put all the code into one file for your assignment, gradient descent and all. I fixed the link. Sorry about that, somehow the files were deleted on the server, perhaps due to the maintenance they were just doing. With the res error, watch out for a missing *, as it sounds like you have something like this: res( blah), and then matlab thinks you're calling a function.
Q: In the background info section to tell us that the food BOB is trying to get to is at a location (2.4, 0.7 , 0.5). I assume this is the actual point coordinates that you want us to use and not just a random example.
I then plotted the data using surf and plotted a starting point with the SAME altitude as the food (zpoint = 0.5) but different x and y values. I used the food location as the endpoint.
My problem: The starting point I plotted is above the terrain (like in the air) and the endpoint that you gave us for the food location is underneath the ground. This seems incorrect b/c BOB should be travelling on the actually terrain, not above and below it, and it is hard to show 2 points that are on different sides of the surface of the terrain. I could alter my starting point to be under the ground like the food point, but that still is not ON the terrain. If we are supposed to be using points exactly on the terrain, then how do we find those? If not, what am I doing wrong?
A: You're problem is just that you're using the wrong points. Use the two points given in the document as 'start' and 'location.' They are the same as the first and last point in BOBposition. Plot them as explained in the assignment document, using plot3 and the line type as '*'. Try to follow along with the code samples (type them in rather than copying and pasting).
Q: I'm having trouble in the computer lab right now with BOBModel. I can get it to run, and then I get 2 values for x. Then I switched the program to run for y.....and it will return 2 values for y. The problem is I'm not sure what to do with these points to have them 'fit the graph'. I was in class today but still am having problems...Help?
A: 1) Come to office hours at the muir woods coffee shop, 1-2 tomorrow for help. I'll also be coming by the CSB115 lab periodically during the day, and I may arrange a help session in the evening as well, but my schedule of the day isn't set yet.
2) those values you get are like the coefficients we got for the least squares homework. Those are the coefficients of the line. So the line is defined as
y_model= x(1) + x(2)*time
where x(1) and x(2) are the coefficients you computed, not the xposition values of the data. Then you have a model, and you can use that to compute y's by putting the time array into the equation I just wrote. The questions then ask you to compare those y's you calculated with the actual data values by E = norm(y_data - y_model);
Q:I understand that we are to plot the results in terms of (res_save), but i am unclear exactly how to go about that. When i enter
plot(res_save(1,:),'r');
a very strange line shows up my plot.
A: I think I can clear up your question. res_save is a record of the residual error in the approximated solution to Ax=b, ie r = Ax - b, and res = r^T * r (so you get one number). The smaller that difference is, the closer you are to solving Ax=b. This plot is not required unless explicitly mentioned in the homework, though you can include it.
In terms of plots, just give what was asked for in the questions. If you want, you might want to plot the actual data and the fit on the same plot (i.e. plug the time array into the equation for the line y = a(1) + a(2)*t, x = .... That will give you a 'sanity check' to see how close you are to fitting your data.
The error is determined quantitatively in this case as E=norm(y_data-y_model) + norm(x_data-x_model). That is your criterion for fitting the data. The residuals are for how close your gradient descent algorithm is to finding the unknown coefficients. It doesn't say anything about the actual fit to the data - if you had data moving up and down wildly, the best linear fit you could get might converge nicely to something very small for res_save, but the straight line can only fit the curved data so closely, res_save doesn't account for that, that is why we look at the difference at every point between the square of the data minus the model fit.
Please come to office hours to discuss it if there is any confusion. Also I'm going to try to arrange a help session for tomorrow night after 6:30 or so. My schedule is still in flux for tomorrow so I'll post something on the web site when I know.
Q: IM having alot of trouble figuring out how to visualize the gradient
decent graph. My code doesnt differ from the stuff that was given but i
dont know why i can't see the graph. Any segestions of points to other
referances?
A: What are you trying to plot is the question to ask yourself. Gradient descent is giving you the coefficients of a line fit like least squares, y = a(1) + a(2) * time, where you are determining a(1) and a(2) so that you fit the y data vs. time or x data vs. time the best possible way you can. So if you wanted to plot the fit vs. the data, you might want to do
plot(time, BOBposition(:,1) )
hold on;
xfit = x(1) + x(2)*time; %where x here is the coefficient you found in gradient descent...
plot(time, xfit,'r')
Please see chapter 5 of the numerical methods book posted on the website. You can also search for gradient descent, least squares, and other questions at www.mathworld.com since they have some useful definitions and short descriptions.
Please come to office hours 1-2 at muir woods tomorrow for more discussion in depth, other questions, etc. Also I'm trying to arrange an evening help session after 6:30pm tomorrow. I spent some time on this in lecture, and if you couldn't make it please get notes from someone as I gave many hints. If you did come and were still confused just come to office hours for direct discussion, or check on the web page for the help session information post.
Q: did you want us to use the points that you gave us in the beginning of the assignment as our endpoint?-- im talking about 2.4, 0.7,0.5
A: Just use the points as they appear in the code example lines. That's best. They are the same as the first and last data point in the BOBposition data.
Q: On the homework on section 4.4.4. I'm doing actually what you describe on the sheet and I keep getting the same errors as follows.
??? Undefined function or variable 'y'.
Error in ==> NLBOBFit>@(x) bobnonlinearfunction(x,t,y) at 9
[estimated_C, n,m, output ]= fminsearch(@(x)
bobnonlinearfunction(x,t,y),initialguess,options);
Error in ==> fminsearch at 175
fv(:,1) = funfcn(x,varargin{:});
Error in ==> NLBOBFit at 9
[estimated_C, n,m, output ]= fminsearch(@(x)
bobnonlinearfunction(x,t,y),initialguess,options);
A: y is the data, so I started this one off with passing in the y data for BOBposition - ie
y=BOBposition(:,2);
Then make the function call. That should clear up your error. Also be sure to clear the variables x and y from the gradient descent algorithm before calling this function.
Q: I'm at the end of 4.4.2 and I don't know exactly how to change matrix A to be for the y data. Ax=b needs to be changed, right? To By=something? What else do I need to change?
A: for the A matrix, when you build it, no change is needed, since time is the same either way.
A = [ones(size(BOBposition(:,1)) t'];
The b array is the change...
b= BOBposition(:,2); % so change the column to 2 instead of 1. That's it!!!
Q: So I did the gradient descent for fitting x, and the output in Matlab was:
A =
6 180
180 7920
Is that correct? If it is, great. If not, what's going on, I did exactly what the assignment asked
A: The A matrix doesn't change because that's what you physically entered, after you multiplied by the transpose of A. What you are looking for is the x that pops out of the gradient descent code. Look at that, and let me know what you see. Yes, computing the y coefficients is the simple change of that one little number:)
Q: Near the middle of page six you state that equations need to be created
for each x, y, and z variable, but after the gradient descent script it
says only y has to be done. I think I'm just confusing myself, because
we're supposed to be doing gradient descent in x and y in regards to z. So
gradient descent doesn't actually have to be done on z does it?
A: Yes you're just finding xdata and ydata fits, no need for z on the gradient descent code. Just find what it asks for. Basically if he's going straight for the food, he's not concerned about z, so we neglect fitting that in the linear case.
Q: For the gradient descent it says we need the start and location
variables, although I don't understand how they actually play into the
gradient decent. We just end up creating an A and b matrix, but never
actually use the start and location specifically. I was looking at the
"steepest_descent.m" code you provided, and it's a somewhat similar
situation. Are the start and location just erroneous?
A: The start and location variables don't actually come into play except for the plots (when plotting the terrain), so yes those aren't absolutely necessary in the GD code. It's just the first and last data point of BOBposition.
Q: The last question is that when I get the gradient descent script working,
the values returned for x and y are a 2 x 1 matrix. Shouldn't there only
be one value returned? Or is it the distance and direction of each
variable? (I pulled that from the lecture 14 notes).
A: For the x fit and y fit of GD, you are, for each, fitting a line for the relationship of xdata vs. time and ydata vs. time. The line is defined by the two coefficients, a(1) and a(2). In the GD code we use x as the coefficient name. So you find x(1) and x(2), and that corresponds to the coefficient you're trying to find for the line:
Y_model = x(1) + x(2)*time
or
X_model = x(1) + x(2)*time
You could just as easily replaced x with a in the GD code for clarity. So the x is the same as computing the a from linear least squares (in the sense of what it represents, not how you compute it):
x = A\b or equivalently a = A\b
I'm trying to break the habits of people getting stuck with intrinsic meanings for variables. It's just a space to fill. But I apologize for any confusion.
oh and the statement I made about the coefficients I just realized could be confusing in one regard specifically - realize there are different coefficients for the x_model fit and the y_model fit (i.e. each has a 2x1 vector that pops of of the GD code which can be different).
Q: In the linear fit, the output for x and y are 2X1 and the BOBposition(:,1)
and BOBposition(:,2) are 6X1 so the dimensions do agree. so the norm
cannot be computed since they dont agree.
A: The output of the linear fits (the 2x1 vectors, in this case I'm calling them a to avoid confusion right now) are the unknown coefficients of the line:
Y_model = a(1) + a(2)*time
Once you have those coefficients you have to put the time array t into the above equation and you get Y_model the size of Y_data. Then you do the norm on Y_model - Y_data. Does that make sense? So you find the line, then compute what the line predicts at each time value, and compare that to the actual recording.
So the x and y coming out of gradient descent, the ones that are 2x1, are not the same as the actual fitted data. They are the coefficients that determine the line fit. So once you have those coefficients you have to do what I said above and compute the fitted model.
Q: I got my gradient descent code to work, but I'm not completely sure what those numbers are though, they're the coefficients right?
A: Yes, coefficients of the line, you've got it,
y_model = a(1) + a(2)* time
So now that you have those, put the time through to compute y_model. Then you'll compare that with the y of the data using the E= norm(y-y_model) equation. The questions ask specifics about the calculations.
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A102785 G.f.: (x-1)/(-2*x^2+3*x^3+2*x-1). 0
1, 1, 0, 1, 5, 8, 9, 17, 40, 73, 117, 208, 401, 737, 1296, 2321, 4261, 7768, 13977, 25201, 45752, 83033, 150165, 271520, 491809, 891073, 1613088, 2919457, 5285957, 9572264, 17330985, 31375313, 56805448 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,5 COMMENTS Inverse binomial transform of A078017. Inversion of A052102. LINKS Index entries for linear recurrences with constant coefficients, signature (2,-2,3). FORMULA a(n+3) = 2a(n+2) - 2a(n+1) + 3a(n), a(0) = 1, a(1) = 1, a(2) = 0 a(n)=sum(k=1..n, sum(i=k..n, (sum(j=0..k, binomial(j,-3*k+2*j+i)*(-2)^(-3*k+2*j+i)*3^(k-j)*binomial(k,j)))*binomial(n+k-i-1,k-1))), n>0, a(0)=1. [From Vladimir Kruchinin, May 05 2011] PROG Floretion Algebra Multiplication Program, FAMP Code: 4jbasekseq[ (+ 'ii' + 'jj' + 'ij' + 'ji' + e)*x) ] where x is defined as 1/4 times the sum of all 16 floretion basis vectors. (Maxima) a(n):=sum(sum((sum(binomial(j, -3*k+2*j+i)*(-2)^(-3*k+2*j+i)*3^(k-j)*binomial(k, j), j, 0, k))*binomial(n+k-i-1, k-1), i, k, n), k, 1, n); [From Vladimir Kruchinin, May 05 2011] (Maxima) makelist(coeff(taylor((x-1)/(-2*x^2+3*x^3+2*x-1), x, 0, n), x, n), n, 0, 32); [Bruno Berselli, May 30 2011] CROSSREFS Cf. A078017, A052102, A077952. Sequence in context: A045221 A046287 A051220 * A260348 A276934 A127493 Adjacent sequences: A102782 A102783 A102784 * A102786 A102787 A102788 KEYWORD nonn AUTHOR Creighton Dement, Feb 11 2005 STATUS approved
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Question
# Let S denotes the set of all real values of x such that (x2016+1)(1+x2+x4+⋯+x2014)=2016x2015. Then
A
Number of elements in S is 2
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B
Number of elements in S is 1
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C
Point (x,2), xS lies inside the parabola Y=X22X+2, XR
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D
Image of the point (x,2) about the line y=x lies on the line x+y=4
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Solution
## The correct options are B Number of elements in S is 1 C Point (x,2), x∈S lies inside the parabola Y=X2−2X+2, X∈R (x2016+1)(1+x2+x4+⋯+x2014)=2016x2015⇒(x+1x2015)(1+x2+x4+⋯+x2014)=2016⇒x+1x2015+x3+1x2013+⋯+x2015+1x=2016⇒(x+1x)+(x3+1x3)+⋯+(x2015+1x2015)=2016 When x>0,x+1x≥2 When x<0,x+1x≤−2 So, L.H.S. = R.H.S. only if x=1 Putting the point (1,2) in the parabola, 2>1−2+2 So, the point (1,2) lies inside the parabola. Image of the point (1,2) about the line y=x is (2,1) which does not lie on the line x+y=4
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# Currency Forward Contracts
I am having a hard time with these types of FRA. I understand the convered interest arbitrage and which currency should be sold or bought but the steps are hard to follow. I don’t follow step 2, can someone please help explain? 1) If fwd price is higher and you sell at mkt price 2) buy 1/(1+rf)^T units of foreign ccy 3) hold position and earn interest 4) at maturity deliver the ccy and get paid the fwd price
steps 2 and 3 combined means you are ending up with 1 Unit of the foreign currency. do you see that? buy 1/(1+rf)^T units of foreign currency. It earns (1+rf)^t of Interest you end up with 1 unit of foreign currency. when you deliver that - in Step 4 - you end up with Forward currency contract amount for the 1 unit of FC.
Thanks CP…i got the unit of foreign ccy. So to calculate the rate of return from this arbitrage, you take the (fwd price/spot price)-1? And is spot price what we got from step 2? Can you please walk me through this question? For example p.54 from text book. Euro trades at \$1.0231, USD risk free is 4% and Euro risk free is 5%. 6 months forward contracts are quoted at \$1.0225. Indicate how you can earn a risk free profite and outline the steps.
Calculated Forward: 1.0231 /euro \* (1.04/1.05)^0.5 = 1.01822 /euro actual forward=1.0225\$ so forward contract is overpriced. So short sell the forward contract . borrow -\>buy Euro at Spot 1=>0.97742 Euro Lend it forward-> 0.97742 * 1.05^0.5 = 1.00156 Euro convert to at the forward rate -\> (real) 1.00156\*1.0225 = 1.02410 your 1\$ which you need to pay back = 1.00 * 1.04^0.5 = 1.0198 so your arbitrage profit = 1.0241-1.0198 = 0.0043 been a long time since I did this. pray tell me I am ok?
CP- i actually used 1eur and got the same answer as you but the book’s answer of return is not factoring in the cost of borrow. So they have (1.0225-.9988)=.0237 as arbitrage profit.
i had the same problem with this EOC and wondered why the arbitrage profit is not 1.0225-1.0182. 1.0182 is the “real” forward rate calculated from the interest rate parity. Can someone please help with this? Thanks!
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# How is the total cost of a car lease calculated?
Contents
In broad terms, you calculate a lease by determining and adding the depreciation fee, plus a monthly sales tax and a financing fee. If you’re looking to calculate your payment manually, here is the formula: Start with the sticker price (MSRP) of the car. Take the MSRP and multiply it by the residual percentage.
## What is the formula for leasing a car?
Here is what that would look like, using our money factor of 0.00125. Step 8. Add the rent charge to the payment you calculated in Step 6 to get your pretax lease payment.
Walk Through a Sample Lease.
Step
3. Equals the residual value = \$13,110
4. Negotiated selling price of car \$21,000
5. Add in fees + \$1,200
## What price is a car lease based on?
Based on the residual percentage, the bank estimates that your vehicle will be worth \$12,200 (\$20,000 × 0.61) at the end of your lease. This is your residual value. Note that we used the MSRP to calculate the residual value, not the selling price. Capitalized cost is the amount being financed.
IT IS IMPORTANT: What is volunteer accident insurance?
## How are finance charges calculated on a leased vehicle?
The money factor is applied to the sum of the net cap cost and the residual value of the car to find the monthly finance charge. Continuing with the example above, use the money factor 0.00333. Multiply this by the sum of the net cap cost and residual as follows: \$40,000 x 0.00333 = \$133.2.
## How is lease money factor calculated?
You can use the lease charge to calculate the money factor with this formula: Money Factor = Lease Charge / (Capitalized Cost * Residual Value) * Lease Term. Once you have the money factor, you can multiply it by 2,400 to convert it to an interest rate.
## How do you calculate lease factor?
Lease Rate Factor Calculation
The lease rate factor is the annual interest rate divided by the number of monthly payments. If the current interest rate is 6 percent, then the lease rate factor in our example is (0.06/60), or 0.0010.
## What is a good lease rate?
Use a rate between 2% and 5% if you have strong credit, between 6% and 9% for average credit and between 10% to 15% for poor credit. Length of the lease: Car leases usually last 36 months, which is how long most extended warranties last.
## How do you calculate lease buyout?
Look for a “buyout amount” or “payoff amount” that will be listed on your monthly leasing statement. This buyout amount is calculated by adding up the residual value of your vehicle at the beginning of the lease, the total remaining payments, and possibly a car purchase fee (depending on the leasing company.)
## How is lease residual value calculated?
Subtract the Depreciated Value from the Original Value
Look up the original value of the car in your lease terms or on the Kelley Blue Book website. Subtract the calculated depreciation value from the original value of the vehicle. This new result is the total residual value of the car.
## How is lease payoff amount calculated?
The payoff amount is calculated by considering the projected residual value of the car plus the amount that you still owe on it, including any interest. For example, if you were to lease a 2014 Buick Enclave 2WD for five years — 60 months — the projected residual value would be \$12,200 at the end of your lease.
## How do you negotiate a car lease?
4 tips for negotiating the best price on a car lease
1. Know the terminology. …
2. Research prices and deals. …
3. Shop multiple dealerships. …
4. Be open to other car models to find the best deal. …
5. Capitalized cost. …
6. Rent charge or money factor. …
7. Mileage allowance.
## How do I calculate a lease payment in Excel?
How to calculate lease payments using Excel in 5 steps
2. Step 2: Enter amounts in the Period and Cash columns. …
3. Step 3: Insert the PV function. …
4. Step 4: Enter the Rate, Nper Pmt and Fv. …
5. Step 5: Sum the Present Value column.
## What is a fair money factor for a car lease?
A decent money factor for a lessee with great credit is typically around 3% to 5%. If you have fantastic credit and you’re offered a lease with a money factor higher than . 0025 (or 6% APR) then it may be worth your time to shop around.
IT IS IMPORTANT: Frequent question: What is regular car insurance?
## What is a good residual value on a lease?
So when you’re shopping for a lease, the first rule of thumb is to look for cars that hold their value better — the ones that have high residual values. Residual percentages for 36-month leases tend to hover around 50 percent but can dip into the low 40s or be as high as the mid-60s.
## How do you calculate a monthly lease for a 3 year lease?
To figure your monthly payments, take the total financed amount of the lease (depreciation, plus taxes, interest, and fees) and divide it by the number of months.
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# A certain car traveled twice as many miles from Town A
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A certain car traveled twice as many miles from Town A [#permalink]
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21 Feb 2012, 13:58
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A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13
B. 13.5
C. 14
D. 14.5
E. 15
MGMAT 1 Q. 11
The explanation given is unclear to me, why weightage ave is not right to apply here? I want to understand conceptually. Can someone help!
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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21 Feb 2012, 14:15
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docabuzar wrote:
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13
B. 13.5
C. 14
D. 14.5
E. 15
MGMAT 1 Q. 11
The explanation given is unclear to me, why weightage ave is not right to apply here? I want to understand conceptually. Can someone help!
Average miles per gallon equals to total miles covered/total gallons used (miles/gallons), so if the distance from A to B is $$2x$$ miles and from B to C is $$x$$ miles then we'll have:
$$average \ miles \ per \ gallon=\frac{total \ miles \ covered}{total \ gallons \ used}=\frac{2x+x}{\frac{2x}{12}+\frac{x}{18}}=\frac{3x}{\frac{2x}{9}}=13.5$$.
P.S. As you can see weighted average concept is exactly what we used here.
Hope it's clear.
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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21 Feb 2012, 14:40
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Lets pick numbers:
Distance A to B: 36 miles (2x) --> 3 gallons
Distance B to C: 18 miles (x) --> 1 gallon
Average: $$\frac{(36+18)}{4}$$ = 13.5
Hence, B
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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21 Feb 2012, 20:58
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docabuzar wrote:
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13
B. 13.5
C. 14
D. 14.5
E. 15
MGMAT 1 Q. 11
The explanation given is unclear to me, why weightage ave is not right to apply here? I want to understand conceptually. Can someone help!
If the concept you do not understand is why (12*2 + 18*1)/3 doesn't work, here you go:
'Distances traveled' (i.e. ratio of 2:1) cannot be the weights here to find the average mileage. The weights have to be 'number of gallons'.
If we change the question and make it:
A certain car used [highlight]twice as many gallons[/highlight] from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
Now you can use (12*2 + 18 *1)/3
Why?
Whenever you are confused what the weights should be (e.g. here should the weights be the distance traveled or should they be gallons of fuel used...), look at the units.
Average required is $$\frac{miles}{gallon}$$. So you are trying to find the weighted average of two quantities whose units must be $$\frac{miles}{gallon}$$.
$$C_{avg} = \frac{C_1*W_1 + C_2*W_2}{{W_1 + W_2}}$$
$$C_{avg}, C_1, C_2 - \frac{miles}{gallon}$$
So $$W_1$$ and $$W_2$$ should be in gallon to get:
$$\frac{miles}{gallon} = (\frac{miles}{gallon}*gallon + \frac{miles}{gallon}*gallon)/(gallon + gallon)$$
Only if weights are in gallons, do we get 'Total miles' in the numerator and 'Total gallons' in the denominator.
We know that Average miles/gallon = Total miles/Total gallons
[highlight]Takeaway: The weights have to be the denominator units of the average.[/highlight]
So what do we do in this question?
What is the distance travelled (i.e.total miles)?
"traveled twice as many miles from Town A to Town B as it did from Town B to Town C"
We know that the ratio of the two distances must be 2:1 or we can say the distances must be 2d and d. Total distance must be (2d + d)
What is the total gallons used?
Fuel used to go from A to B = 2d/12
Fuel used to go from B to C = d/18
So Average miles/gallon = (2d + d)/(2d/12 + d/18)
You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 17 Jan 2012 Posts: 41 GMAT 1: 610 Q43 V31 Followers: 1 Kudos [?]: 146 [0], given: 16 Re: A certain car traveled twice as many miles from Town A [#permalink] ### Show Tags 22 Feb 2012, 03:10 Many Thanks. Just to make myself clearer, we can also say that: Case 1. Car moves A to B @ 12mph and covers 2 miles B to C @ 18 mph and covers 1 mile We want to know the total ave speed, we cant use Weighted averages : (12x2 + 18x1)/3 = 14 mph because the 2 quaintities i.e. Speed units denominator (i.e. hours) does not match with Distance denominator (i.e. nothing). Case 2. Car moves A to B @ 12mph for 2 hrs B to C @ 18 mph for 1 hr We want to know the total ave speed, we can use Weighted averages : (12x2 + 18x1)/3 = 14 mph But in case 2 the units denominators donot match, speed is mph & time is h? What I m missing here! Math Expert Joined: 02 Sep 2009 Posts: 36554 Followers: 7078 Kudos [?]: 93169 [0], given: 10553 Re: A certain car traveled twice as many miles from Town A [#permalink] ### Show Tags 22 Feb 2012, 03:50 docabuzar wrote: Many Thanks. Just to make myself clearer, we can also say that: Case 1. Car moves A to B @ 12mph and covers 2 miles B to C @ 18 mph and covers 1 mile We want to know the total ave speed, we cant use Weighted averages : (12x2 + 18x1)/3 = 14 mph because the 2 quaintities i.e. Speed units denominator (i.e. hours) does not match with Distance denominator (i.e. nothing). Case 2. Car moves A to B @ 12mph for 2 hrs B to C @ 18 mph for 1 hr We want to know the total ave speed, we can use Weighted averages : (12x2 + 18x1)/3 = 14 mph But in case 2 the units denominators donot match, speed is mph & time is h? What I m missing here! I think that the issue was over complicated above: {average rate}={total distance}/{total time}; {average salary}={total salary}/{total # of employees}; {average miles per gallon}={total miles}/{total gallons}; ... Keep it simple. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2134 Kudos [?]: 13648 [0], given: 222 Re: A certain car traveled twice as many miles from Town A [#permalink] ### Show Tags 22 Feb 2012, 04:30 docabuzar wrote: Many Thanks. Just to make myself clearer, we can also say that: Case 1. Car moves A to B @ 12mph and covers 2 miles B to C @ 18 mph and covers 1 mile We want to know the total ave speed, we cant use Weighted averages : (12x2 + 18x1)/3 = 14 mph because the 2 quaintities i.e. Speed units denominator (i.e. hours) does not match with Distance denominator (i.e. nothing). Case 2. Car moves A to B @ 12mph for 2 hrs B to C @ 18 mph for 1 hr We want to know the total ave speed, we can use Weighted averages : (12x2 + 18x1)/3 = 14 mph But in case 2 the units denominators donot match, speed is mph & time is h? What I m missing here! You are right. Case 1 doesn't work but case 2 does. What is the unit of the denominator in average speed? Average speed = miles/hour. So you can use weighted averages when the weights are time (i.e. in hours). That is the reason your case 2 works. (By the way, you yourself said speed is miles/hour and time is in hours. Why do you think the units don't match?) When you want to find weighted average speed, distance cannot be the weights, it has to be time. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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26 May 2013, 04:47
Bumping for review and further discussion.
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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25 Jul 2014, 07:02
Hello from the GMAT Club BumpBot!
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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15 Feb 2015, 14:43
Hello Experts!
I am quiet fascinated with The Bunuel approach... of using algebraic expression.
Kindly assess my approach which I will try next time onward to got a shortcut way
step 1) took LCM of 12 and 18.. came as 36. just multiplied by 10...(to make easy calculation)
step 2) 360 distance between B to C... do 360/18 hence 20 gallons used
step 3) twice distance.. hence 360*2= 720... do as above.. 720/12= 60 gallons used
step 4) total gallons.. 20+60= 80 gallons
step ) total miles= 360+720= 1080 miles
hence.. average of whole journey = 1080/80 which comes to 13.5
Thanks
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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06 May 2016, 13:10
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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12 Aug 2016, 16:13
Hi!
Kindly share your views on my above approach plz..
experts...
thanks
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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14 Aug 2016, 23:57
Celestial09 wrote:
Hello Experts!
I am quiet fascinated with The Bunuel approach... of using algebraic expression.
Kindly assess my approach which I will try next time onward to got a shortcut way
step 1) took LCM of 12 and 18.. came as 36. just multiplied by 10...(to make easy calculation)
step 2) 360 distance between B to C... do 360/18 hence 20 gallons used
step 3) twice distance.. hence 360*2= 720... do as above.. 720/12= 60 gallons used
step 4) total gallons.. 20+60= 80 gallons
step ) total miles= 360+720= 1080 miles
hence.. average of whole journey = 1080/80 which comes to 13.5
Thanks
Celestial
Kudos if it makes sense...
While you approach is not wrong, you have made lots of double and tedious calculations which is quite time consuming. You can do it without multiplying by 10
You got LCM 36. so we will considered total miles traveled under 18 Miles/gallon to be 36. Hence it took 2 gallons
Now, the distance traveled under 12M/ Gallon will be 72 and total gallons used will be 6.
therefore no. of gallons used traveling both route = 2+6 -= 8 and
total miles traveled is 36+72 = 108.
average = 108/8= 13.5 M/Gallon
its much more easier than you steps, which include unnecessary multiplication and double division. Its quite time consuming in GMAT.
Re: A certain car traveled twice as many miles from Town A [#permalink] 14 Aug 2016, 23:57
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here are my expectations for your week 4 paper
# here are my expectations for your week 4 paper - here are...
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here are my expectations for your week 4 paper: Page 1 - APA Cover Page Page 2 - Introductory Paragraph Who was Pythagoras? Pythagoras (c. 580–500 B.C.) was an ancient Greek philosopher who was interested in numbers and their meanings. He discovered the relationships between mathematics and music, proposing that sounds and their relationships with other sounds can be measured using numbers. He also proposed that the Earth is a sphere, that the Earth, Moon, and stars revolve around the Sun, and that astronomy (the study of stars, planets, and heavenly bodies) could be written as mathematical sentences called equations. Pythagoras and his followers used lines, triangles, and squares made out of pebbles to represent numbers. Today Pythagoras is best remembered for the Pythagorean theorem: The square of the length of the hypotenuse (the side of a right triangle opposite the right angle) of a right triangle equals the sum of the squares of the lengths of the triangle's other two sides. What is he famous for? Because pythagoras was the first person to prove the pythagorean theorem correct and his philosophy influence all other philosophers after his death, incluing Plato and Aristotle. His Pythagorean School gained influence and respect during the years around 520 BC in Italy. His Society was spread all throughout Europe and Asia Read more: http://wiki.answers.com/Q/How_did_pythagoras_become_so_famous#ixzz17yENddYx What is the Pythagorean theorem formula? If you have just taken up a basic geometry course, one of the first relations in Euclidean geometry that you will come across is the Pythagorean theorem formula. It is a fundamental theorem related to right angled triangles that is of immense value to geometry and mathematics in general. In this article, I explain what is the formula for Pythagorean theorem. This theorem which details an important property of right angled triangles is named after the Greek mathematician and philosopher Pythagoras who discovered it. However there is evidence that this theorem was known to ancient Indian and even Babylonian mathematicians. An ancient Indian text called the Apastamba Sulba Sutra dating back to about 600 BC has a numerical proof of Pythagorean theorem formula. Many indirect evidences of knowledge of the theorem have been found in ancient Egyptian scrolls and Babylonian artifacts. After that brief discussion of the history of the Pythagorean theorem let us move towards the statement of the theorem and formula. Read more on history of Pythagorean formula . Pythagorean Theorem Formula Explained
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# Question
Hi All,
Thanks
The total perimeter of the Square Paper is actually half of the total perimeter of the 4 rectangles. (by just counting from the picture. 2 sides of each rectangle form the square)
Therefore,
180 / 2 = 90cm
90 / 4 = 22.5cm
0 Replies 1 Like ✔Accepted Answer
Aligning the lower in-between vertical line with that of the upper in-between vertical, there would be 8 similar lengths and 8 similar breadths.
180/8 = 22.5
Ans : 22.5 cm.
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Transforms
Sharing a small page I cooked up to help people explore transformations - both the linear and the non-linear kind by drawing pictures and modifying them using transformations. This hack is in the spirit of the theme of this year’s Infinity festival at Pramati - where we have “Engaging the senses” as a theme. So, engage your senses to grasp the math of transformations.
For the impatient, here is the web page. However, you can also try out a few examples right here.
WARNING: Don’t write production code like the HTML+JS in that page. It is just a quick hack for illustrative purposes.
You’ll see a canvas area and a matrix display at the bottom right. Open up the dev console and you can start putting in your own drawing by setting the global draw() function and using some provided drawing primitives like - line, point, circle, grid and curve. All changes update live.
To dive in …
Yoooos the soource Luuuke!
Examples
A basic one first -
draw = () => { line(P(10,10),P(50,50)); }
tx = Scale(2)
A more fun non-linear one -
tx = (p) => P(p.x + 0.003 * Math.cos(Math.PI*t) * p.y * p.y,
p.y + 0.003 * Math.sin(Math.PI*t) * p.x * p.x)
draw = () => {
grid(30,30);
circle(P(0,0), 140);
}
Concepts
A “transformation” is modelled as a function that maps a given point to another point. You can write your own functions and use them. The inverse transformation of built-in transformations can be accessed with the .inverse property.
• To create a point, use P(x,y). If p is a variable holding a point, then p.x and p.y are x and y coordinates respectively.
Globals
• draw is a zero-argument function that you can set to draw whatever figure you want to .. well, at least ones using the offered primitives.
• tx is a global variable holding the transformation to be applied to your drawing.
In the JS console, just do draw = () => { ... } to set the drawing function.
Also, you can just do tx = Scale(10), for example, to scale your drawing by a factor of 10.
Drawing primitives
• line(P(x1,y1),P(x2,y2)) draws a line from $(x_1,y_1)$ to $(x_2,y_2)$
• circle(P(cx,cy),r) draws a circle centered at $(x,y)$ with radius $r$.
• point(x,y) draws a point at $(x,y)$.
• grid(dx,dy) draws a grid with spacing given by $dx$ and $dy$.
• lattice(xmin,ymin,xmax,ymax,func) calls func over a lattice of points spanning $(x_{min},y_{min})-(x_{max},y_{max})$.
• You can set the pen color for the primitives that follow by setting pen.color = 'blue'
• You can set the pen thickness using pen.thickness = 5.
Built-in transformations
• Linear(mxx,mxy,myx,myy) is a 2D matrix linear transformation. Linear(mxx,mxy,myx,myy).inverse will be the inverse transformation of the first one.
• Translate(dx,dy) shifts the drawing by $(dx,dy)$.
• Rotate(Deg(degrees)) or Rotate(radians) is a rotation about the origin - positive angle = anti-clockwise.
• Scale(s) scales by the factor $s$.
• mul(T1,T2,...) creates the product of the transformations as though they were applied back to front on a point. The product transform’s inverse is also computed.
Params
You can create dynamic parameters using Param like below -
angle = Param("angle", Param.linear(-Math.PI, Math.PI))
tx = Rotate(angle)
draw = () => { line(P(10,10),P(50,50)); }
You’ll see a slider pop up at the top right part of the window, which you can use to dynamically rotate the figure.
You can create simple animations by writing your draw() function to make use of the variable t which gives the current time in seconds relative to when the page started. You can also create your own custom transform function which depends on t to make the global transformation time dependent.
Some of the primitive drawing functions have additional parameters set as properties. For example, line.divisions = 10 will change the number of divisions used to draw a line. Why should a line be broken up into smaller line segments? That will become clear once you start playing around with non-linear transformations.
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## Algebra 2 (1st Edition)
$$7\pm 2\sqrt2$$
We use square roots in order to isolate m and solve the equation: $$(m-7)^2 = 8 \\ m-7 =\pm \sqrt{8} \\ m = 7\pm 2\sqrt2$$
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# TS Grewal Accountancy Class 11 Solutions Chapter 16
## TS Grewal Accountancy Class 11 Solutions Chapter 16 Accounts from Incomplete Records Single Entry System
If you are looking for TS Grewal Accountancy Class 11 Solutions Chapter 16 Accounts from Incomplete Records Single Entry System then you have landed at the right place. This post will explain all about the TS Grewal Accountancy Class 11 Chapter 16 solution and this post provides the solutions of each and every question with related images. We hope this Chapter 16 solution will help you in the preparation for your exam.
Please find the below questions and answers with solutions. You will get all the Class 11 Chapter 16 accounts solutions from this post.
Question 1.
Following information of an accounting year is given:
Opening Capital ₹ 60,000; Drawings ₹ 5,000; Capital added during the year ₹ 10,000 and Closing Capital ₹ 90,000. Calculate the Profit and Loss for the year.
Solution:
Question 2.
Mayank does not keep proper records of his business, he gives you the following information:
Opening Capital – ₹ 1,00,000
Closing Capital – ₹ 1,25,000
Drawings during the year – ₹ 30,000
Capital added during the year – ₹ 37,500
Calculate the profit or loss for the year.
Solution:
Question 3.
Capital of Ganesh Gupta in the beginning of the year was ₹ 70,000. During the year his business earned a profit of ₹ 20,000, he withdrew ₹ 7,000 for his persona use. He sold ornaments of his wife for ₹ 20,000, and invested that amount into the business. Find out his Capital at the end of the year.
Solution:
Capital at the end of the year = Capital in the beginning + Additional Capital + Profit – Drawings
= 70,000 + 20,000 + 20,000 – 7,000
= Rs. 1,03,000
Question 4.
Vikas maintains his books of account on Single Entry System. He provides following information from his books. Find out additional capital
introduced in the business during the year 2017-18.
Opening Capital – ₹ 1,30,000 ; Drawing during the year ₹ 50,000
Closing Capital – ₹ 2,00,000 ; Profit made during the year ₹ 1,00,000
Solution:
Additional Capital = Closing Capital + Drawings – (Opening Capital + Profit) = 2,00,000 + 50,000 – (1,30,000 + 1,00,000)
= 2,50,000 – 2,30,000 = Rs.20,000
Question 5.
Mohan maintains books on Single Entry System. He gives you the following information:
Capital on 1st April, 2017 – ₹ 15,200
Capital on 31st March, 2018 – ₹ 16,900
Drawings made during the year – 4,800
Capital introduced on 1st August, 2017 – 2,000
You are required to calculat the Profit or Loss made by Mohan.
Solution:
Question 6.
Mahesh who keeps his books on Single Entry System sells goods at Cost plus 50%. On 1st April, 2017 his Capital was ₹ 4,00,000 and on 31st March, 2018 it was ₹ 3,50,000. He had withdrawn ₹ 20,000 per month besides goods of the sale value of ₹ 60,000. How much did he earn in 2017-18?
Solution:
Calculation For Cost of Goods Sold:
Sales = COGS + Profit
Cost of Goods Sold = 100
Gross Profit = 50
Sales = 150
Gross Profit = $\frac { 50 }{ 150 }$ or $\frac { 1 }{ 3 }$
Sales = 60,000 x $\frac { 1 }{ 3 }$ = 20,000
COGS = Sales – Gross Profit = 60,000 – 20,000 = 40,000
Drawings = Cash + Cost of Goods Sold
Drawings = 2,40,000 + 40,000 = 2,80,000
Question 7.
Krishan started his business on 1st April, 2017 with a Capital of ₹ 1,00,000. On 31st March, 2018, his assets were :
Cash – ₹ 3,200
Stock – ₹ 34,800
Debtors – ₹ 31,000
Plant – ₹ 85,000
He owed ₹ 12,000 to sundry creditors and ₹ 10,000 to his brother on that date. He withdrew ₹ 2,000 per month for the private expenses. Ascertain his profit.
Solution:
Question 8.
Ram Prashad keeps his books on Single Entry System and from them and the particulars supplied, the following figures were gathered together on 31st March, 2018:
Book Debts ₹ 10,000; Cash in Hand ₹ 510; Stock-in-Trade (estimated) ₹ 6,000; Furniture and Fittings ₹ 1,200; Trade Creditors ₹ 4,000; Bank Overdraft ₹ 1,000; Ram Prashad stated that he started business on 1st April with cash ₹ 6000 paid into bank but stocks valued at ₹ 4,000. During the year he estimated his drawings to be ₹ 2,400. You are required to prepare the statement, showing the profit for the year, after writing off 10% for Depreciation on Furniture and Fittings.
Solution:
Question 9.
Shruti maintains her books of account from Incomplete Records. Her books provide the following information:
She with drew ₹ 500 per month for personal expenses. She sold her Investments of ₹ 16,000 at 5% premium and introduced the amount into business.
You are required to prepare a Statement of Profit or Loss for the year ending 31st March, 2016.
Solution:
Question 10.
Hari maintains her books of account on Single Entry System. His books provide the following information:
His drawings during the year were ₹ 5,000 Depreciate furniture by 10% and provide a reserve for Bad and Doubtful Debts at 10% on Sundry Debtors.
Prepare the statement showing the profits for the year.
Solution:
Question 11.
A commenced business on 1st April, 2017 with a capital of ₹ 10,000. He immediately bought Furniture and Fixtures for ₹ 2,000. On 1st October, 2017, he borrowed ₹ 5,000 from his wife @ 9% p.a. (interest not yet paid) and introduced a further capital of his own amounting to ₹ 1,500. A drew @ ₹ 300 per month at the end of each month for household expenses. On 31st March, 2018 his position was as follows:
Cash in Hand ₹ 2,800; Sundry Debtors ₹ 4,800; Stock ₹ 6,800; Bills Receivable ₹ 1,600; Sundry Creditors ₹ 500 and owing for Rent ₹ 150. Furniture and Fixtures to be depreciated by 10%. Ascertain the profit or loss made by A during 2017-18.
Solution:
Question 12.
Kuldeep, a general merchant, keeps his accounts on Single Entry System. He wants to know the results, of his business on 31st March, 2018 and for that following information is available:
During the year, he had withdrawn ₹ 5,00,000 for his personal use and invested ₹ 2,50,000 as additional cpaital. Calculate his profits on 31st March, 2018 and prepare the Statement of Affairs as on that date.
Solution:
Question 13.
Following information is supplied to you by a shopkeeper:
During the year, he withdrew ₹ 2,500 per month for dometstic purposes. He also borrowed from a friend at 9% a sum of ₹ 20,000 on 1st October, 2017. He has not yet paid the interest. A provision of 5% on debotrs for doubtful debts is to be made.
Ascertain the profit or loss made by him during the period.
Solution:
Question 14.
Vikas is keeping his accounts according to Single Entry System. His capital on 31st December, 2015 was ₹ 2,50,000 and his capital on 31st December, 2016 was ₹ 4,25,000. He further informs you that during the year he gave a loan of ₹ 30,000 to his brother on private account and withdrew ₹ 1,000 per month for personal purposes. He used a flat for his personal purpose, the rent of which @ ₹ 1,800 per month and electricity charges at an average of 10% of rent per month were paid from the business account. During the year he sold his 7% Government Bonds of ₹ 50,000 at 1% premium and brought that money into the business.
Prepare a Statement of Profit or Loss for the year ended 31st December, 2016.
Solution:
Question 15.
Manu started business with a capital of ₹ 4,00,000 on 1st October, 2005. He borrowed from his friend a sum of ₹ 1,00,000. He brought further ₹ 75,000 as capital on 31st March, 2006, his position was:
Cash : ₹ 30,000; Stock : ₹ 4,70,000; Debtors : ₹ 3,50,000 and Creditors : ₹ 3,00,000.
He withdrew ₹ 8,000 per month during this period. Calculate profit on loss, for the period.
Solution:
Question 16.
From the following information relating to the business of Mr. X who keeps books on Single Entry System, ascertaint the profit or loss for the year 2017-18:
Mr. X withdrew ₹ 4,100 during the year to meet his household expenses. He introduced ₹ 300 as fresh capital on 15th January, 2018. Machinery and Furniture are to be depreciated at 10% and 5% p.a. respectively.
Solution:
Question 17.
X, a retailer, has not maintained proepr books of accont but it has been possible to obtain the follwoing details:
Calculate the net profit for this year and draft the Statement of Affairs at the end of the year after noting that:
(a) Shop Fittings are to be depreciated by ₹ 780.
(b) X has drawn ₹ 100 per week for his own use.
(c) Included in the Trade Debtors is an irrecoverable balance of ₹ 270.
(d) Interest at 5% p.a. is due on the loan from Naresh but has not been paid for the year.
Solution:
Question 18.
On 1st April, 2017, X started a business with ₹ 40,000 as his capital. On 31st March, 2018, his position was as follows:
During the year 2017-18, X drew ₹ 24,000. On 1st October, 2017, he introduced further capital amounting to ₹ 30,000. You are required to ascertain profit on loss made by him during the year 2017-18.
(a) Plant is to be depreciated at 10%.
(b) A provision of 5% is to be made against debtors, Also prepare the Statement of Affairs as on 31st March, 2018.
Solution:
Question 19.
C maintains his books according to Single Entry System. Following figures were available from the books for the six months ended 31st December 2017:
(a) He had withdrawn ₹ 200 in the beginning of every month for household purposes.
(b) Depreciation on Plant and Machinery @ 10% p.a.
(c) Further Bad Debts ₹ 5,000 and Provision for Doubtful Debts to be created @ 2%.
(d) During the period, salaries have been prepaid by ₹ 500 while wages outstanding were ₹ 1,000.
(e) Interest on drawings to be reckoned @ 6% p.a.
You are required to prepare the Statement of Profit or Loss for the half year ended 31st December, 2017, followed by Revised Statement of Affairs as on that date.
Solution:
Question 20.
A firm sells goods at a Gross profit of 25% of sales. On 1st April, 2017 the Stock was ₹ 40,000; Purchases were ₹ 1,10,000 and the Stock on 31st March, 2018 was ₹ 30,000. What was the value of Sales?
Solution:
Question 21.
A firm sells goods at Cost plus 25%. Sales to credit customers ($\frac { 3 }{ 4 }$ of total) was ₹1,80,000. His Opening and Closing Stocks were ₹ 20,000 and ₹ 15,000 respectively. Find out the value of Purchases.
Solution:
Calculation For Gross Profit = 2,40,000 × 20% = 48,000
Question 22.
Calculate Stock in the beginning:
Sales – ₹ 80,000
Purchases – ₹ 60,000
Stock at the end – ₹ 8,000
Loss on Cost – $\frac { 1 }{ 6 }$
Solution:
Calculation For Gross Loss = 80,000 × 20% = 16,000
Question 23.
Calculate the Stock at the end:
Stock in the beginning – ₹ 20,000
Cash Sales – ₹ 60,000
Credit Sales – ₹ 40,000
Purchases – ₹ 70,000
Rate of Gross Profti on cost – $\frac { 1 }{ 3 }$
Solution:
Calculation For Gross Profit = 1,00,000 × 25% = 25,000
Question 24.
Calculate the value of CLosing Stock from the following information:
Purchases – ₹ 93,000
Wages – ₹ 20,000
Sales – ₹ 1,20,000
Carriage Outwards – ₹ 3,200
Opening Stock – ₹ 16,000
Rate of Gross Profit 25% on Cost
Solution:
Calculation For Gross Profit = 1,20,000 × 20% = 24,000
Note: Carriage Outward pass the entry on Profit and Loss A/c
Question 25.
Calculate Purchases:
Cost of Goods Sold – ₹ 65,000
Stock in the beginning – ₹ 4,000
Closing Stock – ₹ 5,000
Solution:
Purchases = Cost of Goods sold – Opening Stock + Closing Stock
Purchases = 65,000 – 4,000 + 5,000 = 66,000
Question 26.
Calculate Sales:
Cost of goods sold – ₹ 2,00,000
Rate of Gross Profit 20% on Sales
Solution:
Gross Profit = 2,00,000 × 25% = 50,000
COGS + Gross Profit = Sales
2,00,000 + 50,000 = 2,50,000
Question 27.
Debtors in the beginning of the year were ₹ 30,000, Sales on credit during the year were ₹ 75,000, Cash received from the Debtors during the year was ₹ 35,000, Returns Inward (regarding credit sales) were ₹ 5,000 and Bills Receivable drawn during the year were ₹ 25,000. Find the balance of Debtors at th end of the year, assuming that there were Bad Debts during the year of ₹ 2,000.
Solution:
Question 28.
Creditors on 1st April, 2017 were ₹ 15,000, Purchases on credit were ₹ 30,000, Cash paid to Creditors during 2017-18 was ₹ 20,000, Returns Outward (regarding credit purchases) were ₹ 1,000 and Bills Payable accepted during the year ₹ 10,000. Find the balance of Creditors on 31st March, 2018.
Solution:
Question 29.
Following information is given of an accounting year:
Opening Creditors ₹ 15,000; Cash paid to creditors ₹ 15,000; Returns Outward ₹ 1,000 and Closing creditors ₹ 12,000.
Calculate Credit Purchases during the year.
Solution:
Question 30.
From the following information supplied by X, who keeps his books on Single Entry System, you are required to calculate Total Purchases:
Opening balance of Bills Payable – ₹ 5,000
Opening balance of Creditors – ₹ 6,000
Closing balance of Bills Payable – ₹ 7,000
Closing balance of Creditors – ₹ 4,000
Cash paid to Creditors during the year – ₹ 30,200
Bills Payable discharged during the year – ₹ 8,900
Returns Outward – ₹ 1,200
Cash Purchases – ₹ 25,800
Solution:
Total Purchases = Cash Purchases + Credit Purchases
Total Purchases = 25,800 + 40,300 = Rs 66,100
Question 31.
Cash sales of a business in a year were ₹ 85,000, the Cost of Goods Sold (including direct expenses) was ₹ 97,000 and Gross Profit as shown by the Trading Account for the year was ₹ 1,29,000. Calculate Credit Sales during the year.
Solution:
Gross Profit = Net Sales – Cost of Goods Sold
1,29,000 = Net Sales – 97,000
Net Sales = Rs 2,26,000
Credit Sales = Total Net Sales – Cash Sales
Credit Sales = 2,26,000 – 85,000 = Rs 1,41,000
Question 32.
From the following information, calculate Total Sales made during the period:
Debtors as on 1st April, 2017 – ₹ 20,400
Cash received from debtors during the year (as per Cash Book) – ₹ 60,800
Returns Inward – ₹ 5,400
Debtors as on 31st March, 2018 – ₹ 27,600
Cash Sales (as per Cash Book) – ₹ 56,800
Solution:
Total Sales = Cash Sales + Credit Sales
Total Sales = 56,800 + 75,800 = Rs 1,32,600
Question 33.
Calculate Total Sales from the following information:
Bills Receivables as on 1st April, 2017 – ₹ 7,800
Debtors as on 1st April, 2017 – ₹ 30,800
Cash received on maturity of Bills Receivable during the year – ₹ 20,900
Cash received from Debtors – ₹ 70,000
Bad Debts written off – ₹ 4,800
Returns Inward – ₹ 8,700
Bills Receivable dishonoured – ₹ 1,800
Bills Receivable on 31st March, 2018 – ₹ 6,000
Debtors as on 31st March, 2018 – ₹ 25,500
Cash Sales during the year – ₹ 15,900
Solution:
Total Sales = Cash Sales + Credit Sales
Total Sales = 15,900 + 97,300 = Rs 1,13,200
Question 34.
From the following information, ascertain the opening balance of Sundry Debtors and the closing balance of Sundry Creditors:
Sundry Creditors as on 31st March, 2017 – ₹ 20,600
Sundry Debtors as on 31st March, 2018 – ₹ 37,400
Stock as on 31st March, 2017 – ₹ 26,000
Stock as on 31st March, 2018 – ₹ 24,000
During the year ended 31st March, 2018:
Purchases – ₹ 1,10,000
Discount allowed by creditors – ₹ 800
Discount allowed to customers – ₹ 1,100
Cash paid to sundry creditors – ₹ 95,000
Bills Payable issued by them – ₹ 14,000
Bills Receivable received from customers – ₹ 16,500
Cash received from customers – ₹ 1,30,000
Bills receivable dishonoured – ₹ 1,900
Solution:
Cost of Goods Sold = Opening Stock + Purchases – Closing Stock
Cost of Goods Sold = 26,000 + 1,10,000 – 24,000 = 1,12,000
Gross Profit = $\frac { 30 }{ 70 }$ x 112000 = Rs. 48,000
Sales = Cost of Goods Sold + Gross Profit
Sales = 1,12,000 + 48,000 = Rs 1,60,000
Credit Sales = 1,60,000 – 20,000 = Rs 1,40,000
Question 35.
Roshan, whose accounts are maintained by Single Entry System, acquired a retail business on 1st April, 2017. He had ₹ 40,000 of his own and he borrowed ₹ 20,000 from his wife. He paid ₹ 15,000 for Goodwill ₹ 5,000 for Furniture and ₹ 35,000 for Stock.
Total cash received by him during the financial year from the Debtors was ₹ 2,30,000. His payments were:
Purchases – ₹ 1,56,000
Salary and Wages – ₹ 21,400
Rent:
For business premises – ₹ 5,920
For private house – ₹ 2,960
Payments made for domestic purposes and drawings – ₹ 26,400
At the end of the year, the Stock was ₹ 37,500. He owed ₹ 13,500 to Creditors for goods and his customers owed to him ₹ 15,000. Provide 5% for Depreciation on Furniture, Interest at 5% on wife’s Loan and ₹ 1,000 for Doubtful Debts.
Prepare the Cash Account, the Profit and Loss Account for the year ended 31st March, 2018 and the Balance Sheet at the close of the year.
Solution:
Question 36.
Vijay commenced business as foodgrains merchant on 1st April, 2017 with a capital of ₹ 4,00,000. On the same day, he purchased furniture for ₹ 80,000. From the following particulars obtained from his books which do not conform to Double Entry principles, you are required to prepare the Trading and Profit and Loss Account for the year ended 31st March, 2018 and the Balance Sheet as on that date:
Sales (including Cash Sales ₹ 2,00,000) – ₹ 5,00,000
Purchases (including Cash Purchases ₹ 1,20,000) – ₹ 4,00,000
Vijay’s Drawings (in cash) – ₹ 40,000
Salaries to Staff – ₹ 48,000
Bad Debts written off – ₹ 4,000
Trade Expenses paid – ₹ 16,000
Vijay used goods of ₹ 12,000 for private purposes during the year. On 31st March, 2018, his Debtors amounted to ₹ 1,40,000 and Creditors ₹ 80,000. Stock-in-Trade on that date was ₹ 1,60,000.
Solution:
Question 37.
Following information is obtained from the books of Vinay, who maintained his books of account under Single Entry System:
Vinay banks all receipts and makes payments by means of cheque.
From the above information, prepare Trading and Profit and Loss Account for the year ended 31st March, 2018 and Balance Sheet as on that date.
Solution:
Question 38.
Surya does not keep a systematic record of his transactions. He is able to give you the following information regarding his assets and liabilities:
Following additional information is also avialable for the year ended 31st March, 2018:
Bad Debts during the year were ₹ 900. As regards sale, Surya tells you that he always sells goods at Cost plus 25%. Furniture and Fittings are to be depreciated at 10% of the value in the beginning of the year.
Prepare Surya’s Trading and Profit and Loss Account for the year ended 31st March, 2018 and his Balance Sheet on that date.
Solution:
If you have any questions regarding TS Grewal Accountancy Class 11 Solutions Chapter 16 Accounts from Incomplete Records Single Entry System then feel free to drop a comment and we will try to provide the solution at the earliest.
Updated: January 25, 2021 — 1:09 pm
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# What Percentage Is 40 Minutes Of An Hour?
## What percentage of hour is 15 minutes?
0.250More InformationMinutesFraction of an hour120.200130.217140.233150.25057 more rows•Oct 20, 2020.
## How do you calculate minutes for payroll?
How do I calculate minutes for payroll? To calculate minutes for payroll, you must convert minutes worked to decimal form. Do this by dividing the minutes by 60. Then, multiply your answer by the employee’s hourly rate to get the amount you need to pay for those minutes.
## What percent of a day is 27 minutes?
1 day = (24 × 60) minutes. 1 day = 1440 minute. Hence, the required percentage is 0.069%.
## How much is a 100 minutes?
100 minutes equals 1.67 hours.
## What is 40 minutes of an hour as a decimal?
Minutes to Decimal Hours CalculatorMinutesDecimal Hours370.617380.633390.650400.66756 more rows
## Whats 0.4 of an hour?
This conversion of 0.4 hours to minutes has been calculated by multiplying 0.4 hours by 60 and the result is 24 minutes.
## How much is 1/10th of an hour?
Remember, one tenth of an hour is 6 minutes. This chart illustrates that if an employee punches in early for work at 7:58am, their time will be rounded to 8:00am.
## What percentage of a day is half an hour?
25/12 % of a day is half an hour a shopkeeper bought a watch for280 and sold it for 315.
## What is 1 hour 20 minutes as a decimal?
Time to Decimal Conversion TableTimeHoursMinutes01:15:001.257501:20:001.3333338001:25:001.4166678501:30:001.590115 more rows
## What is .8 of an hour?
Billing Increment Chart—Minutes to Tenths of an HourMinutesTime25-30.531-36.637-42.743-48.86 more rows
## What is .25 of an hour?
Conversion Chart – Minutes to Hundredths of an Hour Enter time in Oracle Self Service as hundredths of an hour. For example 15 minutes (¼ hour) equals . 25, 30 minutes (½ hour) equals . 5, etc.
## What percentage is 20 minutes of an hour?
More InformationMinutesFraction of an hour190.317200.333210.350220.36757 more rows•Oct 20, 2020
## What is .15 of an hour?
Decimal Hours-to-Minutes Conversion ChartMinutesTenths of an HourHundredths of an Hour8.1.149.1.1510.1.1611.1.1855 more rows
## Is 100 minutes an hour and 40 minutes?
This conversion of 100 minutes to hours has been calculated by multiplying 100 minutes by 0.0166 and the result is 1.6666 hours.
## What is 6.75 hours in hours and minutes?
6.75 hours with the decimal point is 6.75 hours in terms of hours. 6:75 with the colon is 6 hours and 75 minutes. .
## What is 1.67 hours and minutes?
1.67 hours is 1 hours, 40 minutes and 12 seconds.
## How do you convert hours to hours and minutes?
How to convert decimal minutes to time format78.6 minutes can be converted to hours by dividing 78.6 minutes / 60 minutes/hour = 1.31 hours.1.31 hours can be broken down to 1 hour plus 0.31 hours – 1 hour.0.31 hours * 60 minutes/hour = 18.6 minutes – 18 minutes.0.6 minutes * 60 seconds/minute = 36 seconds – 36 seconds.More items…
## What is 10 minutes as a percentage of an hour?
Sixty seconds make up a minute, while 60 minutes make up an hour. You can convert minutes into a percentage of an hour using basic division. For example, 30 minutes equals 50 percent of an hour, while 10 minutes equals about 17 percent of an hour.
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## Statistics Tutorial
#### Intro
Being able to draw conclusions based on statistics is a very important component to understanding your results. By viewing the research question, parameter(s) of interest, sample statistic, and p-value, we can conclude what the results of the research were. In this problem, you will be given the above information and it is up to you to write a coherent conclusion paragraph (with complete sentences) that summarizes the results of the research.
#### Sample Problem
Students often use excuses for why they were late to class. A common excuse is that the student’s car had a flat tire. If students are lying about the flat tire, they may pick a certain tire disproportionately.
So the question is: do students pick which tire went flat in equal proportions? It has been conjectured that when students are asked this question and forced to give an answer (left front [LF], left rear [LR], right front [RF], or right rear [RR]) off the top of their head, they tend to answer “right front” more than would be expected by random chance.
To test this conjecture about the right front tire, a recent class of 28 students was asked if they were in this situation, which tire would they say had gone flat. We obtained the following results:
LF: 6; LR: 4; RF:14; RR: 4
The parameter of interest (pi) is the long-run probability that students choose the RF tire. Null hypothesis: pi = 0.25. Alternative hypothesis: pi > 0.25.
Sample statistic: p-hat = 14 students choosing RF / 28 total students = 0.50.
Using a one-proportion t-test, we get a p-value of 0.001 and a z-score of 3.07.
Formulate a conclusion, incorporating relevant information from the above problem. Can we generalize these results to the general population?
#### Solution
Based on this study, we conclude that we have strong evidence against the null hypothesis and we will reject it; we have strong support for the alternative hypothesis as well. The low p-value (< 0.01) and high standardized statistic (> 3.0) suggest that 50% of the students choosing the right front tire by simply guessing from the four tires is extremely unlikely. We therefore also conclude that random chance does not play a factor in students’ choice of tire in this scenario. So, students do not pick which tire went flat in equal proportions but they instead are more likely to choose the right front tire.
We cannot generalize a study of this case to the general population. The scope of this study focuses on a small class of only 28 students. Furthermore, these students could all be from very similar backgrounds or parts of the country and are most likely in a similar age range. We could, however, potentially generalize to students at the school in general, though we cannot be certain without more information about the class and the school population. We would need to know how representative this class is of the school population. Ultimately, because we cannot exclude extraneous variables (such as age, background, etc.), we cannot generalize such a result to the general population due to the low likelihood that the sample is representative of the general population.
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## Jkk
In this new expression for q, there is no longer a denominator with t> in it. Since (1 + d) -> 2 as u -» 1 from either side, we may then conclude that lim q = 2.
Example 3 Given q = (2c + 5)/(v + 1), find lim q. The variable v again
appears in both the numerator and the denominator. If we let v —► + oo in both, the result will be a ratio between two infinitely large numbers, which does not have a clear meaning. To get out of the difficulty, we try this time to transform the given ratio to a form in which the variable v will not appear in the numerator.* This, again, can be accomplished by dividing out the given ratio. Since (2v + 5) is not evenly divisible by (v + 1), however, the result will contain a remainder term as follows:
But, at any rate, this new expression for q no longer has a numerator with v in it. Noting that the remainder 3/(u + 1) -> 0 as v -» + oo, we can then conclude that lim q = 2.
There also exist several useful theorems on the evaluation of limits. These will be discussed in Sec. 6.6.
### Formal View of the Limit Concept
The above discussion should have conveyed some general ideas about the concept of limit. Let us now give it a more precise definition. Since such a definition will make use of the concept of neighborhood of a point on a line (in particular, a specific number as a point on the line of real numbers), we shall first explain the latter term.
For a given number L, there can always be found a number (L — ax) < L and another number (L + a2) > L, where a, and a2 are some arbitrary positive numbers. The set of all numbers falling between (L — ax) and (L + a2) is called the interval between those two numbers. If the numbers (L — a,) and (L + a2) are included in the set, the set is a closed interval, if they are excluded, the set is an open interval. A closed interval between (L — ax) and (L + a2) is denoted by the bracketed expression
[L — at, L + a2] = {q\L — at<q<L + a2) and the corresponding open interval is denoted with parentheses: (6.4) (L - ax, L + a2) = [q \ L - a, < q < L + a2)
* Note that, unlike the t> -> 0 case, where we want to take v out of the denominator in order to avoid division by zero, the v -» oo case is better served by taking v out of the numerator. As v -> oo, an expression containing v in the numerator will become infinite but an expression with v in the denominator will, more conveniently for us, approach zero and quietly vanish from the scene.
Thus, [ ] relate to the weak inequality sign < , whereas ( ) relate to the strict inequality sign < . But in both types of intervals, the smaller number (L — ax) is always listed first. Later on, we shall also have occasion to refer to half-open and half-closed intervals such as (3,5] and [6, oo), which have the following meanings:
(3,5] = {x | 3 < x < 5} [6, oo) = {x | 6 < x < oo}
Now we may define a neighborhood of L to be an open interval as defined in (6.4), which is an interval "covering" the number L* Depending on the magnitudes of the arbitrary numbers ax and a2, it is possible to construct various neighborhoods for the given number L. Using the concept of neighborhood, the limit of a function may then be defined as follows:
As v approaches a number N, the limit of q = g(v) is the number L, if, for every neighborhood of L that can be chosen, however small, there can be found a corresponding neighborhood of N (excluding the point v = N) in the domain of the function such that, for every value of t; in that A'-neighbor-hood, its image lies in the chosen ¿-neighborhood.
This statement can be clarified with the help of Fig. 6.3, which resembles Fig. 6.2a. From what was learned about the latter figure, we know that lim q = L in
Fig. 6.3. Let us show that L does indeed fulfill the new definition of a limit. As the first step, select an arbitrary small neighborhood of L, say, (L — ax, L + a2). (This should have been made even smaller, but we are keeping it relatively large to facilitate exposition.) Now construct a neighborhood of N, say, (N - bx, N + b2), such that the two neighborhoods (when extended into quadrant I) will together define a rectangle (shaded in diagram) with two of its corners lying on the given curve. It can then be verified that, for every value of v in this neighborhood of N (not counting v = N), the corresponding value of q = g(v) lies in the chosen neighborhood of L. In fact, no matter how small an ¿-neighborhood we choose, a (correspondingly small) ¿V-neighborhood can be found with the property just cited. Thus L fulfills the definition of a limit, as was to be demonstrated.
We can also apply the above definition to the step function of Fig. 6.2 c in order to show that neither Lx nor L2 qualifies as lim q. If we choose a very small v->N
neighborhood of L,—say, just a hair's width on each side of L,—then, no matter what neighborhood we pick for N, the rectangle associated with the two neighborhoods cannot possibly enclose the lower step of the function. Consequently, for any value of v > N, the corresponding value of q (located on the lower step) will not be in the neighborhood of Lx, and thus Lx fails the test for a limit. By similar reasoning, L2 must also be dismissed as a candidate for lim q. In fact, in this case no limit exists for q as v -* N. V~'N
* The identification of an open interval as the neighborhood of a point is valid only when we are considering a point on a line (one-dimensional space). In the case of a point in a plane (two-dimensional space), its neighborhood must be thought of as an area, say, a circular area around the point.
0 0
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# Expected number of edges: does $\sum\limits_{k=1}^m k \binom{m}{k} p^k (1-p)^{m-k} = mp$
Find the expected number of edges in $G \in \mathcal G(n,p)$.
Method $1$: Let $\binom{n}{2} = m$. The probability that any set of edges $|X| = k$ is the set of edges in $G$ is $p^k (1-p)^{m-k}$. So the probability that $G$ has $k$ edges is $$\binom{m}{k} p^k ( 1-p )^{m-k}$$
This implies that $$E(X) = \sum_{k=1}^m k \binom{m}{k} p^k (1-p)^{m-k}$$
Method $2$: Choose an indicator random variable $X_e : \mathcal G(n,p) \to \{ 0,1 \}$ as follows:
$$X_e(G) = \begin{cases} 1 & e \in E(G) \\ 0 & e \notin E(G) \end{cases}$$
So $E(X) = \sum_{e \in K_n} E(X_e(G)) = m p$ since each event $e \in E(G)$ and $f \in E(G)$ are independent.
How do you reconcile these answers? I'm looking for either a mistake in reasoning or a direct proof that:
$$\sum_{k=1}^m k \binom{m}{k} p^k (1-p)^{m-k} = mp$$ for $0 < p < 1$.
-
You don't need independence to apply linearity of expectation. That's what makes it so wonderful! Anyway, there's nothing to reconcile. This is a correct proof of the identity you've written down. – Qiaochu Yuan May 5 '13 at 5:12
$$f(x)=( xp+(1-p) )^m = \sum_{k=0}^{m} {m\choose k} p^k(1-p)^{m-k}x^k$$
Differentiating the above equation with respect to $x$ yields
$$\implies mp( xp+(1-p) )^{m-1} = \sum_{k=1}^{m}{m\choose k} k p^k(1-p)^{m-k}x^{k-1}.$$
Subs $x=1$ in the above equation gives the desired result.
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Part 1
Ratio- Comparison of two quantites.
Example:
Circles: squares or Squares: Circles
Rate- A ratio that compares two quantities having different units
Example: \$1.69 per 100g or \$1.69/100g is a rate for purchasing bulk food
also for the units 64 beats/1min.
Question: Justin shovled his lawn and his neighbors. It took him 2 hours he earned \$22 dollars. How lmuch money would he earn in 1 hour?
Proportion- is twoo equivalent -Ratio
-Rate
-Fractions
Part 2-
1. 5 hours to travel 360 kilometers is about 72km/h.
2. Emma saves 28 cents of every dollar that she earns. Emma earned \$75 last week. How much money did Emma save last
week?
Part 3
Way 1 Horizontial
Way 2 Diagonal
Way 3 Vertical
1. Does this seem fair?
No, I dont think it is fair.
2. With what you know about proportion look and read what is in the image above. Does it seem just and fair?
No, I still don't it's fair.
3. Why have you made this choice?
Because the homeless guy got sentenced for 15 years in jail for just stealing \$100, but then turned his self in the next for being guilty. As when the other guy gets sentenced for only 4o months when he stole \$3 billion.
4.What would you have done if you were the judge?
If I were the judge, I would've sentenced the homeless guy for only 1-2 months. As for the other guy I would've sentenced him for more than 15 years in jail. Or I would reduced if he returned the money back which he probably didn't.
Part 6
Kid In the woods
My Question for this picture is How tall is the kid and how tall is the tree and what kind of tree it is.
So I searched the Average size of a 13-14 year old boy 9because he looks my age).
And it is 5'4-5'7 and I'm gonna choose 5'6(1.69m) because I see alot of boys taller than me or my height. So that answers my first question and which is gonna help me with my next question.
So my next question is how tall the trees are.
So the trees are about 3 of my thumbs. The boy is 1 of my thumbs.
and the boy is about 5'6 so then. 5.6x3=16.8.
so then it is 16 feet and 8 inches tall.
The trees are approx. 16'8ft tall.
And the tree is Birch tree because I've seen in millions of times when I go camping and use to always ask what the tree is called to my family.
Yours truly,
Kiara Foster 8-16
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# Search by Topic
#### Resources tagged with Addition & subtraction similar to Three by Three:
Filter by: Content type:
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### There are 255 results
Broad Topics > Calculations and Numerical Methods > Addition & subtraction
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### All Seated
##### Stage: 2 Challenge Level:
Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties?
### Two Dice
##### Stage: 1 Challenge Level:
Find all the numbers that can be made by adding the dots on two dice.
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Sealed Solution
##### Stage: 2 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
### Two Egg Timers
##### Stage: 2 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
##### Stage: 1 and 2 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Finding Fifteen
##### Stage: 2 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Buying a Balloon
##### Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Page Numbers
##### Stage: 2 Short Challenge Level:
Exactly 195 digits have been used to number the pages in a book. How many pages does the book have?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### On Target
##### Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### Hubble, Bubble
##### Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### Oh! Harry!
##### Stage: 2 Challenge Level:
A group of children are using measuring cylinders but they lose the labels. Can you help relabel them?
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Open Squares
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, focuses on 'open squares'. What would the next five open squares look like?
### Robot Monsters
##### Stage: 1 Challenge Level:
Use these head, body and leg pieces to make Robot Monsters which are different heights.
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
##### Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### How Old?
##### Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Build it Up
##### Stage: 2 Challenge Level:
Can you find all the ways to get 15 at the top of this triangle of numbers?
### Build it up More
##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
### The Dice Train
##### Stage: 2 Challenge Level:
This dice train has been made using specific rules. How many different trains can you make?
### Shapes in a Grid
##### Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
### Train Carriages
##### Stage: 1 and 2 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
### X Is 5 Squares
##### Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Seven Square Numbers
##### Stage: 2 Challenge Level:
Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number.
### Six Is the Sum
##### Stage: 2 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Dart Target
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
### Today's Date - 01/06/2009
##### Stage: 1 and 2 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Mrs Beeswax
##### Stage: 1 Challenge Level:
In how many ways could Mrs Beeswax put ten coins into her three puddings so that each pudding ended up with at least two coins?
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### Two Primes Make One Square
##### Stage: 2 Challenge Level:
Can you make square numbers by adding two prime numbers together?
### Money Bags
##### Stage: 2 Challenge Level:
Ram divided 15 pennies among four small bags. He could then pay any sum of money from 1p to 15p without opening any bag. How many pennies did Ram put in each bag?
### Five Coins
##### Stage: 2 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
### Routes 1 and 5
##### Stage: 1 Challenge Level:
Find your way through the grid starting at 2 and following these operations. What number do you end on?
### Cuisenaire Environment
##### Stage: 1 and 2 Challenge Level:
An environment which simulates working with Cuisenaire rods.
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### It's All about 64
##### Stage: 2 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
### Domino Join Up
##### Stage: 1 Challenge Level:
Can you arrange fifteen dominoes so that all the touching domino pieces add to 6 and the ends join up? Can you make all the joins add to 7?
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# Simplified III: The Goat Problem
“The solution often turns out more beautiful than the puzzle.”
— Richard Dawkins
The goat problem, or Monty Hall problem, named after the host of the game show “Let’s Make a Deal” confuses the heck out of people. Here’s the original formulation from Marilyn vos Savant:
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
As counter-intuitive as it might seem, the correct answer is — yes, it is!
Everyone (unless they knew this problem already) including software developers and physicists that I confronted with this answer, shook their heads in disbelieve. They are, however, in good company. According to this Wikipedia article:
Many readers of vos Savant’s column refused to believe switching is beneficial despite her explanation. After the problem appeared in “Parade”, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant’s predicted result (Vazsonyi 1999).
Fascinating, isn’t it? The problem (or rather the solution) is so hard to grasp, yet it’s so easy to follow this simplified™ explanation:
– If you initially picked a goat and you switch, you’ll win.
– If you initially picked a car and you switch, you’ll lose.
– It’s twice as likely that you initially picked a goat.
– Thus, it’s twice as likely to win when you switch!
Still not convinced? I wrote a little simulation in Python. Here’s the result after one million experiments:
The first number is the total number of experiments. The second number is the percentage of games won when the contestant stayed, while the third number shows the percentage of games won when the contestant switched.
# Simplified II: The Weasel That Teaches Evolution
“Weaseling out of things is important to learn. It’s what separates us from the animals… except the weasel.”
— Matt Groening
It’s Towel Day again! What a great opportunity to reflect upon Life, the Universe, and Everything. In this installment of the “Simplified” series, I want to tackle nothing less than Darwin’s theory of evolution. Why? Because I seriously believe that the world would be a better place if people finally understood it.
One common misconception, for example, is that some falsely believe that evolution occurs in a linear fashion, from single-celled organisms to homo sapiens. Such individuals, like the taxi driver that I mentioned in a previous post, think that they can smash the theory of evolution by posing this cunning question:
“If we humans are really decedents of animals like apes and dogs, why are there still apes and dogs around?”
Just in case this reasoning sounds conclusive to you as well: In reality, there is no single line of evolution, it’s rather a tree with many, many branches. On these branches, the universe performs experiments and decides which species survive, through natural selection. Consequently, humans did not evolve from apes, they are rather cousins whose common ancestor was neither ape nor human.
Another fallacy is Fred Hoyle’s famous Boeing 747 analogy. Hoyle was a brilliant scientist, no doubt, yet he refused to accept that complex life-forms can emerge by chance:
“A junkyard contains all the bits and pieces of a Boeing-747, dismembered and in disarray. A whirlwind happens to blow through the yard. What is the chance that after its passage a fully assembled 747, ready to fly, will be found standing there?”
A variant of this argument is the infinite monkey theorem: a monkey typing random letters at a typewriter will never be able to produce Shakespeare’s works.
In 1986, evolutionary biologist Richard Dawkins set out to dispel such misunderstandings by implementing a computer program that works—in Dawkins’ own words—like this:
“It […] begins by choosing a random sequence of 28 letters, […] it duplicates it repeatedly, but with a certain chance of random error – ‘mutation’ – in the copying. The computer examines the mutant nonsense phrases, the ‘progeny’ of the original phrase, and chooses the one which, however slightly, most resembles the target phrase, METHINKS IT IS LIKE A WEASEL.”
Here’s a more detailed explanation of his weasel program: it starts with a string of 28 random letters. Next, it creates N offspring strings by copying the original 28 random letters N times. When being copied, the chance of a letter being changed into another (random) letter is P. Now that we have a set of N new strings, we compare each string letter by letter against “METHINKS IT IS LIKE A WEASEL” (a quote from Shakespeare’s Hamlet, by the way) and pick the one with the most character matches (the highest match score). This one is deemed the “fittest” and kept as the survivor of the first generation; the other N-1 strings are discarded. We repeat the whole process by creating again N offspring from the survivor, then pick a new survivor and so on until the survivor finally matches “METHINKS IT IS LIKE A WEASEL”.
Let’s choose N to be 100 (one hundred offspring per generation) and P to be 5%, as in Dawkins original experiment. How many generations would it take until we finally reach “METHINKS IT IS LIKE A WEASEL”? Just guess a number, I’ll wait here…
Answer: about 50 generations! To me, this number was so ridiculously small that I decided to implement the weasel program myself. Please check it out here. Through command-line parameters, you can change the values of N and P, as well as the target phrase. Toying around with this program is so eye-opening, you suddenly realize that evolution is possible, that complexity can emerge by mutation and survival of the fittest.
Interestingly, if you set P to a high value (say 20%), you are likely to never arrive at the target phrase. A high value of P simulates an early universe, or any environment with a high natural radioactivity level. In one experiment where I set P to 20%, it took 800 generations, while a P value of 1% reached the target phrase after only 90 generations.
I also played with much longer target phrases with a length of more than 150 letters. What I’ve found is that the longer the phrase gets, the more detrimental a high mutation probability becomes. If you think about it, that’s not surprising, as the likelihood that already matching letters are replaced again with non-matching letters increases. Conclusion: the evolution of complex life-forms requires an environment that provides the right mutation probability—neither too low, nor too high. But while a low mutation probability can always be compensated by time (evolution will just progress slower), a high mutation probability stifles evolution entirely.
Anyway, while the weasel program is a blatant simplification of real life, I think it’s a great tool for demonstrating that random variation combined with non-random cumulative selection can produce complex structures. This is what evolution is all about; not monkeys hacking away at keyboards and winds blowing through airplane junkyards.
Share and Enjoy!
# Simplified I: Idiot Tax
“If you can’t explain it to a six-year-old, you don’t understand it yourself”
— Albert Einstein
The opening quote is attributed to Albert Einstein, even though it’s unlikely that he used these exact words. Nevertheless, the idea behind it has always resonated with me: If you explain something in easy words the person at the receiving end can immediately grasp it and is enlightened forever. But reducing a topic to its essence and explaining it from the vantage point of somebody else is even well worth the time for the one at the giving end: As always, the teacher learns the most.
I recently held an embedded programming workshop for kids and what I took home is this: if you pick the right examples and metaphors, even 12-year-olds can understand topics like pulse-width modulation. With this post, however, I want to address adults; specifically, I want to talk about something that has puzzled me for quite some time: why do people, even highly educated people, take part in lotteries?
A common knee-jerk reply is that everyone wants to get rich quick, without earning it. But that’s not the answer I am looking for. Let me refine my question: why do people play lotteries even though they know that the odds of winning are infinitesimally small?
Let’s take the German lottery system as an example. In order to become a millionaire, you have to get six numbers right, out of a total of 49. According to their official website, the odds to get these six numbers right are 1 to 15.537.573, or roughly 1 to 15 million.
Even though our gut feeling tells us that this is quite unlikely, we completely underestimate how unlikely it really is. Why? My guess is that it’s because we hear about figures that are in the millions every day. This guy is a millionaire, and that one even a billionaire. Government spends millions here and billions there. Still, we are nothing more than remotely acquainted with large numbers; we constantly fail to truly and fully comprehend them.
Let’s use an approach that has helped me quite a lot and treat big numbers like distances. What the human species has done for thousands of years is walk, so we naturally have a good grasp on distances. Let’s pretend we have a distance of 15 million centimeters.
15 million centimeters divided by 100 gives us 150 thousand meters, which equals 150 kilometers. People are especially good at understanding kilometers. 150 kilometers means roughly a two-hour car ride on a regular country road. That’s quite a distance, isn’t it?
Further, imagine a single M&M chocolate piece. It’s diameter is roughly one centimeter, give or take. Put one M&M after another, on a single straight line that is 150 kilometers long and you get what the number 15 million looks like. Somewhere along that line, there is that one special M&M where they printed a “\$” sign on the back, instead of the usual “m”. So you drive along this 150 kilometer line with your car, stop at some random point and flip a single M&M of your choice, hoping to see a “\$” sign on it’s back.
Now, how much money would you invest in such a bet? I don’t know about you, but I wouldn’t even invest a single cent. While it’s not impossible to win, it’s very, very improbable. Any money would surely be gone.
This is why the enlightened call lotteries “idiot tax”. It’s the tax that people—out of greed and ignorance—pay voluntarily.
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#### How to Find the Slope
What is a slope? The slope represents the change in the y coordinates over the change in the x coordinates between two points on a line. In other words, slope = change in y / change in x
Standard form of a linear equation.
Ax + By = C
Slope intercept form of a linear equation
Y= mx + b (m represents the slope; b represents the y-intercept)
------------------------------------------------------------------------------------------------------------
HOW TO FIND THE SLOPE GIVEN AN EQUATION
1) Change equation to slope intercept form by isolating Y
2) Observe equation for “ m” which is the slope
Example 1:
Y = 3 X + 5 ======= > Already in slope intercept form
Y = mx + b
Slope = 3 or 3/1 or m = 3 or m = 3/1; normally it is written as m = 3
Example 2:
8x + 4y = 12
8x – 8x + 4y = 12 – 8x ===== > Subtract 8x from both sides
4y = 12 – 8x
4y/4 = 12/4 – 8x/4 ======= > Divide both sides by 4
Y = 3 – 2x
Y = -2x + 3 ============== > Slope Intercept Form
Y= mx +b
Thus slope = -2 or m = -2
------------------------------------------------------------------------------------------------------------
FINDING THE SLOPE GIVEN TWO POINTS: (X1, Y1) and (X2, Y2)
As usual, there is a formula for this task ==== > m = Y2 – Y1 / X2 – X1
Remember, slope represents the change in the y- coordinates which is the numerator over the change in the x- coordinates which is the denominator.
The word “change” tells us to find the difference or to subtract the first coordinate from the second coordinate.
Example 3:
(3, 1) and (5, 4)
Let’s identify the coordinates.
X1 = 3
Y1 = 1
X2 = 5
Y2 = 4
Next, plug into the formula
M = 4 - 1 / 5 - 3
M = 3 / 2
Note: Something, you may not have considered. Take a moment and look at the slope formula again. m = Y2 – Y1 / X2 – X1
Notice Y2 is first in the numerator and X2 is first in the denominator. However, it’s okay to subtract the second coordinate from the first coordinate, but you must be consistent and do it for the numerator and denominator. Let’s try it.
M = Y1 – Y2 / X1 – X2
M = 1 – 4 / 3 - 5
M = -3 / -2
M = 3/2
Now, all you need to do is practice. There’s a link below for online practice. If you want an easy to understand workbook with helpful hints, consider the following to practice finding the slope and other Pre-Algebra concepts.
Mastering Essential Math Skills PRE-ALGEBRA CONCEPTS (Mastering Essential Math Skills)
You Should Also Read:
Practice - How to Find the Slope of a Line
Book Review - Mastering Essential Math Skills
Find the Equation of a Line
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# [seqfan] Re: Generally known (to those who know of such things)
Jonathan Post jvospost3 at gmail.com
Fri Nov 20 20:16:47 CET 2009
```"there are more primes that are 3 mod 4 than 1 mod 4." -- keep
reading. Use keywords "prime race"
Derbyshire, J. Prime Obsession: Bernhard Riemann and the Greatest
Unsolved Problem in Mathematics. New York: Penguin, pp. 125-126, 2004.
Sloane, N. J. A. Sequences A038691, A096628, A096629, and A096630 in
"The On-Line Encyclopedia of Integer Sequences."
Weisstein, Eric W. "Chebyshev Bias." From MathWorld--A Wolfram Web
Resource. http://mathworld.wolfram.com/ChebyshevBias.html
On Fri, Nov 20, 2009 at 10:52 AM, Victor S. Miller
<victorsmiller at gmail.com> wrote:
> I don't quite understand the notation in your sum:
>
> Do you mean that u(t) = sum(k=1 to t) k*p1[k]
>
> Where p1[k] is the k-th prime =1 mod 4?
>
> In any case this may be related to "Chebyshev's bias" there are more
> primes that are 3 mod 4 than 1 mod 4.
>
> Victor
>
> Sent from my iPhone
>
> On Nov 20, 2009, at 11:39 AM, "Meeussen Wouter \(bkarnd\)" <wouter.meeussen at vandemoortele.com
> > wrote:
>
>> I don't understand, again, why the primes (p1) of form 4n+1 differ in
>> behaviour from those (p3) of form 4n+3 in the following:
>>
>> u=Sum=1..p1; k (p1-k) ) is strictly increasing in function of p1,
>> while
>> v=Sum(k=1..p3; k (p3-k) ) is not (in function of p3).
>>
>> v is descending from v(37) to v(38)
>> {4, 28, 66, 190, 322, 558, 946, 1316, 1888, 2278, 2982, 3476, 3652,
>> 5768, 5992, 8636, 9170, 10008, 12382, 13366, 15698,
>> 16826, 20628, 21492, 22788, 26314, 26786, 32026, 33132, 37872, 39566,
>> 40752, 47892, 54114, 55608, 61766, 71082, 70464
>>
>> I checked 'u' way up to 10,000 and found no 'drops'.
>>
>>
>> Wouter.
>> ===============================
>> This email is confidential and intended solely for the use of the
>> individual to whom it is addressed.
>> If you are not the intended recipient, be advised that you have
>> received this email in error and that any use, dissemination,
>> forwarding, printing, or copying of this email is strictly prohibited.
>> You are explicitly requested to notify the sender of this email that
>> the intended recipient was not reached.
>>
>>
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
```
More information about the SeqFan mailing list
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# Arithmetic Operators in Visual Basic
Let’s now discuss the different kinds of operators under the Arithmetic function
There are 3 groups of operators in Visual Basic: we have the arithmetic operators, relational operators and the logical operators, but this article will only discuss about the arithmetic operators.
Arithmetic Operators
Addition is used to add two or more numbers (basic math). But in visual basic + sign is also used to add the value of two or more variables (numbers and text).
In our example above we: the compute SUM button will compute the sum of the Value 1 and Value 2, for example you will input 1 and 4, the result would be 5.
The add value button will combine the value of the two text field, for example you will input Hi and everybody the result would be Hi everybody.
Subtraction (-)
Subtraction operator is used to get the difference of the two numbers.
On the example above the user enter 10 on the Value 1 and 5 on the Value 2, the result will display if the user will click the Subtract button and the result for that is 5.
If you will enter a number in Value 2 which is higher to the Value 1 the result will be negative.
Multiplication (*)
Multiplication is a repeated addition. For example 2 * 3 is similar to 2 + 2 + 2. The result of multiplying two or more numbers is called a product. The multiplication process is represented by an asterisk.
You can also multiply a whole number and a decimal.
Division (/) (\)
Division in simplest form is the process of splitting the group into equal parts. In visual basic there are two kinds of division: the real division and the integer division.
Real division is represented by (/) a forward slash sign this process will retain the decimal point of the quotient. Example: 10 / 4 = 2.5
In integer division (division with no remainder) it is represented by (\) a backslash. It will ignore the decimal of the quotient and return only a whole number. Example 10 \ 4 =2
The Exponent Operator (^)
In visual basic the exponent operator is represented by ^ caret. It includes the base number and the exponent (to the power of). Example: 102 = 100
Modulo the remainder operator (Mod)
The Mod operator will get and display the remainder of the number after division. Example: 10 / 3, the remainder is 1.
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# Find the limit (x^2+x-6)/(x+3) when x approach -3.
limit (x^2+x-6)/(x+3) when x approaches -3
first we need to find the value of the function when x=-3
then, lim (-3^2 -3 -6)/-3+3 = 0/0
0/0 means that this method has failed because both fuctions has a root of -3
Now we try to factor the function:
limit (x^2+x-6)/(x+3) = lim...
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limit (x^2+x-6)/(x+3) when x approaches -3
first we need to find the value of the function when x=-3
then, lim (-3^2 -3 -6)/-3+3 = 0/0
0/0 means that this method has failed because both fuctions has a root of -3
Now we try to factor the function:
limit (x^2+x-6)/(x+3) = lim (x+3)(x-2)/(x+3 = lim (x-2)
then lim(x-2) as x --> -3 is (-3-2) = -5
Approved by eNotes Editorial Team
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# Is $\frac{x^2 + 2x}{x}$ a polynomial? [duplicate]
Is $$\dfrac{x^2 + 2x}{x}$$ a polynomial?
• One argument against it being a polynomial would be that the domain doesn't include zero, and a polynomial has always domain = $\mathbb{R}$ – Guillermo Mosse Jan 5 '17 at 14:42
• @SeñorBilly A polynomial does not "always" have domain $\mathbb{R}$. Remember, domains are specified, not intrinsic. – MathematicsStudent1122 Jan 5 '17 at 14:46
• @TonyK Definitely. If the expression is considered as an element of $\mathbb{R}(x)$ (the field of fractions of the polynomial ring $\mathbb{R}[x]$), then it is certainly a polynomial, because we can “cancel out the $x$”. As a function of a real variable, it is not a polynomial function, because its domain is not $\mathbb{R}$. – egreg Jan 5 '17 at 14:51
• @user1952009 You're comparing apples and oranges. – egreg Jan 5 '17 at 14:53
• @user1952009 Not really: 9/3 = 3 is always true; $x^2/x=x$ is only correct if $x \ne 0$; since the left-hand side is undefined for $x=0$. – StackTD Jan 5 '17 at 15:04
Some people would say that the rational number $7/1$ is not really equal to the integer $7$, but merely canonically identified with it. But (after reaching a certain level of sophistication) mathematicians say that $7/1$ and $7$ are indeed equal.
Some people would say that the rational function $$\frac{x^2+2x}{x} \tag{*}$$ is not really equal to the polynomial $$x+2, \tag{**}$$ but merely canonically identified with it. But (after reaching a certain level of sophistication) mathematicians say that (*) and (**) are indeed equal.
• It may depend on the context, but there is some subtlety to this analogy (as pointed out in the comments). If you, after canonical identification, indeed would say they are equal, then what about (what happens at) $x=0$...? – StackTD Jan 5 '17 at 15:14
Well, it's a polynomial in the variable $t=\tfrac{x^2+2x}{x}$... But you probably mean a polynomial in the (real?) variable $x$. What precise definition of polynomial are you using?
I would say no, because it is not of the form $$a_0+a_1x+a_2x^2+\ldots+a_nx^n$$ for any $n \in \mathbb{N}$ and real numbers $a_i$ ($0 \le i \le n$).
Note that you cannot just simplify $$\frac{x^2+2x}{x} = x+2$$ as this equality is only valid for non-zero $x$, so not for all $x$.
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Home / Reasoning Tricks / Tips to Solve Schedule Day& Date Question in Reasoning Section
# Tips to Solve Schedule Day& Date Question in Reasoning Section
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SCHEDULE DAY& DATE
IMPORTANT POINTS TO KNOW:
• In this type of questions a particular day/date in a particular year is given
• Either a day before or after the date is to found out
• Using this information we have to find out the day which falls on the given date in the question
• A normal year consists of 365 days
• Every fourth year consists of 366 days i.e., 1 day extra known as the leap year
• The extra day in the leap year is 29thof February
• When the number of days are converted into number of weeks we may get some remainder days known as odd days
• Normal year = 365 days = 52 weeks + 1 day
o Here the remaining 1 day = odd day
#### Get Best Maths Short Tricks Book With 2500+ Questions (50% Off)> Click Here
• Similarly, Leap year = 366 days = 52 weeks + 2 days
o Here there are 2 odd days in a leap year
• A month consists of 30, 31, 28 or 29 days depending on name of the month
• 31 days months are = January, March, May, July, August, October, December
• 30 days months are = April, June, September, November
• February consists of 28 days in a normal year and 29 days in a leap year
• A year can be determined as a leap year if it is divisible by 4
o Leap year = [year/4]
• There are 7 days in a week
• The number of odd days are crucial to find the day of the date given in the question
TABLE1:IF THE DAY TO BE FOUND IS AFTER THE REFERENCE DATE GIVEN IN THE QUESTION:
ODD DAY DAY OF THE WEEK 0 Previous Day 1 Given/Same day 2 Same day + 1 3 Same day + 2 4 Same day + 3 5 Same day + 4 6 Same day + 5 7 Same day + 6
TABLE 2:IF THE DAY TO BE FOUND IS BEFORE THE REFERENCE DATE GIVEN IN THE QUESTION:
ODD DAY DAY OF THE WEEK 0 Next Day 1 Given/Same day 2 Same day – 1 3 Same day – 2 4 Same day – 3 5 Same day – 4 6 Same day – 5 7 Same day – 6
QUESTION1:
The 1st day of the year 1998 was Thursday. If the birthday of Arun falls on 26th June, then on which day in 1998 was his birthday?
GIVEN:
1st January 1998 = Thursday
SOLUTION:
First we have to check whether 1998 is a normal or a leap year
1998 is not divisible by 4 i.e., it is a normal year with February = 28 days
Number of days to be calculated from 1stJanuary 1998 to 26th June 1998
January + February + March + April + May + June
= 31 + 28 + 31 + 30 + 31 + 26
= 177 days
Now to find the number of odd days convert the days into weeks
[177/7] = 25 weeks + 2 days
Therefore odd days = 2
Hence from table 1, 2 odd days = same day + 1
àThursday + 1 = Friday
Therefore,26th June 1998 was a Friday
QUESTION2:
The Independence Day was celebrated on Friday, 15th August in 1996. What was the 1stday of 1996?
GIVEN:
15th August 1996 = Friday
SOLUTION:
Here the day prior to the reference date is to be calculated
1st we have to find out whether 1996 is a normal/leap year
[1996/4] = 499
Hence, 1996 is a leap year = February = 29 days
Now the number of days from 1stJanuary 1996 to 15th August 1996 is to calculated
January + February + March + April + May + June + July + 15th August
= 31 + 29 + 31 + 30 + 31 + 30 + 31 + 15
= 228 days
To find the number of odd days
[228/7] = 32 weeks + 4 days
From Table 2,
4 odd days = Same day – 3 = Friday – 3 = Tuesday (3 days before Friday)
Therefore1stJanuary 1996 was a Tuesday
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# math
posted by on .
a 20-ounce bag of popcorn costs \$2.80. if the unit price stays the same, how much does a 35-ounce bag of popcorn cost?
• math - ,
20/2.80 = 35/x
Cross multiply and solve for x.
• math - ,
\$2.80/20 = x/35
Divide \$2.80 by 20 and multiply by 35.
• math - ,
20/2.80=35/x
20x=35*2.80
20x=98
divided by 20 on both sides
x=4.90?
• math - ,
Right.
• math - ,
thank you.
• math - ,
You're welcome.
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USE CODE: chait6
``` find square root
(x-y)4+2(x4+y4)-2(x2+y2)(x-y)2
```
6 years ago
Share
``` Hello Ajinkya,
lets say (x-y)4+2(x4+y4)-2(x2+y2)(x-y)2 as 'A'
A= (x-y)4+2(x4+y4)-2(x2+y2)(x-y)2
Taking the (x-y)2 common from 1st term and the 3 rd term in 'A' , we have,
A = 2(x4+y4) + (x-y)2((x-y)2-2(x2+y2))
We know that 2(x2+y2) = (x-y)2 + (x+y)2, on substituting this in 'A', we have
A = 2(x4+y4) +(x-y)2((x-y)2-(x-y)2-(x+y)2) , cancelling out + and - (x-y)2 ,we get
A = 2(x4+y4) -(x-y)2(x+y)2
we know that (a-b)*(a+b) = a2-b2 ; so (x-y)2(x+y)2 = (x2-y2)2
therefore, we are left with A = 2(x4+y4) - (x2-y2)2
Again we know that (x2-y2)2 + (x2+y2)2= 2(x4+y4)
on replaching 2(x4+y4) in A with above terms we get ,
A=(x2-y2)2 + (x2+y2)2-(x2-y2)2
We get A = (x2+y2)2
We are asked to find the sqrt of A , which will be x2+y2.
Best of luck ajinkya.
You can ask us any questions as we are dedicated IITians group specially here to answer your queries.
Muralidhar
```
6 years ago
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Ajay 6 months ago
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Ajay 6 months ago
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solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
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mycroft holmes one month ago
a cos A = b cos B 2R sin A cos A = 2R sin B cos B sin 2A = sin 2B Either A = B (isoceles or equilateral) or 2A = 180 o – 2B so that A+B = 90 o .(Right-angled)
mycroft holmes one month ago
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More Questions On Algebra
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# Calculate antiderivative of y=3x*(e raised to 3x)
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You should evaluate the integral of the function `y=3x*e^(3x).` Since you detect that the function y is a product of two simpler functions, `3x` and `e^(3x), ` you should use integration by parts.
Remember the formula of integration by parts:
u*v = int udv + int vdu
Put `u = 3x =gt du = 3dx`
Put `dv = e^(3x) dx=gt v = (e^(3x))/3`
Therefore you will have:
`3x*(e^(3x))/3 = int 3x*(e^(3x))dx + int (3*(e^(3x))/3) dx`
Separate the integral you should evaluate from the rest of terms.
`int 3x*(e^(3x))dx = 3x*(e^(3x))/3 - int (e^(3x)) dx`
`int 3x*(e^(3x))dx = 3x*(e^(3x))/3 - (e^(3x))/3 + c`
`int 3x*(e^(3x))dx = (e^(3x))/3* (3x-1) + c`
The antiderivative of the function y: `int 3x*(e^(3x))dx = (e^(3x))/3* (3x-1) + c`
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Anneross
41
# What is the solution to the equation? 5(8 + 7) = 5 • x + 5 • 7 A. x = 1 B. x = 5 C. x = 7 D. x = 8
$5(8+7)=5\cdot x+5\cdot7\\\\x=8\\\\Answer:D\\\\5\cdot x+5\cdot7=\underline{5}\cdot x+\underline{5}\cdot7=\underline{5}\cdot(x+7)=5\cdot(8+7)\iff x=8$
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# Integrals of Trigonometric Expressions Help
(not rated)
By — McGraw-Hill Professional
Updated on Aug 31, 2011
## Introduction to Integrals of Trigonometric Expressions
Trigonometric expressions arise frequently in our work, especially as a result of substitutions. In this section we develop a few examples of trigonometric integrals.
The following trigonometric identities will be particularly useful for us.
I. We have
The reason is that
cos 2x = cos2x − sin2x = [1 − sin2x ] − sin2x = 1 − 2 sin2x .
II. We have
The reason is that
cos 2x = cos2x − sin2x
sin2x = cos2x − [1 − cos2x ] = 2 cos2x − 1.
Now we can turn to some examples.
#### Example 1
Calculate the integral
∫ cos2x dx .
#### Solution 1
Of course we will use formula II . We write
#### Example 2
Calculate the integral
∫ sin3x cos2x dx .
#### Solution 2
When sines and cosines occur together, we always focus on the odd power (when one occurs). We write
sin3x cos2x = sinx sin2x cos2x = sin x (1 − cos2x ) cos2x
= [cos2x − cos4x ] sinx .
Then
∫ sin3x cos2 dx = ∫ [cos2x − cos4x ] sin x dx .
A u -substitution is suggested: We let u = cos x , du = − sin x dx . Then the integral becomes
Resubstituting for the u variable, we obtain the final solution of
You Try It : Calculate the integral
∫ sin23x cos53x dx .
Calculate
#### Solution 3
Substituting
into the integrand yields
Again using formula II, we find that our integral becomes
Applying formula II one last time yields
You Try It : Calculate the integral
You Try It : Calculate the integral
Integrals involving the other trigonometric functions can also be handled with suitable trigonometric identities. We illustrate the idea with some examples that are handled with the identity
#### Example 4
Calculate
∫ tan3x sec3x dx .
#### Solution 4
Using the same philosophy about odd exponents as we did with sines and cosines, we substitute sec 2 x − 1 for tan 2 x . The result is
∫ tan x (sec2x − l) sec3 x dx .
We may regroup the terms in the integrand to obtain
∫ [sec4 x − sec2 x ] sec x tan x dx .
A u -substitution suggests itself: We let u = sec x and therefore du = sec x tan x dx . Thus our integral becomes
Resubstituting the value of u gives
Calculate
#### Solution 5
We write
Letting u = tan x and du = sec2 x dx then gives the integral
You Try It : Calculate the integral
Further techniques in the evaluation of trigonometric integrals will be explored in the exercises.
Find practice problems and solutions for these concepts at: Methods Of Integration Practice Test.
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# ANALYSIS
Derived from the Latin word calor (meaning heat), and the Greek term metry (to measure), Calorimetry is the science of measuring the amount of heat. This means that all calorimetric techniques are therefore based on the measurement of heat that may be generated (in an exothermic process), consumed (endothermic), or simply dissipated by a sample. Experiment 302 focuses on the former two heat lost and heat gained. It should also be noted that calorimetry makes use of the law of heat exchange, which states that in an isolated system, the amount of heat given up by the hotter body is equal to the amount of heat absorbed by the colder body. In this case, the isolated system is the calorimeter and its contents. This relationship (referring to the numerical value of heat) thus gives the equation: Qloss = Qgained Equation 1
But since their signs will be opposite (lost = (-) & gained = (+)), this equation can be rewritten as: Qloss + Qgained = 0 Equation 1.1
Another equation to take note of is the equation for heat, involving the specific heat of a body (c), as shown below. Equation 2 Where: Q = heat (cal) m = mass (g) T = change in temperature, Final Temperature Initial Temperature (C)
Materials Used for the Duration of the Experiment The first part of the experiment utilized these first two equations, established the relationship between them and derived an equation in order to solve for the experimental value of the specific heats of both aluminum and copper. The resulting equation is as follows:
From the equation above, the values of certain data required can be inferred. Among these include: Water: mass, specific heat, change in temperature Calorimeter: mass, specific heat, change in temperature Metal: mass, change in temperature
After measuring these values, the specific heats of the two sample metals were calculated separately (in two trials). Refer to the table of data for Part 1. These values were then compared to the actual/accepted values listed in the table of specific heats. The results yielded moderately high percentages of error 19.78% and 25.21% respectively. These were most likely caused by erroneous readings from the electronic scale, and the thermometer. Part 1. Determining the Specific Heat of Metals Trial 1. Aluminum 45.6 g Mass of metal, mm 46.4 g Mass of calorimeter, mc 233.33 g Mass of water, mw Initial Temperature of metal, tom 71C Initial Temperature of Calorimeter, toc 28C Initial Temperature of Water, tow 28C Final Temperature of Mixture, tmix 30C Experimental Specific Heat of Metal, cm 0.2604 cal/g-C Actual Specific Heat of Metal, cm 0.2174 cal/g-C 19.78% Percentage of Error
Trial 2. Copper 19.9 g 46.4 g 150 g 72C 29C 29C 30C 0.1167 cal/g-C 0.0932 cal/g-C 25.21%
Substance Aluminum Beryllium Brass Copper Ethanol Ice Iron Lead Mercury
c, specific heat capacity 0.2174 0.4760 0.0917 0.0932 0.5800 0.5017 0.1123 0.0310 0.0329
## 0.2100 0.0560 1.000
The second part of this experiment dealt with the experimental determination of the latent heat of fusion of ice (Lf). The latent heat of a substance refers to the amount of heat needed to change the phase of unit mass without any change in temperature. The latent heat of fusion is used for situations that involve change of phase from solid to liquid, or vice versa. On the other hand, the latent heat of vaporization (Lv) is used for phase changes from liquid to gas or vice versa. The equations for heat involving latent heat are as follows: Equation 3 ....... Note: The signs (+/-) depend on what kind of process is taking place (endothermic/exothermic) Using equations 1-3, an equation wherein Lf is isolated can be derived, and the resulting equation is shown below.
Again, the values needed to completely solve for Lf can be inferred, and these are: Water: mass, specific heat, change in temperature Calorimeter: mass, specific heat, change in temperature Ice: mass, change in temperature After getting these values, experimentally, the latent heat of fusion of ice was computed (in two trials). Again, these were compared to the actual specific latent heat of fusion of ice, which is 80 cal/g.
The percentage errors yielded were 13.78% and 2.31% for trials 1 and 2, respectively.
Part 2. Latent Heat of Fusion of Ice Mass of calorimeter, mc Mass of water, mw Mass of mixture, mmix Mass of ice, mi Initial Temperature of Ice, toi Initial Temperature of Calorimeter, toc Initial Temperature of Water, tow Final Temperature of Mixture, tmix Experimental Latent Heat of Fusion, Lf Actual Specific Latent Heat of Fusion, Lf Percentage of Error Trial 1 46.4 g 229.8 g 257.3 g 27.5g 0 C 28C 28C 19C 69.98 cal/g 80 cal/g 13.78% Trial 2 46.4 g 196.4 g 215.4 g 18.8g 0 C 20C 20C 12C 81.85 cal/g 80 cal/g 2.31%
It should be noted that multiple erroneous or inaccurate readings from the thermometer can be disastrous to the experiment, as temperatures of the samples, the calorimeter, and the equilibrium temperature play a large part in all of the equations used for this experiment. This was what most likely caused the unusually high percentage errors of the results yielded.
CONCLUSION
As an experiment highlighting the use of calorimetry and the concept of heat, E302 focused on using an isolated system (a calorimeter) in order to solve for the experimental values of two properties of the samples given. In particular, performance of experiment 302 involved the calculation of the specific heat of two sample metals, particularly aluminum and copper, and the determination of the latent heat of fusion of ice. The first part dealt with the specific heat of the two metals. After gathering the relevant data, the following equation was then used to determine c m:
This value would then be treated as the experimental value of c and then compared to the accepted value listed on the table of specific heats. The second part of the experiment was all about computing for the experimental specific latent heat of fusion of ice, wherein two trials were performed. Again, the required data were first measured/gathered, and then the following equation, derived from the law of heat exchange, was used:
Like the first part, this value would be compared to the actual value and the percentage of error was calculated.
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > fcofo GIF version
Theorem fcofo 5424
Description: An application is surjective if a section exists. Proposition 8 of [BourbakiEns] p. E.II.18. (Contributed by FL, 17-Nov-2011.) (Proof shortened by Mario Carneiro, 27-Dec-2014.)
Assertion
Ref Expression
fcofo ((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) → 𝐹:𝐴onto𝐵)
Proof of Theorem fcofo
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 simp1 904 . 2 ((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) → 𝐹:𝐴𝐵)
2 ffvelrn 5300 . . . . 5 ((𝑆:𝐵𝐴𝑦𝐵) → (𝑆𝑦) ∈ 𝐴)
323ad2antl2 1067 . . . 4 (((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) ∧ 𝑦𝐵) → (𝑆𝑦) ∈ 𝐴)
4 simpl3 909 . . . . . 6 (((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) ∧ 𝑦𝐵) → (𝐹𝑆) = ( I ↾ 𝐵))
54fveq1d 5180 . . . . 5 (((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) ∧ 𝑦𝐵) → ((𝐹𝑆)‘𝑦) = (( I ↾ 𝐵)‘𝑦))
6 fvco3 5244 . . . . . 6 ((𝑆:𝐵𝐴𝑦𝐵) → ((𝐹𝑆)‘𝑦) = (𝐹‘(𝑆𝑦)))
763ad2antl2 1067 . . . . 5 (((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) ∧ 𝑦𝐵) → ((𝐹𝑆)‘𝑦) = (𝐹‘(𝑆𝑦)))
8 fvresi 5356 . . . . . 6 (𝑦𝐵 → (( I ↾ 𝐵)‘𝑦) = 𝑦)
98adantl 262 . . . . 5 (((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) ∧ 𝑦𝐵) → (( I ↾ 𝐵)‘𝑦) = 𝑦)
105, 7, 93eqtr3rd 2081 . . . 4 (((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) ∧ 𝑦𝐵) → 𝑦 = (𝐹‘(𝑆𝑦)))
11 fveq2 5178 . . . . . 6 (𝑥 = (𝑆𝑦) → (𝐹𝑥) = (𝐹‘(𝑆𝑦)))
1211eqeq2d 2051 . . . . 5 (𝑥 = (𝑆𝑦) → (𝑦 = (𝐹𝑥) ↔ 𝑦 = (𝐹‘(𝑆𝑦))))
1312rspcev 2656 . . . 4 (((𝑆𝑦) ∈ 𝐴𝑦 = (𝐹‘(𝑆𝑦))) → ∃𝑥𝐴 𝑦 = (𝐹𝑥))
143, 10, 13syl2anc 391 . . 3 (((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) ∧ 𝑦𝐵) → ∃𝑥𝐴 𝑦 = (𝐹𝑥))
1514ralrimiva 2392 . 2 ((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) → ∀𝑦𝐵𝑥𝐴 𝑦 = (𝐹𝑥))
16 dffo3 5314 . 2 (𝐹:𝐴onto𝐵 ↔ (𝐹:𝐴𝐵 ∧ ∀𝑦𝐵𝑥𝐴 𝑦 = (𝐹𝑥)))
171, 15, 16sylanbrc 394 1 ((𝐹:𝐴𝐵𝑆:𝐵𝐴 ∧ (𝐹𝑆) = ( I ↾ 𝐵)) → 𝐹:𝐴onto𝐵)
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 97 ∧ w3a 885 = wceq 1243 ∈ wcel 1393 ∀wral 2306 ∃wrex 2307 I cid 4025 ↾ cres 4347 ∘ ccom 4349 ⟶wf 4898 –onto→wfo 4900 ‘cfv 4902 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-io 630 ax-5 1336 ax-7 1337 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-8 1395 ax-10 1396 ax-11 1397 ax-i12 1398 ax-bndl 1399 ax-4 1400 ax-14 1405 ax-17 1419 ax-i9 1423 ax-ial 1427 ax-i5r 1428 ax-ext 2022 ax-sep 3875 ax-pow 3927 ax-pr 3944 This theorem depends on definitions: df-bi 110 df-3an 887 df-tru 1246 df-nf 1350 df-sb 1646 df-eu 1903 df-mo 1904 df-clab 2027 df-cleq 2033 df-clel 2036 df-nfc 2167 df-ral 2311 df-rex 2312 df-v 2559 df-sbc 2765 df-un 2922 df-in 2924 df-ss 2931 df-pw 3361 df-sn 3381 df-pr 3382 df-op 3384 df-uni 3581 df-br 3765 df-opab 3819 df-mpt 3820 df-id 4030 df-xp 4351 df-rel 4352 df-cnv 4353 df-co 4354 df-dm 4355 df-rn 4356 df-res 4357 df-ima 4358 df-iota 4867 df-fun 4904 df-fn 4905 df-f 4906 df-fo 4908 df-fv 4910 This theorem is referenced by: fcof1o 5429
Copyright terms: Public domain W3C validator
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Close
## What degree is 30 percent?
Table of Common Slopes in Architecture
14.04° 1 : 4 25%
15° 1 : 3.73 26.8%
26.57° 1 : 2 50%
30° 1 : 1.73 57.7%
## What is the central angle of 30 percent?
Total percentage = 100….Pie Chart.
Item Expenditure (in percentage) Central Angle
Cement 20% (²/₁₀₀ × 360)° = 72°
Steel 10% (¹/₁₀₀ × 360)° = 36°
Labour 25% (²⁵/₁₀₀ × 360)° = 90°
Miscellaneous 30% (³/₁₀₀ × 360)° = 108°
## How do you convert percentages into degrees for a pie chart?
Answer. To convert the degree into percentage in pie chart: As we know to find the percentage we divide the number by total of whatever amount and multiply it by 100, now in this case degree needs to be converted into percentage.
## How do you find the degrees in a pie chart?
Divide the count for each variable by the total count and then multiply this figure by 360 to determine the angle in degrees. When the pie chart is drawn, the slice of pie for Group A will have an angle of 196.36°; as such, it will represent just over half of the full circle.
## How do you turn data into a pie chart?
Excel
2. Click Insert > Insert Pie or Doughnut Chart, and then pick the chart you want.
3. Click the chart and then click the icons next to the chart to add finishing touches:
## Does a pie chart have to equal 100?
Pie charts are designed to show parts of a whole, so any sum below or above 100% doesn’t represent the entire picture. Finally, when it comes to legends, pie charts don’t generally need one.
## What is a pie chart used for?
A pie chart shows how some total amount is divided among distinct categories as a circle (the namesake pie) divided into radial slices. Each category is associated with a single slice whose size corresponds with the category’s proportion of the total.
## What can I use instead of a pie chart?
Stacked Bar Charts are the closest linear equivalent to Pie Charts, in terms of both one-to-one mapping and layout. They may be the best alternatives to Pie charts. A single-series Pie chart with N slices is actually equivalent with N series of Full 100% Stacked Bars, each with one single value.
## What makes a good pie chart?
Pie charts work best if you only have a few values – four max. If you have more than four shares, consider a stacked column or stacked bar chart. Not only will it look less cluttered, but also the labelling will be tidier. Pie charts might be unnecessary if you only want to show two values.
## When should you make a pie chart?
A pie chart is best used when trying to work out the composition of something. If you have categorical data then using a pie chart would work really well as each slice can represent a different category. A good example of a pie chart can be seen below.
## When should you avoid a pie chart Doughnut chart?
You should avoid using pie charts when: Your data values are not distinctly separated; data analysis using a pie chart becomes restricted when dealing with data points of similar sizes. You need to compare data for more than one metric. You need to showcase specific data values and facilitate a part-to-part comparison.
## What are the disadvantages of a pie chart?
• do not easily reveal exact values.
• Many pie charts may be needed to show changes over time.
• fail to reveal key assumptions, causes, effects, or patterns.
• be easily manipulated to yield false impressions.
## What makes a bad pie chart?
From a design point of view, a pie chart takes up far too much space to convey a set of data compared to other options. In addition, the labels don’t line up, so the result becomes cluttered and hard to read — strike three against pie charts, as they often make the data more complicated than before.
## Why is Tufte flat wrong pie chart?
Tufte is wrong to make an assertion about pie charts based on his own context (the analysis and presentation of complex data) and use broad strokes to apply that to domains where he has no expertise (presenting and selling ideas in the boardroom). Pie charts have earned their place in your business presentations.
## Why are 3D charts bad?
This is not an original idea. This argument is really about how pie charts are bad, and adding 3D makes them even worse. The underlying point is that 3D tends to skew how we perceive the data. You may think you are adding depth and visual intrigue, but you are actually making it harder to understand.
## Are 3D pie charts misleading?
Pie chart. The usage of percentages as labels on a pie chart can be misleading when the sample size is small. Making a pie chart 3D or adding a slant will make interpretation difficult due to distorted effect of perspective. Bar-charted pie graphs in which the height of the slices is varied may confuse the reader.
## When should a 3D chart be used?
Another use for a 3d chart is to emphasise important information. Following on from my previous blog on why two charts are often better than one, adding a 3d chart as an auxiliary chart can bring attention to noteworthy trends or values comparisons (NOT individual values).
## When would you use a 3D plot?
Use a 3D surface plot to see how a response variable relates to two predictor variables. A 3D surface plot is a three-dimensional graph that is useful for investigating desirable response values and operating conditions. A surface plot contains the following elements: Predictors on the x- and y-axes.
## How do you explain a 3D plot?
3D scatter plots are used to plot data points on three axes in the attempt to show the relationship between three variables. Each row in the data table is represented by a marker whose position depends on its values in the columns set on the X, Y, and Z axes.
## How do you plot a 3D function?
1. Draw x, y, and z axes.
2. Roughly determine the domain of the function you will be graphing.
3. Draw the box enclosing the three planes just drawn.
4. Calculate the slice curves for x = 1, 0, -1 and draw them in on the appropriate planes.
5. Draw the slice curves for y = 1, 0, -1 onto the appropriate planes.
## How do you plot a 3D surface?
Basic steps
1. Establish the domain by creating vectors for x and y (using linspace, etc.)
2. Create a “grid” in the xy-plane for the domain using the command meshgrid.
3. Calculate z for the surface, using component-wise computations.
4. Plot the surface. The main commands are mesh(x,y,z) and surf(z,y,z)
## How do you make a 3d function in Python?
Three-Dimensional Plotting in Matplotlib
1. from mpl_toolkits import mplot3d.
2. fig = plt. figure() ax = plt. axes(projection=’3d’)
3. fig = plt. figure() ax = plt. axes(projection=’3d’) ax.
4. ax. view_init(60, 35) fig. Out[7]:
5. fig = plt. figure() ax = plt. axes(projection=’3d’) ax.
6. ax = plt. axes(projection=’3d’) ax.
7. theta = 2 * np. pi * np.
8. In [12]: ax = plt.
## How do you make an interactive 3d plot in Python?
Steps
1. Create a new figure, or activate an existing figure.
2. Create fig and ax variables using subplots method, where default nrows and ncols are 1, projection=’3d”.
3. Get x, y and z using np. cos and np.
4. Plot the 3D wireframe, using x, y, z and color=”red”.
5. Set a title to the current axis.
6. To show the figure, use plt.
## What is 3d surface?
A 3D surface model is a digital representation of features, either real or hypothetical, in three-dimensional space. Some simple examples of 3D surfaces are a landscape, an urban corridor, gas deposits under the earth, and a network of well depths to determine water table depth.
## Is Plotly Python free?
Yes. Plotly for Python is free and open-source software, licensed under the MIT license. It costs nothing to install and use. You can view the source, report issues or contribute using our Github repository.
## How do you make an interactive plot in Python?
You can make a plot in matplotlib, add interactive functionality with plugins that utilize both Python and JavaScript, and then render it with D3. mpld3 includes built-in plugins for zooming, panning, and adding tooltips (information that appears when you hover over a data point).
## How do I make a Matplotlib plot interactive?
To configure the integration and enable interactive mode use the %matplotlib magic:
1. In [1]: %matplotlib Using matplotlib backend: Qt5Agg In [2]: import matplotlib.pyplot as plt.
2. In [3]: fig, ax = plt. subplots()
3. In [4]: ln, = ax. plot(range(5))
4. In [5]: ln. set_color(‘orange’)
5. In [6]: plt. ioff()
6. In [7]: plt. ion()
## What does Panda do in Python?
pandas is a software library written for the Python programming language for data manipulation and analysis. In particular, it offers data structures and operations for manipulating numerical tables and time series.
## How do you read a 3D scatter plot?
Interpret the key results for 3D Scatterplot
1. Data points that tend to rise together suggest a positive correlation.
2. Data points that tend to rise as other data points tend to decline suggests a negative correlation.
3. Outliers fall far from the main group of data points.
2019-08-28
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# What is a sentence that uses a prepositional phrase as an adverb?
## What is a sentence that uses a prepositional phrase as an adverb?
The ballerina danced across the floor. The object of the preposition is “floor”. The preposition “across” is telling us the relationship between floor and “danced”. “Danced” is a verb, so the prepositional phrase is an adverb phrase.
## What is the difference between a term and a phrase?
As nouns the difference between phrase and term is that phrase is a short written or spoken expression while term is limitation, restriction or regulation.
How did you classify a polynomial from not a polynomial?
Answer: An algebraic expression is not a polynomial when there are square roots, negative powers, and variables in the denominator of any fractions….
What do you call a 4 term polynomial?
Polynomials can be classified by the number of terms with nonzero coefficients, so that a one-term polynomial is called a monomial, a two-term polynomial is called a binomial, and a three-term polynomial is called a trinomial. The term “quadrinomial” is occasionally used for a four-term polynomial.
### How did you classify a polynomial?
Answer. Answer: Polynomial is being categorized according to the number of terms and the degree present. Based on the value, one term is called as monomial (when n = 1), two-degree polynomial (when n = 2) and three-degree polynomial (when n = 3)….
### How do you classify the number of terms?
Example: The polynomial is classified by the number of terms as:
1. Monomial – One term – 3x.
2. Binomial – Two Term – 7a-5.
3. Trinomial – Three Term –
What are the classification of polynomials according to the degree?
Table 10.2 Classifying a Polynomial Based on Its Degree
Degree Classification Example
0 constant 2×0 or 2
1 linear 6×1 + 9 or 6x + 9
2 quadratic 4×2 – 25x + 6
3 cubic x3 – 1
Is 5x a polynomial?
Different types of polynomials Monomials – these are polynomials containing only one term (“mono” means one.) 5x, 4, y, and 5y4 are all examples of monomials. Trinomials – a trinomial is a polynomial that contains three terms (“tri” meaning three.)…
#### What is the degree of 9x?
Each term in a polynomial has what’s called a degree, or a value based on the exponent attached to its variable. The degree of 9×2 is 2, for example.
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## Drawing Conical Objects
The last of the four basic forms is the cone. Probably the first object that pops into your mind is an ice cream cone. But there are bottles, glasses, lampshades, bowls, and many other man-made objects with shapes based upon a cone. There are also countless creations of nature—sea shells, flowers, and trees—with conical shapes.
When you begin to draw conical objects, remember that a cone is a solid mass that tapers uniformly from a circular base to a point. Look for this form first in the object you're drawing, whether it's long and thin like a beer glass, or short and broad like a lampshade.
### SYMMETRY OF THE CONE
The best way to draw a symmetrical cone is to begin with a center line. Then draw the ellipse at right angles to this center line. Mark the place on the center line where the tip of the cone should be, depending on its height. Having established the base and the tip, it's a simple matter to run two diagonal lines from the tip to the ends of the ellipse to form the basic cone. (See the demonstration at right.) Now you can begin to add the details that pertain to your particular object.
1. The artist began by drawing the entire geometric cone. Notice the line drawn through the cone's center. The artist used that line to help him establish the proper proportions and thus, the cone's symmetry.
2. Having established the dimensions of the cone, the artist departed from the geometric form by adding detail. The detail transformed the geometric shape into a recognizable object, an ice cream cone.
This sea shell has two cones forming its construction. In order to achieve its proper proportions, you must ignore, at first, the shell's rather elaborate detail. Here the artist has indicated the two cone forms and their relative positions within the large sea shell.
### DRAWING EVERYTHING
Draw as many conical forms as you can find. When you're finished with the last demonstration in this section, you'll have achieved a great deal. It's really a fine accomplishment to be able to draw—in their correct proportions—all the objects suggested here. If you've mastered the four basic forms, you can draw anything in existence. Imagine!
This sea shell has two cones forming its construction. In order to achieve its proper proportions, you must ignore, at first, the shell's rather elaborate detail. Here the artist has indicated the two cone forms and their relative positions within the large sea shell.
+2 0
## How To Become A Professional Pencil Drawing Artist
Realize Your Dream of Becoming a Professional Pencil Drawing Artist. Learn The Art of Pencil Drawing From The Experts. A Complete Guide On The Qualities of A Pencil Drawing Artist.
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### Responses
• marisa
How to sketch conical objects?
8 years ago
• Aapo
How to draw aniME CONE SHELLS?
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How to draw seashells?
8 years ago
• Olli-Pekka
How to draw a realistic shell?
8 years ago
• daavid
How does a seashell form?
8 years ago
How to draw seashells step by step?
7 years ago
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How to draw a seashell?
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How to draw a seashell step by step?
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3 years ago
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+0
# Algebra -- Please quickly solve -- (not to be rude)
0
58
3
1) Alice and Bob each have a certain amount of money. If Alice receives n dollars from Bob, then she will have 3 times as much money as Bob. If, on the other hand, she gives n dollars to Bob, then she will have 2 times as much money as Bob. If neither gives the other any money, what is the ratio of the amount of money Alice has to the amount Bob has?
2) At the Dessert Island Pastry Cafe, there are 2 cakes for every 7 donuts and 11 brownies for every 3 cakes. What is the ratio of brownies to donuts?
Guest Oct 9, 2017
edited by Guest Oct 9, 2017
edited by Guest Oct 9, 2017
edited by Guest Oct 9, 2017
Sort:
#1
0
Its been 1800 secounds!!!!!
and I need to go in like 10 secounds ......
i need you to compose quicker!!!!
not to be rude
Guest Oct 9, 2017
#2
+76972
+2
Let the amount of Alice's original money = A
Let the amount of Bob's original money = B
So
(A + n )* 3 = (B - n) → 3A + 3n = B - n → 3A - B = -4n (1)
(A - n) = 2 * ( B + n) → A - n = 2B + 2n → A - 2B = 3n (2)
Multiply the first equation by 3 and the second equation by 4 and add them together
9A - 3B = -12n
4A - 8B = 12n
13A - 11B = 0 add 11B to both sides
13A = 11B divide both sides by 13, B
A / B = 11 / 13
At the Dessert Island Pastry Cafe, there are 2 cakes for every 7 donuts and 11 brownies for every 3 cakes. What is the ratio of brownies to donuts?
2 cakes : 7 donuts → 6 cakes : 21 donuts
3 cakes : 11 brownies → 6 cakes : 22 brownies
22 brownies : 21 donuts
CPhill Oct 9, 2017
#3
0
CPhill number 1 is wrong!!!!! how could you do this to a fellow guest?????
Guest Oct 10, 2017
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Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
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# Wire Rope Slings
## Sling angles and influence on capacity.
Slings angles affects ropes capacities.
If angle - alpha - is measured between
• the vertical line (as with gravity force), and
• the rope or wire
the relative capacity compared to a vertical straight lifting is reduced with reduction factor as indicated below.
f = cos(α) (1)
where
f = reduction factor
α = angle between vertical line and rope (degrees)
Angle
- α -
(degrees)
Reduction Factor
- f -
0 1.000
10 0.985
20 0.940
30 0.866
40 0.766
50 0.643
60 0.500
70 0.342
### Example - Capacity of a Single Rope or Wire
The capacity of a single rope that follows a vertical line is 100% since the reduction factor is 1.
If the weight of a body is W - the load in the wire is
F = W (2)
where
F = force in rope (N, lbf)
W = m g = weight of body (N, lbf)
m = mass of body (kg, slugs)
g = acceleration of gravity (9.81 m/s2, 32.17 ft/s2)
For a body with mass 100 kg the load in the rope can be calculated
F = (100 kg) (9.81 m/s2)
= 981 N
= 9.8 kN
### Example - Capacity of Two Ropes (or Wires)
#### Two wires or ropes follows the vertical line
The capacity of two wires that follows the vertical line is 100% since the reduction factor is 1.
If the weight of a body is W - the load in each wire is
F = W / 2 (3)
For a body with weight 1000 N the load in each rope can be calculated as
F = (1000 N) / 2
= 500 N
= 0.5 kN
#### Two wires - or ropes - with angle 30o to the vertical line
The capacity of two wires with angle 30o to the vertical line is 86.6% since the reduction factor is 0.866.
If the weight of a body is W - the load in each wire is
F = (W / 2) / cos(30o)
= (W / 2) / f
= (W / 2) / 0.866
= 0.577 W (4)
For a body with weight 1000 N the loads in the ropes can be calculated
F = 0.577 (1000 N)
= 577 N
= 0.58 kN
### Wire Rope Slings Calculators
The calculators below can be used to calculate wire rope forces. Note that mass (kg) and not weight (N) is used as input.
mass (kg)
length - a (m)
height - h (m)
mass (kg)
length - a (m)
height - h (m)
mass (kg)
length - a (m)
length - b (m)
height - h (m)
## Related Topics
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• Statics - Loads - forces and torque, beams and columns.
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Temperature
oC
oF
Length
m
km
in
ft
yards
miles
naut miles
Area
m2
km2
in2
ft2
miles2
acres
Volume
m3
liters
in3
ft3
us gal
Weight
kgf
N
lbf
Velocity
m/s
km/h
ft/min
ft/s
mph
knots
Pressure
Pa (N/m2)
bar
mm H2O
kg/cm2
psi
inches H2O
Flow
m3/s
m3/h
US gpm
cfm
6 24
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# Do not use the for loop to calculate the first number greater than 0 in each line
1 vue (au cours des 30 derniers jours)
slevin Lee le 19 Nov 2022
Modifié(e) : Jan le 19 Nov 2022
x=randn(10,10)
...
##### 2 commentairesAfficher AucuneMasquer Aucune
Jan le 19 Nov 2022
This sounds like a homework question. So please show, what you have tried so far and ask a specific question about Matlab. The forum will not solve your homework, but assist you in case of questions concerning Matlab.
slevin Lee le 19 Nov 2022
Modifié(e) : Jan le 19 Nov 2022
x=randn(10,10)
x1=x>0
x2=cumsum(x1,2)
x3=x2==0
x4=sum(x3,2)+1
x5=x(:,x4).*eye(10,10)
sorry!
I just want to see if there's a better way
Connectez-vous pour commenter.
### Réponse acceptée
Jan le 19 Nov 2022
Modifié(e) : Jan le 19 Nov 2022
x = randn(10, 10)
y = x.'; % Required for the last step y(y3==1)
y1 = y > 0
y2 = cumsum(y1)
% Now a row can contain multiple 1's. Another CUMSUM helps:
y3 = cumsum(y2)
y(y3 == 1)
Another way:
x = randn(10, 10)
m = x > 0
y = cumsum(x .* m, 2)
min(y, [], 2)
Or with beeing nitpicking for the formulation "no for loop":
result = zeros(1, 10);
m = x > 0;
k = 1;
while k <= 10
row = x(k, :);
result(k) = row(find(row(m(k, :)), 1));
k = k + 1;
end
And another method:
[row, col] = find(x > 0);
first = splitapply(@min, col, row);
x(sub2ind(size(x), 1:10, first.'))
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens
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### Plus de réponses (1)
DGM le 19 Nov 2022
Modifié(e) : DGM le 19 Nov 2022
Here's one way.
x = randn(10,10)
x = 10×10
0.5355 -1.4904 -0.9662 -0.7693 0.6789 0.6031 0.2375 -1.0743 0.6916 -0.8738 -0.0884 0.5313 -0.2913 1.0506 -1.1930 0.1070 0.2047 -0.2512 2.0512 0.1435 -0.1235 0.5840 1.3542 -0.3692 1.1048 -0.4211 -2.1762 -0.3553 0.8296 -2.1670 -0.6151 0.7344 -1.1325 -1.9521 -1.5595 0.8968 -0.8484 -2.5253 -0.1668 1.8909 1.0192 -0.4541 0.9363 -0.1635 -1.7753 -0.2266 1.1185 2.0062 -1.3746 -0.3283 -0.2484 -1.0466 1.2285 0.4551 -1.1110 -1.1021 0.3378 0.2945 -0.2886 1.0569 1.4228 -1.4178 0.5278 0.1037 0.7001 -0.3782 0.3081 1.4122 0.6484 -1.0982 -0.0277 -1.6063 -0.8334 -0.2916 -1.0505 -0.6081 -0.4252 -1.0511 0.4218 0.2551 -0.2180 0.5246 0.2145 1.4641 2.2411 -1.4396 -0.2848 -0.4727 2.5402 2.0645 0.5302 0.6045 -0.7767 1.2057 -0.9240 0.0234 1.1590 -0.0401 1.3279 0.5901
[~,idx] = min(x<=0,[],2) % note the logical inversion
idx = 10×1
1 2 2 2 1 3 1 9 2 1
firstnz = x(sub2ind(size(x),(1:size(x,1))',idx)) % get the values from the array
firstnz = 10×1
0.5355 0.5313 0.5840 0.7344 1.0192 1.2285 1.4228 0.4218 0.5246 0.5302
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens
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# Notation confusion about time derivative of a vector in a rotating frame
As far as I can tell, this question, or similar ones, have been asked a number of times:
Derivation of the time-derivative in a rotating frame of refrence
Time derivatives in a rotating frame of reference
Time derivative in rotating frame
The time derivatives of vectors in rotating frames
That said, I have read all of these posts, and remain confused, I think largely by notation. My goal is to understand the derivation for (Eq. 0):
$$\left( \frac{d\vec{A}}{dt} \right)_S = \left( \frac{d\vec{A}}{dt} \right)_{S'}+\left( \vec{\omega} \times \vec{A}\right)$$
Where $$\vec{A}$$ is any random vector, S is an inertial frame, S' is a non-inertial, rotating frame at angular velocity $$\vec{\omega}$$ with respect to S.
In particular, I am confused in many notations by things like:
1. $$(\vec{A})_S$$
2. $$\left(\frac{d\vec{A}}{dt}\right)_S$$
What do they mean?
Here is my best guess, but please assist me if I have made mistakes:
$$\vec{A}$$ represents some vector quantity (say position vector) that has a meaningful expression with respect to both the coordinates of frame S, and frame S', but $$\vec{A}$$ is a geometric object that exists independant of any frame. So one could certainly write: (Eq. 1) $$\vec{A} = A_i \hat{e_i} = A_i' \hat{e_i}'$$ where $$\hat{e_i}$$ are the basis vectors in frame S, and $$\hat{e_i}'$$ are the basis vectors in frame S'. Eq. 1 says "The geometric object $$\vec{A}$$ may be written in two different bases".
On the other hand $$(\vec{A})_S$$ is a completely different thing, this is not a geometric object, but a mathematical object that holds the components of geometric object $$\vec{A}$$ when written in the basis of frame S. Thus: $$(\vec{A})_S = vector(A_1,A_2,A_3)$$ and $$(\vec{A})_{S'} = vector(A_1',A_2',A_3')$$ and, as mathematical objects $$(\vec{A})_{S'} \ne (\vec{A})_S$$ because their components are not the same. (But $$(\vec{A})_{S'} = R(\vec{A})_S$$ where R is an appropriate transformation matrix)
Here we see that vectors like $$\vec{A}$$ are geometric objects, and vectors like $$(\vec{A})_S$$ are mathematical objects, corresponding to the expression of $$\vec{A}$$ in frame S. Thus the two are incompatible, this is one confusion of mine, because both of these types of objects are used in the equation at the top I am trying to understand (Eq. 0). I believe this is meaningless.
Similarly, $$\frac{d\vec{A}}{dt}$$ represents the geometric object of the time derivative of $$\vec{A}$$ which may be expressed in either frame S or S'. (i.e. $$\frac{d\vec{A}}{dt} = \left(\frac{d\vec{A}}{dt}\right)_i \hat{e_i} = \left(\frac{d\vec{A}}{dt}\right)_i' \hat{e_i}'$$). However, we now need to be careful, because the question "what is $$\frac{dA_i \hat{e_i}}{dt}$$ or $$\frac{dA_i' \hat{e_i}'}{dt}$$?" becomes more complicated, because the answer depends on which frame we are observing the object from. Somehow? I think? I'm wildly confused at this point.
Any insights would be useful. Help understanding Q. 1 and 2, as well as the below A and B as well as any insight into the paragraphs I tried to write above.
A) Is this statement true? $$\left( \frac{d\vec{A}}{dt} \right)_S = \frac{d((\vec{A})_S)}{dt}$$. What does it mean?
B) Does $$\frac{d}{dt}$$ have any meaning? Or must we write $$(\frac{d}{dt})_S$$ (i.e. with reference to one frame)
• Closely related: physics.stackexchange.com/q/515944/226902 Commented Jul 27, 2023 at 13:09
• If notations are confusing, then work out your own notation. There are multiple ways to describe rotating derivatives, all of which are correct. Commented Jul 27, 2023 at 15:06
• Related if not Duplicate : Velocity in a turning reference frame. Commented Jul 27, 2023 at 20:26
Consider a rotating object and some vector quantity $$\vec{A}$$. Now, we can decompose the vector into basis vectors as follows:
$$\vec{A} = A^i(t) e_i(t)$$
where both the components and the basis vectors are endowed with time dependence as the object is rotating since we are in an inertial frame. If we take the time derivative, we have to apply the leibniz rule as follows
$$\frac{d\vec{A}}{dt} = \frac{A^i(t)}{dt} e_i(t) + A^i(t) \frac{de_i(t)}{dt} = \dot{A}^i e_i + \vec{\omega} \times \vec{A}$$
Now this holds for a frame where $$\textbf{we know the object is rotating}$$. However, imagine you were in that rotating frame. Unlike the other case, you do not rotate, everything else rotates. Therefore, the basis that was thought to be rotating and time dependent is now time-independent for you. Therefore, when you take derivative of the vector $$\vec{A}$$, you will only get
$$\frac{d\vec{A}}{dt} = \frac{A^i(t)}{dt} e_i(t)$$
Therefore, we can write
$$\left (\frac{d\vec{A}}{dt} \right)_S = \left (\frac{d\vec{A}}{dt} \right)_{S'} + \vec{\omega} \times \vec{A}$$
Consider two senarios. An observer {S} fixed at the center of rotation observing the vector $$\vec{A}$$ and how it changes with time, and an observer {S'} co-rotating with the rotating frame at the same location, and their observation of vector $$\vec{A}$$.
The equation presented relates the first observation to the second observation.
$$\underbrace{ \left( \dot{ \vec{A} } \right)_S }_{\text{change by fixed observer}} = \underbrace{\left( \dot{ \vec{A} } \right)_{S'}}_{\text{change by co-rotating observer}} + \vec{\Omega} \times \underbrace{ \left( \vec{A} \right)_S}_{\text{vector by fixed observer}}$$
where $$\vec{\Omega}$$ is the rotational velocity of the rotating frame, expressed in the fixed frame.
You’re perfectly justified in being confused. The notation isn’t great, and this is made worse by the fact that everything is happening in $$\Bbb{R}^3$$, so it is very easy to get confused by a vector, its components with respect to one basis, its components with respect to a different basis, and also the unnatural use of the cross product (I know you didn’t object to this, but this is confusing (atleast for me when I was learning)). Let us therefore speak in general terms, so as to limit the chances for such confusions. Also, for the most part I won’t talk about two systems; I’ll just talk about one and do things generally so that for the second, you just apply the analogous equations.
Let $$V$$ be an $$n$$-dimensional real (or complex) vector space. Let $$I\subset\Bbb{R}$$ be an open interval, and $$\xi_1,\dots,\xi_n:I\to V$$ be differentiable maps such that for each $$t\in I$$, $$(\xi_i(t))_{i=1}^n$$ is an ordered basis for $$V$$; we may write this as simply $$\xi(t)$$. Now, let us define an operator $$\frac{d_{\xi}}{dt}$$:
Definition 1. (‘Time derivative relative to $$\xi$$’)
With the notation $$V,\xi$$ as above, for each differentiable map $$f:I\to V$$, let us first denote the components of $$f$$ relative to the $$\xi$$ basis as $$f_{\xi}^1,\dots,f_{\xi}^n:I\to\Bbb{R}$$, meaning that for each $$t\in I$$, we have $$f(t)=f_{\xi}^i(t)\xi_i(t)$$ (Einstein summation used). Next, we shall define the operator $$\frac{d_{\xi}}{dt}$$ as follows: its action on a function $$f$$ is defined to be \begin{align} \frac{d_{\xi}f}{dt}\bigg|_t:=\frac{d f_{\xi}^i}{dt}\bigg|_t\,\xi_i(t). \end{align}
The motivation for this definition is that someone who “moves along with the basis $$\xi$$” sees it as fixed so this is how they would describe the derivative of $$f$$. Anyway, this should not be confused with the derivative $$f’=\frac{df}{dt}$$ (see below).
Next comes the definition of (generalized) angular velocity
Definition 2. (Angular velocity endomorphism)
With $$V,\xi$$ defined as above, note that for each $$t\in I$$, we have the velocity vectors $$\dot{\xi_i}(t)\equiv\frac{d\xi_i}{dt}(t)$$ for $$i\in\{1,\dots, n\}$$. So by linear algebra there is a unique linear transformation $$V\to V$$ which sends the basis elements $$\xi_1(t),\dots,\xi_n(t)$$ to $$\dot{\xi_1}(t),\dots, \dot{\xi_n}(t)$$. Let us denote this unique linear transformation $$\Omega^{(\xi)}(t):V\to V$$, and call this the angular velocity at time $$t$$ of the basis $$\xi$$. If there is no chance for confusion, we shall simply write $$\Omega$$ instead of $$\Omega^{(\xi)}$$. So, to emphasize, the defining property of $$\Omega^{(\xi)}$$ is that for each $$t\in I$$ and each $$i\in\{1,\dots, n\}$$, \begin{align} \frac{d\xi_i}{dt}\bigg|_t\equiv\dot{\xi_i}(t)&=\Omega^{(\xi)}(t)[\xi_i(t)]. \end{align}
Also, the only reason I put “generalized” in brackets is because so far I have not restricted myself to $$3$$ dimensions, and I did not assume I had an inner product on $$V$$, and I did not assume that the basis is orthonormal.
With these two definitions in place, we see that for any differentiable function $$f:I\to V$$, each $$t\in I$$, \begin{align} \frac{df}{dt}\bigg|_t&=\frac{d}{dt}\bigg|_t\left(f_{\xi}^i\,\xi_i\right)\\ &=\frac{df_{\xi}^i}{dt}\bigg|_t\xi_i(t)+f_{\xi}^i(t)\frac{d\xi_i}{dt}\bigg|_t\tag{product rule}\\ &=\frac{d_{\xi}f}{dt}\bigg|_t+f_{\xi}^i(t)\,\Omega^{(\xi)}(t)[\xi_i(t)]\tag{definitions 1,2}\\ &= \frac{d_{\xi}f}{dt}\bigg|_t+\Omega^{(\xi)}(t)[f_{\xi}^i(t)\xi_i(t)]\tag{\Omega^{(\xi)}(t) is linear}\\ &= \frac{d_{\xi}f}{dt}\bigg|_t+\Omega^{(\xi)}(t)[f(t)] \end{align} This gives us our sought-after equation (omitting the $$t$$ from the notation): \begin{align} \frac{df}{dt}&=\frac{d_{\xi}f}{dt}+\Omega^{(\xi)}(\cdot)[f(\cdot)].\tag{*} \end{align} This equation tells us how the derivative of an arbitrary function $$f:I\to V$$ can be expressed as a sum of two parts: the first part is the “naive interpretation by an otherwise oblivious observer” and the second part “fixes the observer’s blunder”. So, really, $$(*)$$ is nothing but the product rule plus notation/definition.
Special Cases.
Now let’s talk about some special cases.
• The first and most important is of course if $$\frac{d\xi_i}{dt}=0$$ identically for each $$i$$, because then $$\Omega^{(\xi)}(t)=0$$ for all $$t\in I$$. In words, if the basis is not changing in time, then it has no angular velocity. As a result, from $$(*)$$, we have $$\frac{df}{dt}=\frac{d_{\xi}f}{dt}$$. In particular, if we now have two bases, $$e$$ and $$\xi$$, where $$e$$ is not time-dependent, then \begin{align} \frac{df}{dt}&=\frac{d_ef}{dt}=\frac{d_{\xi}f}{dt}+\Omega^{(\xi)}(\cdot)[f(\cdot)]. \end{align}
• Another special case to keep in mind is when we have $$V=\Bbb{R}^3$$, and the $$\xi_i$$’s are orthonormal. In this case, the linear map $$\Omega^{\xi}(t):\Bbb{R}^3\to\Bbb{R}^3$$ is skew-adjoint, meaning that it has a skew-symmetric matrix representation relative to the standard basis, and hence there is a unique vector $$\omega^{(\xi)}(t)\in\Bbb{R}^3$$, the angular velocity vector, such that for all $$v\in\Bbb{R}^3$$, $$\Omega^{(\xi)}(t)[v]=\omega^{(\xi)}(t)\times v$$ is given by the cross product. In this case, equation $$(*)$$ becomes \begin{align} \frac{df}{dt}=\frac{d_{\xi}f}{dt}+\omega^{(\xi)}\times f. \end{align}
• Putting the two remarks above together, we see that in $$\Bbb{R}^3$$, if we have an orthonormal (time-dependent) basis $$\xi=\{\xi_1,\xi_2,\xi_3\}$$ and a time-independent basis $$e=\{e_1,e_2,e_3\}$$, then for any differentiable function $$f:I\to\Bbb{R}^3$$, \begin{align} \frac{df}{dt}=\frac{d_ef}{dt}=\frac{d_{\xi}f}{dt}+\omega^{(\xi)}\times f. \end{align} It is this second equality which is written in your post and in all textbooks.
Next, just so we directly answer everything:
• $$(\vec{A})_S$$ typically means exactly what you wrote: take the vector $$\vec{A}$$, figure out its components relative to a basis defining $$S$$, stick those components into a 3-tuple. However as you hopefully agree with my presentation, you don’t really need this notation.
• for your questions regarding the various derivatives, see my definition 1, and my first bullet point above. In particular, in light of the notation $$(\vec{A})_{S’}$$ (your point 1) the notation $$\left(\frac{d\vec{A}}{dt}\right)_{S’}$$ is indeed confusing (especially since $$S’$$ is non-inertial) because it does not mean “write $$\frac{d\vec{A}}{dt}=\alpha^i\hat{e}_i’$$ and consider the tuple $$(\alpha^1,\alpha^2,\alpha^3)$$”, i.e there is a wonderful conflict of notation with the bracket and subscript $$(\cdot)_{S’}$$. The intended meaning is what I wrote in definition 1. This is why I wrote the operator as $$\frac{d_{\xi}}{dt}$$, to emphasize that it is a differential operator built using the basis $$\xi$$, and not that you take the time derivative then extract the components relative to that basis.
• For your question (A), the components of $$\frac{df}{dt}$$ relative to the $$\xi$$ basis is different from the components of $$\frac{d_{\xi}f}{dt}$$ relative to the $$\xi$$ basis (there’s the extra angular velocity term which needs to be accounted for).
• For your question (B), hopefully I’ve answered it already in my definition 1, and in my special cases above.
Calculating accelerations.
You didn’t ask, but let’s work out second derivatives using this language. We start from equation (*), and apply $$\frac{d}{dt}$$ to both sides (and of course assume everything is twice differentiable): \begin{align} \frac{d^2f}{dt^2}&=\frac{d}{dt}\left(\frac{d_{\xi}f}{dt}+\Omega^{(\xi)}[f]\right)\\ &=\frac{d}{dt}\left(\frac{d_{\xi}f}{dt}\right)+\frac{d}{dt}\left(\Omega^{(\xi)}[f]\right)\\ &=\left(\frac{d_{\xi}}{dt}\left(\frac{d_{\xi}f}{dt}\right)+\Omega^{(\xi)}\left[\frac{d_{\xi}f}{dt}\right]\right) + \frac{d\Omega^{(\xi)}}{dt}[f]+\Omega^{(\xi)}\left[\frac{df}{dt}\right]\\ &=\left(\frac{d_{\xi}}{dt}\right)^2f+\Omega^{(\xi)}\left[\frac{d_{\xi}f}{dt}\right]+ \frac{d\Omega^{(\xi)}}{dt}[f]+ \Omega^{(\xi)}\left[\frac{d_{\xi}f}{dt}+\Omega^{(\xi)}[f]\right]\\ &=\frac{d_{\xi}^2f}{dt^2} + 2\Omega^{(\xi)}\left[\frac{d_{\xi}f}{dt}\right] +\frac{d\Omega^{(\xi)}}{dt}[f] +\left(\Omega^{(\xi)}\circ\Omega^{(\xi)}\right)[f] \end{align} Keep in mind that the square brackets $$[\cdot]$$ mean you evaluate the endomorphism on the given vector (after plugging in the time $$t$$ everywhere). So, we see that the second derivative of a function $$f$$ equals a sum of a whole bunch of terms: the “second derivative relative to $$\xi$$”, the coriolis acceleration term, the Euler acceleration term, and the centrifugal acceleration term respectively.
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Sequences in the triangle and the fourth dimension
This is the second in a series of guest posts by David Benjamin, exploring the secrets of Pascal’s Triangle.
Sequences in the diagonals
There are many sequences of numbers to be found in Pascal’s triangle. The Natural numbers occur in the second diagonal, running in either direction, and the next two diagonals after that contain other important sequences:
The triangular numbers are part of the sequence of polygonal numbers, all of which can be displayed geometrically by a series of dots. All sets of polygonal numbers begin with $1$ – representing the degenerate polygon, which isn’t really a polygon.
The third diagonal in Pascal’s triangle contains the set of triangular numbers.
Two congruent right-angled triangles can be joined to create a square. In a similar way, summing successive pairs of triangular numbers creates the set of square numbers and so the sequence of square numbers can also be considered to be contained in the third diagonal.
{$1 + 3, 3 + 6, 6 +10, 10 +15, 15 + 21,…$} = {$1, 4, 9, 16, 25, 36,…$}
Yang Hui discovered the formula to find the sum of the triangular numbers. The $nth$ term for the sequence is $\frac{n(n+1)}{2}$ and the sum of the first $n$ terms is $\frac{n(n+1)(n+2)}{6}$, where each term or sum is found by substituting $n = 1, 2, 3, 4,..$
More Square numbers
Square numbers can also be found in the triangle by calculating the product of the six numbers surrounding any internal number:
$4\times 1\times 1\times 6\times 15\times 10=3 \hspace{0.1cm} 600=60^2$
$84\times 210\times 330\times 165\times 45\times 36=1 \hspace{0.1cm} 556 \hspace{0.1cm} 006 \hspace{0.1cm} 760 \hspace{0.1cm} 000=1 \hspace{0.1cm} 247 \hspace{0.1cm} 400^2$
$1\times 7\times 28\times 36\times 9\times 1=63 \hspace{0.1cm} 504=252^2$
Higher dimensional numbers
The fourth diagonal of the triangle contains the sequence of tetrahedral numbers, part of the set of pyramidal numbers. They can be represented in 3 dimensions and are created from the triangular numbers.
$1$ on top with $3$ underneath, then $6$, then $10$,….They are displayed below as a stack of balls representing a triangular-based prism. They can also be generated by the power series $\displaystyle\frac{x}{(x-1)^4}=x+4x^2+10x^3+20x^4+35x^5+56x^6+…$. Conway and Guy in their wonderful book, The Book of Numbers, wrote that the tetrahedral numbers are all even except every fourth one, which is odd.
The fifth diagonal of the triangle contains the pentatope numbers: $1, 5, 15, 35, 70, 126, 210,..$ Pascal called them the triangulo-triangular numbers. Pentatope numbers are the sums of the tetrahedral numbers which are the sums of the triangular numbers. The $nth$ term for the sequence is $\frac{n(n+1)(n+2)(n+3)}{24}$ which tells us that the product of four consecutive numbers is divisible by $24$. The power series to generate them is $\frac{x}{(x-1)^5}=x+5x^2+15x^3+35x^4+70x^5+126x^6+…$.
Wolfram Mathworld describes the pentatope as the simplest regular figure in four dimensions, representing the four-dimensional analog of the solid tetrahedron. It is also called the 5-cell, since it consists of five vertices, or pentachoron. The pentatope is the four-dimensional simplex.
The power series $\frac{x}{(x-1)^n}$ generate the diagonals of Pascal’s triangle for $n=1,2,3,4,5,…$
Some other patterns
In Part 1 we saw that the numbers in each row sum to the powers of $2$. If we concatenate the numbers in the first five rows of the triangle then we obtain the powers of $11$:
Now with some reverse concatenation and summing some pairs of digits we obtain further powers…
$1 5 10 10 5 1 \rightarrow 1 (5 + 1) (0 + 1) 0 5 1 \rightarrow 161051 = 11^5$
$1 6 15 20 15 6 1 \rightarrow 1 (6 + 1) (5 + 2) (0 + 1) 5 6 1 \rightarrow 1771561 = 11^6$
$1 7 21 35 35 21 7 1 \rightarrow 1 (7 + 2) (1 + 3) (5 + 3) (5 + 2) 1 7 1 \rightarrow 19487171 = 11^7$
$1 8 28 56 70 56 28 8 1 \rightarrow 1 (8 + 2) (8 + 5) (6 + 7) (0 + 5) (6 + 2) 8 8 1$
Working from the right, adding the brackets and carrying any $10$ as a $1$ to the sum of the next bracket yields
$214 \hspace{0.1cm} 358 \hspace{0.1cm} 881 = 11^8$
Primes and Pascal
When the second number in each row is a prime number, the symmetry of the triangle means the penultimate number in that row is the same prime and it turns out that all the numbers between the two primes are multiples of that prime.
$55 = 5 \times 11$, $165 = 15 \times 11$, $330 = 30 \times 11$, $462 = 42 \times 11$
We conclude Part 2 with the finite hockey stick sequences:
Start with any $1$ and move down the diagonal any number of rows. The sum of those numbers is equal to the number on the next row on the reverse diagonal
$1 + 4 + 10 + 20 = 35$, $1 + 7 + 28 + 84 + 210 = 330$, $1 + 8 + 10 + 36 = 45$.
In the next part, we’ll consider another famous sequence with an unexpected connection to Pascal’s triangle – the Fibonacci numbers.
• David Benjamin
Teacher of mathematics [1975-2018], Author [Heinemann, Stanley Thornes], Key contributor and mathematics editor [teachitmaths], Mathematics software producer [Virtual Image]
One Response to “Sequences in the triangle and the fourth dimension”
1. Stuart Goodall
Some remarkable discoveries are uncovered here:
The product of the six numbers surrounding any internal number produces a square number!
I particularly liked how power series are used to generate the diagonals of Pascal’s triangle
The concatenation section is astonishing albeit slightly mind-boggling!
Multiples of the prime they are sandwiched between & Hockey stick sums – all v surprising.
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Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.
2. Originally Posted by alexmahone
Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.
Unfortunately I can't provide you with an algebraic solution.
In the attachment you see the only possible dimensions of the quadrilateral with approximative values.
Maybe you can use the drawing to control your exact result.
(The radius is of course $r = \dfrac h2 \approx 0.87$
3. Hello, alexmahone!
I have a start on this problem, but haven't finished it.
Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD.
$AB = 3CD$, and the area of the quadrilateral is 4.
If a circle can be inscribed in the quadrilateral, find its radius.
Code:
D - - - x - - - C
- o-------*-*-*---o
: | * : *\
: | * : *
: |* :r *
: | : \
: * : *\
2r * * * \
: * : * \
: | : \
: |* :r * \
: | * : * \
: | * : * \
- o-------*-*-*---------------o
A - - - - - - 3x - - - - - B
$ABCD$ is a trapezoid with two right angles.
Its height is $2r$, its bases are $AB = 3x,\:CD = x.$
The area of a trapezoid is: . $A \;=\;\frac{h}{2}(b_1 + b_2)$
So we have: . $A \;=\;\frac{2r}{2}(x + 3x) \:=\:4 \quad\Rightarrow\quad rx \:=\:1$
Now where can I find another equation?
4. Originally Posted by alexmahone
Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.
I hope that I now found the solution:
For convenience I set CD = t.
Your quadrilateral is a trapezium with t and 3 t as the lengths of the parallel sides. The area is calculated by:
$A = \dfrac{t+3t}2 \cdot h = 4$ which yield $h = \dfrac2t$ and therefore $r = \dfrac1t$
The length of $x = t- \dfrac1t$ and
$y = 3t- \dfrac1t$
Since $FB = 2t$ you can calculate the sides of the right triangle CFB:
$\left(\dfrac2t\right)^2 + (2t)^2=\left(t-\dfrac1t+3t-\dfrac1t\right)^2$
Expand the brackets and collect like terms:
$\dfrac4{t^2}+4t^2=16t^2-16+\dfrac4{t^2}~\implies~12t^2=16~\implies~\boxed{ t=\sqrt{\dfrac43}}$
Re-substitute this value and you'll get $r = \dfrac12 \cdot \sqrt{3}$
EDIT: Too late
5. Originally Posted by earboth
I hope that I now found the solution:
For convenience I set CD = t.
Your quadrilateral is a trapezium with t and 3 t as the lengths of the parallel sides. The area is calculated by:
$A = \dfrac{t+3t}2 \cdot h = 4$ which yield $h = \dfrac2t$ and therefore $r = \dfrac1t$
The length of $x = t- \dfrac1t$ and
$y = 3t- \dfrac1t$
Since $FB = 2t$ you can calculate the sides of the right triangle CFB:
$\left(\dfrac2t\right)^2 + (2t)^2=\left(t-\dfrac1t+3t-\dfrac1t\right)^2$
Expand the brackets and collect like terms:
$\dfrac4{t^2}+4t^2=16t^2-16+\dfrac4{t^2}~\implies~12t^2=16~\implies~\boxed{ t=\sqrt{\dfrac43}}$
Re-substitute this value and you'll get $r = \dfrac12 \cdot \sqrt{3}$
EDIT: Too late
Thanks so much!
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# AREA OF TRIANGLE AND QUADRILATERAL WITH VERTICES WORKSHEET
(1) Find the area of the triangle formed by the points
(i) (1,–1), (–4, 6) and (–3, –5) Solution
(ii) (-10, -4) (-8, -1) and (-3, -5) Solution
(2) Determine whether the sets of points are collinear?
(i) (-1/2, 3), (-5, 6) and (-8, 8) Solution
(ii) (a, b+c), (b, c+a) and (c, a+b) Solution
(3) Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p ’. Solution
Solution
(4) In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2, 3), (4, a) and (6, –3) Solution
(ii) (a, 2 – 2a), (–a + 1, 2a) and (–4–a, 6–2a) Solution
(5) Find the area of the quadrilateral whose vertices are at
(i) (–9, –2), (–8, –4), (2, 2) and (1, –3) Solution
(ii) (–9, 0), (–8, 6), (–1, –2) and (–6, –3) Solution
(6) Find the value of k, if the area of a quadrilateral is 28 sq.units, whose vertices are (–4, –2), (–3, k), (3, –2) and (2, 3) Solution
(7) If the points A(-3, 9) , B(a, b) and C(4, -5) are collinear and if a + b = 1 , then find a and b. Solution
(8) Let P(11,7) , Q(13.5, 4) and R(9.5, 4) be the mid- points of the sides AB, BC and AC respectively of triangle ABC . Find the coordinates of the vertices A, B and C. Hence find the area of triangle ABC and compare this with area of triangle PQR Solution
(9) In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.
(10) A triangular shaped glass with vertices at A(-5,-4) , B(1,6) and C(7,-4) has to be painted. If one bucket of paint covers 6 square feets, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied. Solution
(11) In the figure, find the area of (i) triangle AGF (ii) triangle FED (iii) quadrilateral BCEG
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Solved Problems on Binomial Expansion
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### GCSE items that can be used to define successful outcomes for peer and self assessment activities
Assessment of practical work
## Using Exam pro items to support successful outcomes
Learners will be able to test their progress against learning outcomes using questions taken from past AQA GCSE questions.
### Q1 Foundation
AQA-PH1FP-QP-Jun12 Q1 The appliances shown on the next slide transfer electrical energy to other types of energy.
The vacuum cleaner is designed to transfer electrical energy to kinetic energy.
a)Name the three other appliances which are also designed to transfer electrical energy to kinetic energy.
Q1 cont. b) Which two of the following statements are true?
•A Appliances only transfer part of the energy usefully.
•B The energy transferred by appliances will be destroyed.
•C The energy transferred by appliances makes the surroundings warmer.
•D The energy output from an appliance is bigger than the energy input.
Q1 mark scheme a) Washing machine, fan, drill 1 mark for each
3
b) A C 1 mark for each
2
Q2 Foundation Exampro QM94R505
The diagram shows what happens to each 100 joules of energy from crude oil when it is used as petrol in a car.
The widths of the arrows show exactly how much energy is transferred in each particular way.
(a) Complete the diagram by adding the correct energy value alongside each arrow. (3)
(b) Calculate how efficient the car engine is at transferring the energy
from petrol
into useful movement.
(2)
(c) Two students are discussing the diagram.
The first says that
none
of the energy released from the crude oil is really lost.
The other says that
all
of the energy released from the crude oil is really lost.
What do you think?
Q2 mark scheme
(a) A = B = 20, C = 60
3
(b)
evidence of or
20%/0.2 gains 1 mark
but
25%/0.25
or
figure consistent with B in part (a)
gains 2 marks 2
(c)
ideas that
“lost” energy is (also) transferred (in ways we don’t want) total amount of energy (before and after any transfer) remains the same energy becomes (progressively) more spread out/less easy to transfer again this applies to
all
the energy/even to the energy usefully transferred
## Q3 higher
AQA-PH1HP-QP-Jan12 Q 6 The table on the next slide gives data about two types of low energy bulb.
(a) Both types of bulb produce the same useful power output.
(a) (i) Calculate the useful power output of the CFL.
Use the correct equation from the Physics Equations Sheet.
Useful power output = ................................................. W
(2 marks)
Q3 contd.
(a) (ii)
Calculate the efficiency of the LED bulb.
Use the correct equation from the Physics Equations Sheet.
Efficiency = ......................................................
(1 mark)
Q3 contd.
6 (b) Sketch and label a Sankey diagram for the CFL
(2 marks)
6 (c) LED bulbs are expensive. This is because of the large number of individual electronic LED chips needed to produce sufficient light from each bulb.
6 (c) (i) Use the data in the table to evaluate the cost effectiveness of an LED bulb compared to a CFL.
(2 marks)
6 (c) (ii) Scientists are developing brighter and more efficient LED chips than those currently used in LED bulbs.
Suggest one benefit of developing brighter and more efficient LED chips.
(1 mark)
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# A Young-Laplace Equation
1. Jul 25, 2016
### joshmccraney
Hi PF!
Here I'm trying to derive the Young-Laplace equation, which states $$\Delta P = \gamma \left(\frac{1}{R_1} + \frac{1}{R_2} \right)$$ where $\Delta P$ is change in pressure, $\gamma$ is a proportionality constant, and $R$ is a radius of curvature, both of which are orthogonal to the other and work so as to parameterize a fluid surface. The derivation follows:
Given a surface of fluid we know (how?) that the change in surface energy $d E_s$ equals the change of work done $dW$. Notice $dW = \Delta P dV = \Delta P x y dz$ (cartesian coordinates). Now $d E_s = \gamma dA$. Then $A = xy$ and $A' = (x+dx)(y+dy) = xy +ydx + x dy$ ignoring higher order infinitesimals. Then $dA = ydx + x dy$. Suppose the surface $dA$ is parameterized by two radii of curvature $R_1$ and $R_2$. Then, by similar triangles, we have $$\frac{R_1+dz}{R_1} = \frac{x+dx}{x} \implies \\ dx = \frac{x dz}{R_1}.$$ Substitute this into the expression for $dA$ to arrive at $$dA = \frac{yxdz}{R_1} + x dy.$$ Identical logic applied to $dy$ implies the final relation for area $$dA = xy dz \left( \frac{1}{R_1}+\frac{1}{R_2} \right) \implies \\ d E_s = xy dz \gamma \left( \frac{1}{R_1}+\frac{1}{R_2} \right).$$ Since $dW = d E_s$ we then have $$\Delta P = \gamma \left( \frac{1}{R_1}+\frac{1}{R_2} \right)$$ where the $xy dz$ term cancels.
My question is, is the following relation correct: $d E_s = \gamma d A$ and $d W = \Delta P dV = \Delta P dA dz$ which implies $\gamma = \Delta P dz$, which, when compared with the above result, yields $1/dz = 1/R_1 + 1/R_2$. Is this still correct?
2. Jul 25, 2016
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# 812. Largest Triangle Area
## Description
Given an array of points on the X-Y plane points where points[i] = [xi, yi], return the area of the largest triangle that can be formed by any three different points. Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]]
Output: 2.00000
Explanation: The five points are shown in the above figure. The red triangle is the largest.
Example 2:
Input: points = [[1,0],[0,0],[0,1]]
Output: 0.50000
Constraints:
• 3 <= points.length <= 50
• -50 <= xi, yi <= 50
• All the given points are unique.
## Solutions
• class Solution {
public double largestTriangleArea(int[][] points) {
double ans = 0;
for (int[] p1 : points) {
int x1 = p1[0], y1 = p1[1];
for (int[] p2 : points) {
int x2 = p2[0], y2 = p2[1];
for (int[] p3 : points) {
int x3 = p3[0], y3 = p3[1];
int u1 = x2 - x1, v1 = y2 - y1;
int u2 = x3 - x1, v2 = y3 - y1;
double t = Math.abs(u1 * v2 - u2 * v1) / 2.0;
ans = Math.max(ans, t);
}
}
}
return ans;
}
}
• class Solution {
public:
double largestTriangleArea(vector<vector<int>>& points) {
double ans = 0;
for (auto& p1 : points) {
int x1 = p1[0], y1 = p1[1];
for (auto& p2 : points) {
int x2 = p2[0], y2 = p2[1];
for (auto& p3 : points) {
int x3 = p3[0], y3 = p3[1];
int u1 = x2 - x1, v1 = y2 - y1;
int u2 = x3 - x1, v2 = y3 - y1;
double t = abs(u1 * v2 - u2 * v1) / 2.0;
ans = max(ans, t);
}
}
}
return ans;
}
};
• class Solution:
def largestTriangleArea(self, points: List[List[int]]) -> float:
ans = 0
for x1, y1 in points:
for x2, y2 in points:
for x3, y3 in points:
u1, v1 = x2 - x1, y2 - y1
u2, v2 = x3 - x1, y3 - y1
t = abs(u1 * v2 - u2 * v1) / 2
ans = max(ans, t)
return ans
• func largestTriangleArea(points [][]int) float64 {
ans := 0.0
for _, p1 := range points {
x1, y1 := p1[0], p1[1]
for _, p2 := range points {
x2, y2 := p2[0], p2[1]
for _, p3 := range points {
x3, y3 := p3[0], p3[1]
u1, v1 := x2-x1, y2-y1
u2, v2 := x3-x1, y3-y1
t := float64(abs(u1*v2-u2*v1)) / 2.0
ans = math.Max(ans, t)
}
}
}
return ans
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
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# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Rachel's Problem:
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### There are 162 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Mod 3
##### Age 14 to 16 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
### Number Rules - OK
##### Age 14 to 16 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Elevenses
##### Age 11 to 14 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### A Biggy
##### Age 14 to 16 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
##### Age 11 to 14 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Take Three from Five
##### Age 14 to 16 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Perfectly Square
##### Age 14 to 16 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
### Even So
##### Age 11 to 14 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Sprouts Explained
##### Age 7 to 18
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### Archimedes and Numerical Roots
##### Age 14 to 16 Challenge Level:
The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
### Unit Fractions
##### Age 11 to 14 Challenge Level:
Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation.
### More Mathematical Mysteries
##### Age 11 to 14 Challenge Level:
Write down a three-digit number Change the order of the digits to get a different number Find the difference between the two three digit numbers Follow the rest of the instructions then try. . . .
### To Prove or Not to Prove
##### Age 14 to 18
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### Multiplication Square
##### Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
##### Age 11 to 14 Challenge Level:
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .
### Pythagorean Triples II
##### Age 11 to 16
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### For What?
##### Age 14 to 16 Challenge Level:
Prove that if the integer n is divisible by 4 then it can be written as the difference of two squares.
### Pythagorean Triples I
##### Age 11 to 16
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
### Sixational
##### Age 14 to 18 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### Common Divisor
##### Age 14 to 16 Challenge Level:
Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.
### N000ughty Thoughts
##### Age 14 to 16 Challenge Level:
How many noughts are at the end of these giant numbers?
### Some Circuits in Graph or Network Theory
##### Age 14 to 18
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
##### Age 7 to 14 Challenge Level:
I added together some of my neighbours house numbers. Can you explain the patterns I noticed?
### The Great Weights Puzzle
##### Age 14 to 16 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
### Greetings
##### Age 11 to 14 Challenge Level:
From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . .
### What Numbers Can We Make?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### What Numbers Can We Make Now?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Advent Calendar 2011 - Secondary
##### Age 11 to 18 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### A Long Time at the Till
##### Age 14 to 18 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### Iffy Logic
##### Age 14 to 18 Challenge Level:
Can you rearrange the cards to make a series of correct mathematical statements?
### More Number Sandwiches
##### Age 11 to 16 Challenge Level:
When is it impossible to make number sandwiches?
### Triangular Intersection
##### Age 14 to 16 Short Challenge Level:
What is the largest number of intersection points that a triangle and a quadrilateral can have?
### Classifying Solids Using Angle Deficiency
##### Age 11 to 16 Challenge Level:
Toni Beardon has chosen this article introducing a rich area for practical exploration and discovery in 3D geometry
### The Frieze Tree
##### Age 11 to 16
Patterns that repeat in a line are strangely interesting. How many types are there and how do you tell one type from another?
### The Triangle Game
##### Age 11 to 16 Challenge Level:
Can you discover whether this is a fair game?
### Magic Squares II
##### Age 14 to 18
An article which gives an account of some properties of magic squares.
### Picturing Pythagorean Triples
##### Age 14 to 18
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
### Yih or Luk Tsut K'i or Three Men's Morris
##### Age 11 to 18 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### A Knight's Journey
##### Age 14 to 18
This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.
### Impossible Sandwiches
##### Age 11 to 18
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
### Calculating with Cosines
##### Age 14 to 18 Challenge Level:
If I tell you two sides of a right-angled triangle, you can easily work out the third. But what if the angle between the two sides is not a right angle?
### Same Length
##### Age 11 to 16 Challenge Level:
Construct two equilateral triangles on a straight line. There are two lengths that look the same - can you prove it?
### Why 24?
##### Age 14 to 16 Challenge Level:
Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.
### Mouhefanggai
##### Age 14 to 16
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
### Angle Trisection
##### Age 14 to 16 Challenge Level:
It is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square.
### Postage
##### Age 14 to 16 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Ratty
##### Age 11 to 14 Challenge Level:
If you know the sizes of the angles marked with coloured dots in this diagram which angles can you find by calculation?
### Always Perfect
##### Age 14 to 16 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Kite in a Square
##### Age 14 to 16 Challenge Level:
Can you make sense of the three methods to work out the area of the kite in the square?
### Fitting In
##### Age 14 to 16 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .
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# What is the difference between sample space and random variable?
A random variable is a set of possible values from a random experiment.
It then takes the example of flipping a coin, defines $Heads=0$ and $Tails=1$, and says that $X = \{0, 1\}$ is a random variable.
Next, it defines sample space as
A random variable's set of values is the sample space.
It then takes the example of throwing a dice, and states that the sample space is $\{1, 2, 3, 4, 5, 6\}$.
So, both terms are defined to be a set of outcomes for an experiment and, as a result, I got confused and couldn't differentiate between them.
What is the difference between sample space and random variable?
I've consulted WIkipedia too and although I can understood the article on Sample Space but the article on Random Variable appears too technical and I couldn't comprehend it.
• Intuitively ( you say you don't like technical papers) a random variable is more than just the outcomes, it is the outcomes and the probability of each outcome ( not fully correct because I would have to introduce events)
– user83346
Commented Feb 27, 2017 at 8:58
• @fcop from what I recall from high school, event is any element of the power set of the set of all the outcomes. Commented Feb 27, 2017 at 14:18
• In the language of my post at stats.stackexchange.com/a/54894/919, the sample space is a set of tickets in a box and a random variable is a consistent way of writing numbers on those tickets. (Math sites rarely make this distinction because to their writers the world is built of sets. In the real world, though, a "value from an ... experiment" is a rich, complicated thing: it could be a tissue sample, a photo of a star field, or a collection of handwritten answers on a survey. It often does not start out as a numerical object: the numbers have to be added by means of measurement.)
– whuber
Commented Feb 27, 2017 at 14:22
• Well it is not necessarily the power set of the sample space; it should be a sigma-algebra on the sample space. Each element of the sigma-algebra should be measurable. As @whuber says the random variable is then a map from this sigma-algebra (the tickets) in the real set (the consistent way of writing numbers on the tickets). Because the tickets are measurable, (some) subsets of the real set will also be.
– user83346
Commented Feb 27, 2017 at 16:45
From statistical inference by Casella and Berger,
Definition 1.1.1 The set, $S$, of all possible outcomes of a particular experiment is called the sample space for the experiment.
So sample space can be thought of as all possible observations one could make from a particular experiment. A sample space for a coin toss is a set $\{H, T\}$; a sample space for rolling a six-sided die is a set $\{1, 2, 3, 4, 5, 6\}$.
Definition 1.4.1 A random variable is a function from a sample space $S$ into the real numbers
so random variable can be thought of as a function. The notation used for random variable is an uppercase letter. So if we have a random variable that maps sample space to real numbers, we have $$X: S \to \mathbb{R}$$
No one really expresses random variables this way; instead, it's often denoted as $X$.
If that random variable $X$ is a set of possible values from a random experiment, then $$X: S \to S$$ so random variable is an identity function.
A sample space is the SET of values a random variable can take.
You can think of random variable as an unopened box. This unopened box contains each member of the sample space with some probability.
In your example of a dice roll, the sample space is {1,2,3,4,5,6}. The random variable that represents a roll of the dice has a probability of 1/6 of taking on each of these 6 values.
• You have the concept of sample space right I wouldn't characterize the random variable as an unopened box. It is the value of the side of the die that you roll and each of the possible values has a probability associated with it such that the sum of the probabilities for each side is 1. Commented Oct 22, 2017 at 5:02
• This is similat to what I say in my comment supra
– user83346
Commented Oct 22, 2017 at 7:45
• Although often (especially in elementary texts) sample spaces are conflated with values of random variables, this doesn't generalize well. For more complicated situations it is essential to separate the concepts of "element of the sample space" and "possible value of a random variable." Please see my comment to the question.
– whuber
Commented Oct 22, 2017 at 16:32
• @whuber Could you explain in which situations it is essential to separate both concepts? why can't we just take the values of the random variable as the elements of a sample space and so avoid the extra layer of complexity in the model? Thanks. Commented Aug 6, 2020 at 19:55
• @FCardelle Dealing with sequences of variables and their possible interrelationships becomes trickier. So does analyzing (and even defining) stochastic processes. Separating the sample space from the random variables is not an additional layer of complexity: it's a huge conceptual and mathematical simplification.
– whuber
Commented Aug 6, 2020 at 19:59
Amazonian's answer is wrong. The sample space IS NOT the set of values a random variable can take. The sample space is the domain upon which a random variable is defined. The example Amazonian gave is one example of a random variable whose values happen to be the elements in the sample space.
A random variable is a function that assigns a value to every element in the sample space. There is nothing stopping you from defining a random variable X on the sample space S = {1, 2, 3, 4, 5, 6} where X = 10 * s for s in S. Then the values that X can take are {10, 20, 30, 40, 50, 60}.
• Although I am sympathetic with this point of view (see my other comments in this thread), the reason the situation is not so clear-cut is that one can always create a random variable with the given distribution by defining it as the identity function on its support and endowing the support with a suitable probability distribution. Thus, it's going a little too far to assert Amazonian is wrong. Your answer could be found misleading, too, in strongly suggesting the sample space cannot be the range of the random variable (although I realize it doesn't quite come out and say that).
– whuber
Commented Oct 1, 2020 at 20:37
• The sample space should not be defined as the set of values a random variable can take. This seems to be a misleading definition that’s all over the internet. The sample space is the set of all possible outcomes of a random experiment. Commented Oct 1, 2020 at 21:23
• The reason I object to Amazonian's answer is because it attempts to define the sample space and does so both incorrectly and backwards. If we want to define what a dog is, we don’t do so by giving the definition “a dog is a dachshund”. Instead, we say “a dog is a member of the family Canidae”. Can a dog be a dachshund? Sure, but that's an irrelevant piece of information that shouldn't be in the definition. Commented Oct 1, 2020 at 21:26
• The statement “a dog is a dachshund” is not claiming that “some dogs are dachshunds”. It is claiming “all dogs are dachshunds”, which is clearly wrong. It is also misleading because it goes backwards (the concept of ‘dog’ does not rely on the concept of ‘dachshund’). Commented Oct 1, 2020 at 21:30
• I did also make it a point to explicitly say that the elements in the sample space and the values taken by the random variable CAN be the same, by pointing out Amazonian’s example. If Amazonian wasn't trying to give a definition of the sample space (which seems highly unlikely), it's true that the statement "if s is a sample space, then it is the set of values a random variable can take" is not a strictly false statement. But it is highly misleading and misses the point of the original question.That's why I called Amazonian's answer wrong. Commented Oct 1, 2020 at 21:36
When the sample space consists
• exclusively of real numbers (as in your examples),
there is no difference between the sample space and the random variable.
The difference arises in other situations, because
the sample space may be a set of arbitrary elements, e.g. $$\{\color{red} {\text{red}}, \color{green} {\text{green}}, \color{blue} {\text{blue}}\}$$.
With such freedom of possible outcomes it is difficult to work. So someone invented to work not with arbitrary elements, but only with real numbers.
To reach it, the first thing is to map such elements to real numbers, e.g.
\begin{aligned} \color{red} {\text{red }} &\mapsto 6.72\\ \color{green} {\text{green}} &\mapsto -2\\ \color{blue} {\text{blue}} &\mapsto 19.5 \end{aligned}
And that mapping is called a random variable.
So now there is a difference, because
• the sample space is $$\{\color{red} {\text{red}}, \color{green} {\text{green}}, \color{blue} {\text{blue}}\}$$, whereas
• the random variable is that mapping, often freely interpreted as the set $$\{\color{red} {6.72}, \color{green} {-2}, \color{blue} {19.5}\}$$.
Note:
Why is that mapping so important?
Because we got rid of problems how to operate with arbitrary things — after choosing such a mapping we may perform calculation with numbers.
I believe the confusion comes from the choice of examples used. For sake of clarity lets use instead the example of the throw of 2 dices. Then, in that case the sample space would be: {(1,1),(1,2),(2,1),(1,3),(3,1),...,(5,6),(6,5),(6,6)}. A random variable is just a function having this set as domain and the Reals as codomain (in more advanced statistics one could use other more abstract codomains). So, for example, one could define a random variable as being the sum of the results of the dice. So the different result one could obtain are: {2,3,4,5,6,7,8,9,10,11,12}, and the probability of each one occurring is derived directly from the sample space and the probability associated with each element of the set (in this particular case each element has the same probability, but in general it could not be).
From a formal perspective a sample space is "more fundamental" than a random variable. There can be no random variable without the specification of the elements and its correspondent probabilities of the sample space of its domain.
From a practical perspective, a random variables give you 2 advantages over simple sample spaces: (i) they are numerical. That means that one can represent, for example, a coin toss {H,T} into the numerical values {1,0}, and then apply all sort of numerical tools that cannot be applied over qualitative elements. (ii) it makes easier to deal with elements with different probabilities. In the example given we easily created a distribution with different probabilities based on a simple calculation over a sample space with elements of equal probability. Trying to create the given result artificially could be confusing and laborious.
It is common to just use a random variable without the explicit specification of its sample space.
For flipping a coin:
• The sample space is {head, tail},
• the random variable is {0, 1}.
For rolling a die:
• The sample space is {1, 2, 3, 4, 5, 6},
• the random variable is {1, 2, 3, 4, 5, 6}, too,
because the sample space already is the set of numbers only.
In other words, the sample space is the set of arbitrary elements, while the random variable is the set of numbers.
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How to drive a car without a pedal
Mechanical advantages and the way we drive the vehicle are all things we can do without.
But how do we drive a machine without a motor?
A mechanical advantage is a concept that gives us more control of the vehicle.
An advantage, like mechanical wave definition (MWD), is a way to define a function that can be measured or calculated in advance.
For example, when we want to know the mechanical advantage of a wheel, we calculate its resistance and use this to determine its mechanical properties.
This calculation gives us the mechanical wave.
It is also known as a mechanical equation.
When we have an equation that describes a function, it is known as an equation of motion (EV).
This equation can be written as: The physical advantage of the object is the difference between the distance the object has to travel from its center of mass and the distance it would have to travel if it were stationary.
Mechanical wave definition has been described by two different authors, and it is often called the mechanical equation of inertia.
Mechanical advantages of motor design have been described in terms of the force produced when the wheels are turned.
The force of the wheel can be described by its mechanical resistance.
A wheel has a mechanical resistance when the wheel has to turn at the same speed as the vehicle that drives it.
This force is proportional to the wheel’s radius.
When the wheel turns, the wheel also turns at the speed it is turning.
This is the mechanical turning.
Mechanical advantage and mechanical wave definitions are often combined.
For instance, we can combine mechanical advantages and mechanical waves to make the mechanical equivalent of the power of a motor.
A motor that is at least as powerful as a human being can produce a force equal to the force of a human at the wheels’ radius.
This may seem surprising, but the reason why this is so is that we think of human beings as being more powerful than the speed of light.
The human body weighs about 25 times as much as the wheel.
This means that we would be able to drive the human body with the same acceleration that we could drive the wheel at the wheel radius.
However, we would have only a mechanical advantage if we could use the wheel as a propulsion source.
For that to be true, we need to be able the human-powered vehicle to be propelled at the maximum speed it can travel.
For a human-driven vehicle to use the wheels to propel itself, we must also have the power to use them to move the wheels.
When a human drives the wheel, the wheels have to turn in a circular path.
This circular path is called the gearbox.
If the wheels turn faster than the gear box can move them, the gear-box will stop working and the vehicle will not move.
This prevents the vehicle from going faster than its acceleration.
If we can increase the speed at which the wheels can move in order to increase the acceleration, then we can get a mechanical disadvantage of the wheels and a mechanical wave of the gears.
This mechanical advantage or mechanical wave defines the maximum rate of change of the gear ratio.
When this occurs, the mechanical and mechanical-wave terms become equal.
This can be useful because when a motor or a generator is driving a generator, the generator may use the gear ratios to accelerate the generator to greater speeds.
This speed increase will increase the power and torque that the generator generates, which will increase our speed.
The speed increase can be as great as the speed the generator can operate at.
For this to occur, we will need to increase our mechanical advantage.
When mechanical wave-definition is used to define the mechanical advantages of a system, we may define the advantage as the difference in power between the mechanical output and the power that is produced by the mechanical input.
For the example above, we define the value of mechanical advantage as being the difference of the motor’s output torque at the gear level and the motor output torque as the ratio of the generator’s output power at the generator level to the generator output power.
When using mechanical wave concepts, we often want to define mechanical advantages in terms or terms of mechanical wave, which can then be used to describe the mechanical or mechanical-type of a component.
Mechanical-type components are often referred to as motor-type or gear-type systems.
We may then describe a system as having a mechanical-element or a motor-element.
The mechanical-instrument may be a motor, a generator or an inverter, or it may be either a motor and generator, or a inverter and generator.
For more information on mechanical-types, see our article on motor-types.
Mechanical, mechanical wave and mechanical advantage Mechanical-wave definitions often include a mechanical or a mechanical derivative, which is the measurement of a value of a parameter by the mathematical formula of an object with respect to its environment.
The mathematical expression is known by its form: whereis the equation for the value, and
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Question
# Three positive numbers form an increasing GP. If the middle term of the GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is (IIT JEE Main 2014)
A
2+3
B
3+2
C
23
D
2+3
Solution
## The correct option is D 2+√3Let a, ar, ar2 are in GP. If the middle term of the GP is doubled, then new numbers are in AP. ⇒a,2ar,ar2 are in AP. ⇒4ar−a+ar2 ⇒r2−4r+1=0 ⇒r−2±√3 Since the series is an increasing GP so, ⇒r−2+√3
Suggest corrections
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# Fractions2
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### Fractions2
1. 1. Learning OutcomesMultiply mixed numbers with a whole number.Divide fractions with :a whole number.a fractionDivide mixed numbers with :a whole number.a fraction.
2. 2. Multiply mixed numbers with a whole number. Example 1 : 1 × 50 Convert to Step 1 : improper 3 × 50 fraction 2 Step 2 : Multiply the 3 3x50 numerator by × 50 the whole 2 2 number Carry out the division Step 3 : 150 150 ÷2 2 Answer = 75 Cancellation Alternative : method 25 × 50 = 75
3. 3. Example 2: 1 × 100 Convert to improper Step 1 : fraction × 100 Step 2 : Multiply the numerator by × 100 the whole number Carry out the division Step 3 : ? ÷4 Answer = ?
4. 4. Divide fractions with : a whole number and a fraction Example 1 : Find the quotient ÷ 5 Change the ÷ to x Inverse the divisor 5 5= Step 1 : 1 5 ÷ 3 1 Multiply numerator Step 2 : by numerator, 1 1 1 denominator by × = 3 5 15 denominator Example 2 : Find the quotient ÷ ÷ Change the ÷ to x Inverse the divisor Step 1 : 1 2 ÷ 3 9 Step 2 : 3 × = = Carry out the plan by cancellation = 1 method *Don’t forget to convert improper fraction to a
5. 5. Example 3 : Find the quotient ÷ 6Change the ÷ to x Inverse the divisor Step 1 : ÷ Step 2 : × = = Answer in the simplest formExample 4 : Find the quotient ÷ ÷Change the ÷ to x Inverse the divisor Step 1 : ÷ Step 2 : × =
6. 6. Divide mixed numbers with : a whole number and a fraction Convert into improper fraction Example 1 : Find the quotient 2 ÷5 Change the ÷ to × Inverse the divisor 5 5= Step 1 : 9 5 ÷ 4 1 Multiply numerator Step 2 : by numerator, 9 1 9 denominator by × = 4 5 20 denominator Example 2 : Find the quotient ÷ 4 ÷ Convert into improper fraction Change the ÷ to x Inverse the divisor Step 1 : 40 8 ÷ 9 9 Step 2 : 5 Carry out the plan 40 9 by cancellation × 9 8 method =5
7. 7. Example 3 : Find the quotient 2 ÷ 6Change the ÷ to x Inverse the divisor Step 1 : ÷ 7 Step 2 : 1 × = Example 4 : Find the quotient 2 ÷ ÷Change the ÷ to × Inverse the divisor Step 1 : ÷ 7 Step 2 : 1 × =
8. 8. Solve problems involving multiplication and division. Example : Material Quantity Price Yellow beads 1 RM5 per packet 2 packets 2 1 Mina buys 2 packets of yellow beads. Calculate the total cost of the beads. 2 STEP 1: 1 2 × RM5 2 Understand the problem and underline the key words Convert mixed number intoSTEP 2 : improper fractions 5 × RM5 2STEP 3: RM 25 = RM12.50 2
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maths
posted by on .
If a= 3 +2 root 2 , then find the value of a^2 - 1/a^2 ????
• maths - ,
a = 3+2√2
a^2 = 9 + 12√2 + 8 = 17+12√2
1/a^2 = (17-12√2)/(289-288) = 17-12√2
a^2 + 1/a^2 = 34
• maths - ,
sorry I didn't see the minus sign. But it's easy to fix. What do you get?
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# Find F'(x) Given That F(x)= 15/(4+ln(7x)) What Is
AnonymousFrom: -Posts: -Votes: - Find f'(x) given that f(x)= 15/(4+ln(7x)) what is the domain?
MathTeacherFrom: USPosts: 3Votes: 3 F(x)= 15/(4+ln(7x)) f'(x) = -15*7*(1/7x)/(4+ln(7x)^2 = -15/(x(4 + ln(7x))^2 4 + ln(7x) = 0 Therefore x = e-4/7 = 2.6*10^-3 Therefore Domain is all real no. except x = 0 , 2.6*10^-3
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## What is Differential Equations:?
Differential Equation is a mathematical equation of an unknown function of one or several variables which relates the values of the function & its derivatives of different orders.
Consider a differential equation x+ y = 0
‘y’ is the dependent variable & ‘x’ is the independent variable.
## Order of a Differential Equation:
Order of a Differential Equation is the highest order derivative of the dependent variable with respect to the independent variable.
## Degree of a Differential Equation:
Degree of a Differential Equation is the highest power of the highest order derivative of the Differential Equation.
Example:
xy + x - y = 0
Ans:
The highest order of the derivative is 2.
The degree is 1 as the power of is 1.
## Types of Differential Equation:
### 1) Ordinary Differential Equation
A differential Equation is called an ordinary differential equation if it has ordinary derivatives in it.
### 2) Partial Differential Equation
A differential equation is called a partial differential equation if it has differential derivatives in it.
All types of problems & concepts are on these topics are explained & solved by experts at Transtutors.com who provide Assignment and Homework Help for the same.
### 3) Linear Differential Equation:
A linear Differential Equation can be written as follows:
+ + …….+y = g (x) where is the nth order derivative of ‘y’ with respect to ‘x’
Here & g (x) are continuous functions & 0.
If we divide the above equation by we get
+ + ……+ + y =
Let = q (x), = , ….., = , =
+ , + ……+ + y = q (x)…………………..(1)
This equation is called a Linear Differential Equation which is non-homogenous if
q (x) 0.
If q (x) = 0 then the equation is
+ + ……+ + y = 0 is a homogenous linear equation.
In the above equation (1) if , ,……, are constants like = for all i.
+ + ……+ + y = q (x) then it is called Linear Non homogenous Differential equation with constant coefficients.
+ + ……+ + y = 0 then it is called Linear homogenous Differential equation with constant coefficients.
## Linear Homogenous Differential Equation:
If & are solutions of the Linear homogenous Differential equation with constant coefficients
+ + ……+ + y = 0 then for any constants &
+ is also a solution of the equation.
Assignment and Homework Help on Differential Equations can be found in Transtutors.com
Note:
1) If = 0 in the above solution then is a solution.
2) If ,,,…..,are solutions of the above Differential Equation then + + +…..+ is also a solution of the equation.
3),,,….., are n functions & ,,…..,are constants then the expression + + +…..+ is a linear combination of ,,,…..,.
4) The n functions,,,….., are said to be linearly dependent if there exist constants ,,….., not all zero simultaneously such that
+ + +…..+ = 0.
If + + +…..+ is satisfied only when
= 0,= 0,…..,= 0 then the functions ,,,….., are said to be linearly independent.
Let D = be the operator. Then
+ + ……+ + y = 0 can be written as
y+ y + ……+ Dy + y = 0
f (D) y = 0 where f (D) = + + ……+ D +
## Auxiliary Equation:
Consider the homogenous Linear Differential Equation with constant coefficients
+ + ……+ + y = 0 which can be written as
y+ y + ……+ Dy + y = 0…………………….(1)
f (D) y = 0
(+ + ……+ D + )(y) = 0
Exponential Shift rule:
Consider f (D)() = f (m) .
If m is the root of the equation f (m) = 0 then
f (m) = 0
So f (D)() = 0
Hence y = is a solution of the equation in (1).
f (m) = 0 is called the auxiliary equation or characteristic equation.
If the given Differential Equation is y+ y + ……+ Dy + y = 0 then
y+ y + ……+ my + y = 0 is the auxiliary equation.
### Real Non Repeated Roots:
When the auxiliary equation f (m) = 0 gives a solution for m whose values are distinct then m = ,,,…...
Then the solution is y = + + +…..+
This solution is called the general solution.
All the information & Assignment and Homework Help can be found in Transtutors.com.
### Real Repeated Roots:
When the auxiliary equation f (m) = 0 gives a solution for m whose values are not distinct then m = ,,,…... Here is repeated twice. Hence the coefficient of is a linear polynomial. If the value is repeated thrice then the coefficient of will be a quadratic polynomial & so on.
Then the solution is y = (+ x) + +…..+
This solution is called the general solution.
## Linear Non Homogenous Differential Equation:
Consider the non homogenous Linear Differential Equation with constant coefficients
+ + ……+ + y = q (x) which can be written as
y+ y + ……+ Dy + y = q (x)…………………….(1)
f (D) y = q (x)
(+ + ……+ D + )(y) = q (x)
Then considering the auxiliary equation & solving it
= + + +…..+ is a general solution of the homogenous equation f (D) y = 0.
Next to find the total solution we need to find the particular solution of f (D) = q (x)
Then y = + is the solution of (1).
is found by considering f (D) = q (x)
= q (x)
## 4) First Order Differential Equation:
A first order Differential Equation is of the form
1) + p (x) y = q (x)
or 2) = f (x, y)
or 3) + p (x) y = q (x)
p (x) & q (x) are functions which can be a polynomial or a constant.
In first order Differential Equations there are different ways of finding the solution..
Equations of type (1) can be solved by using Integrating Factor.
Equations of type (2) can be solved by using variable separable form or by
Euler’s Method.
Equations of type (3) are called Bernoulli’s Differential Equation which can be can be solved by using the method of substitution & solving further using the above methods.
## 5) Second Order Differential Equation:
Any second order Linear Differential Equation is of the form
p (x) + q (x) + r (x) y = g (x)
p (x) + q (x) + r (x) y = g (x)
If g (x) = 0 then this is a homogenous Linear second order Differential Equation.
If g (x) 0 then this is a non homogenous Linear second order Differential Equation.
If p (x), q (x) & r (x) are constants then this is Linear Second Order Differential Equation with constant Coefficients.
a + b + c y = g (x) is the Linear Non Homogenous Second Order Differential Equation with constant Coefficients.
a + b + c y = 0 is Linear Homogenous Second Order Differential Equation with constant Coefficients.
The Auxiliary Equation for the second order Linear Differential Equation is
a+ bm + c = 0
It has two roots say & . This gives two solutions = ,= .
The solutions can be obtained by the above method & other methods like variation of parameters, Method of undetermined Coefficients & Reduction of order.
## Related Topics
All Science/Math Topics
More Q&A
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# Superposition Theorem for Electric Circuits
The Superposition Theorem is a powerful method used in electrical engineering to analyze linear, time-invariant circuits with multiple sources (voltage or current sources). The theorem states that the response (voltage or current) in any branch of a linear circuit having multiple independent sources can be determined by considering the individual effect of each source acting alone, while all other sources are turned off (voltage sources are replaced by short circuits, and current sources are replaced by open circuits). The total response in the circuit is the algebraic sum of the responses caused by each independent source.
The Superposition Theorem helps simplify complex circuit analysis by breaking down the problem into smaller, more manageable parts.
To apply the Superposition Theorem, follow these steps:
1. Identify all independent voltage and current sources in the circuit.
2. For each independent source, turn off (neutralize) all other sources:
• Replace independent voltage sources with short circuits (a wire, which has zero resistance).
• Replace independent current sources with open circuits (a break in the circuit, which has infinite resistance).
3. Analyze the circuit with only one source acting at a time, and calculate the desired response (voltage or current) in the circuit. Use appropriate circuit analysis techniques, such as Ohm’s Law, Kirchhoff’s Laws, or node-voltage analysis.
4. After analyzing the circuit with each independent source acting alone, sum up the individual responses to find the total response in the circuit.
Keep in mind that the Superposition Theorem is only applicable to linear, time-invariant circuits. It does not apply to circuits with non-linear components, such as diodes or transistors, or circuits with time-varying components, such as switches.
The Superposition Theorem is especially useful when analyzing circuits with multiple sources and simplifies the analysis by focusing on one source at a time. This approach helps identify the contribution of each source to the overall response of the circuit.
## Other Circuit Theorems
Circuit theorems are essential tools for analyzing and simplifying complex electrical circuits. These theorems help engineers and technicians find equivalent circuits, solve for unknown quantities, and optimize circuit performance. Some of the most important circuit theorems include:
The primary purpose of this project is to help the public to learn some exciting and important information about electricity and magnetism.
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## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Calculus is a very important branch of mathematics. The different branches of mathematics have been
enriched a lot by the application of calculus.
1) Differential calculus
· Real number
· Function
· Limit
· Continuity
· Differentiation
· Second order derivative
2) Integral calculus
· Indefinite integral
· Method of substitution
· Integration by parts
· Integrals of some special form of functions
· Definite integral
3) Differential equation
· Order and degree of differential equations
· Differential equations of the first order and of the first degree
· Linear second order differential equations with constant coefficients
4) Application of calculus
· Significance of derivatives
· Tangent and normal
· Maxima and minima
· Definite integral as an area
· Velocity and acceleration
Few examples of advance calculus problem: -
Example 1: - Find the derivatives of y w.r.t. x when y = (x + 1) (x – 1)
Solution: - Differentiating both sides w.r.t. x,
dy / dx = (x + 1) d / dx ( x – 1) + (x – 1) d/dx (x + 1)
= (x + 1) [ d / dx (x) – d / dx (1)] + (x – 1) [d/dx (x) + d/dx (1)]
= (x + 1) (1 – 0) + (x – 1) (1 – 0)
= (x+1) + (x -1)
= x + 1+ x – 1
= 2x
Therefore dy /dx = 2x
Example 2: Evaluate ∫ x ^4 dx
Solution: -
∫ x ^4 dx = x^ (4 + 1) / (4+ 1)
= x^5 / 5
Therefore ∫ x ^4 dx = x^5 / 5
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# Trace operation on dynamic equation: physical meaning?
Suppose we have Heisenberg equation of motion for some observable $A$,
$$i\hbar\frac{dA}{dt}= -[H,A]$$ since the trace of any finite dimensional commutator structure vanish(not something like $[x,p]=i\hbar$ ),
$$Tr\left(\frac{dA}{dt}\right)=\frac{i}\hbar~Tr([H,A])=0$$
so my question is, what does $Tr(\frac{dA}{dt})=0$ indicate?
• $A$ and $H$ are typically not finite dimensional, the energy spectrum is typically infinite, so you cannot use commutativity of Trace.
– user7757
Jun 12, 2014 at 19:22
• @ramanujan_dirac I am not talking about [x,p]=ih, in many cases people are using heisenberg equation of motion, and applied to finite dimensional system. Jun 12, 2014 at 19:27
## 1 Answer
If the trace of an operator with a complete set of eigenvectors is well defined, then it is equal to the sum of the eigenvalues of the operator.
Time evolution of an operator is a unitary transformation, which leave the eigenvalues of an operator unchanged. This means if $a_i$ is an eigenvalue of an operator $A$
$$\frac{\mathrm{d}a_i}{\mathrm{d}t} = 0$$
and so clearly
$$\mathrm{Tr}\left(\frac{\mathrm{d}A}{\mathrm{d}t}\right) = \sum_i\frac{\mathrm{d}a_i}{\mathrm{d}t} = 0$$
So physically this result is to do with the fact that although the system may evolve with time, the spectrum of allowed results for a measurement of $A$ does not.
• "Time evolution of an operator is a unitary transformation", I don't think this's always true, right? Jun 13, 2014 at 5:38
• We can write $A = U^\dagger A_0U$, where $A_0$ is time independent. The solution to the Heisenberg equation for $U$ is then $U = \exp\left(-\imath\frac{Ht}{\hbar}\right)$ which can be shown to be unitary for Hermitian $H$ Jun 13, 2014 at 9:05
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` ANTIC VOL. 2, NO. 7 / OCTOBER 1983`
ARITHMETRIX
And now the amazing Atari!
by JERRY WHITE
Requires 16K RAM
Here is a simple ATARI BASIC program that demonstrates five arithmetic tricks. Arithmetrix can be used to fool your friends and make them think that your computer can read their minds.
Following is a sample run of the program, presented step by step. In each case, press [RETURN] after typing. Just read the screen, and answer the prompts as follows.
You will be asked to select a positive whole number. For all but the fourth case, use the number five for our first sample run.
(1) You are asked to multiply your number (minus one) by your number (plus one). We have chosen the number five so our number (minus one) equals four and our number (plus one) equals six. We multiply six by four and enter the result 24. The program tells us that our original number is five.
(2) You are asked to multiply your number by three, then tell the program if the result is odd or even. Our number is five, and five times three equals 15, so type ODD. We are asked to add one to the result (16), divide by two (eight), multiply by three (24), subtract six (18), then divide by nine and disregard any remainder (two). When you enter the result of two, the program will again tell you that your original number is five.
(3) You are asked to divide your original number by three, then enter the remainder. Five divided by three is one with a remainder of two, so type two. You are then asked to divide your original number by four and enter the remainder of one, then divide by five and enter the remainder of zero. Once again, the program has enough information to figure out your original number.
(4) This time the computer will predict your result. You are asked to select a three-digit number, and make sure that the difference between the first and last digits is greater than one. For this example, we will use the number 123. The first result is obtained by reversing the order of our number (321). The second result is obtained by finding the difference between our original number and the first result (198). Result number three is obtained by reversing the digits of the second result (891). You are asked to add the second result to result number three then press [RETURN]. Like magic, your computer will tell you that the final result is 1089.
(5) In the final algorithm the computer again predicts the answer. Choose a starting number, perhaps five. To obtain the first result we multiply by six (30), add 36 (66), divide by two (33). To obtain result number two we multiply our original number by three (15). Subtract result number two from the first result, press [RETURN], and your smart little machine will tell you that your final result is 18.
The program listing is divided into sections, each labeled with REM statements. It should be obvious that there isn't any magic at all. It's all BASIC logic.
The display routine reads from DATA statements and centers each new line on the screen. Since a comma indicates a new line of DATA, the "@" character was used in the data wherever a comma is to be displayed on the screen. The display routine then searches each line and converts each "@" to ",". My thanks to Sid Hyman for the idea for Arithmetrix.
Jerry White is Technical Consultant and regular contributor to ANTIC. His programs are available from several software companies, notably Adventure International and Educational Software.
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## Recent
recent
[email protected]
31. An equilateral triangle of sides 3 inch each is given. How many equilateral triangles of side 1 inch can be formed from it?
a)18 b) 10 c) 9 d)18
32. If A/B = 3/5,then 15A = ?
a) 9B b)18A c)9A d)27A
33. Each side of a rectangle is increased by 100% .By what percentage does the area increase?
a)250% b) 300% c)200% d)400%
34. Perimeter of the back wheel = 9 feet, front wheel = 7 feet on a certain distance, the front wheel gets 10 revolutions more than the back wheel .What is the distance?
a) 315 feet. b) 320 feet c) 410 feet d) 540 feet
35. Perimeter of front wheel =30, back wheel = 20. If front wheel revolves 240 times. How many revolutions will the back wheel take?
a)340 times b)360 times c)400 times d)450 times
36. 20% of a 6 litre solution and 60% of 4 litre solution are mixed. What percentage of the mixture of solution
a) 36% b)40% c)24% d)20%
37. City A's population is 68000, decreasing at a rate of 80 people per year. City B having population 42000 is increasing at a rate of 120 people per year. In how many years both the cities will have same population?
a)150 years b) 130 years c)120 years d)100 years
38. Two cars are 15 kms apart. One is turning at a speed of 50kmph and the other at 40kmph . How much time will it take for the two cars to meet?
a)1 hours b)2 hours c)5 hours d) 3/2 hours
39. A person wants to buy 3 paise and 5 paise stamps costing exactly one rupee. If he buys which of the following number of stamps he won't able to buy 3 paise stamps.
a) 12 b)8 c) 9 d)10
40. Which of the following fractions is less than 1/3
(a) 22/62
(b) 15/46
(c) 2/3
(d) 1
31.c
32.a
33.b
34.a
35.b
36.a
37.b
38.d
39.c
40.b
Basic Quantitative aptitude--four Reviewed by BANK INTERVIEWS on 4:28:00 PM Rating: 5
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# Base Ten
## Base
Add: ozufapeg29 - Date: 2020-12-12 13:42:47 - Views: 8660 - Clicks: 4988
Base Ten Blocks are especially useful in providing students with ways to physically represent the concepts of place value and addition, subtraction, multiplication, and division of whole numbers. This video is targe. Base Ten Blocks virtual manipulatives is to help students learn mathematical concepts including addition, subtraction, number sense, place value and counting. This group of lessons is based on a Math game show theme. Log base 10, also known as the common logarithm or decadic logarithm, is the logarithm to the base 10. This essentially means that you can only use ten unique digits (0 to 9) in each place of a base ten number.
Level 1: Base Ten Numbers. 10&39;s and 1&39;s are represented by rods and blocks. In base 10, each digit in a number represents the number of copies of that power of 10. Numbers are explained and built using blocks to represent place value. EXAMPLE: base numbers and number systems. It is a helpful way to keep track of our money, as we can exchange pennies for dimes and dimes for dollars. 10 Base 10 Blocks (all) Title: Microsoft Word - base10_all. The base-5 to base-10 conversion table and conversion steps are also listed.
To convert a natural logarithm to base-10 logarithm, divide by the conversion factor 2. This is an introduction to the base 10 system for whole numbers. Base Ten Blocks (also known as "Base 10 Blocks" and "Place Value Blocks") is an online mathematical manipulative that helps students learn addition, subtraction, number sense, place Base Ten value and counting. • the Base-10 number system is known as the decimal system and has 10 digits to show all numbers 0,1,2,3,4,5,6,7,8,9 using place value and a decimal point to separate whole numbers from decimal fractions.
Following this example, the binary number 10 is 2 in our (base-10) system. In base numbering system, numbers are represented using digits (0-9) and basic latin alphabet letters (from "A" to "Z" = 26 letters). Natural antilogs may be represented by symbols such as: InvLn, Ln^(-1), e^x, or exp. As a fun bonus, children can select different snake styles when they answer 10 questions correctly! The common logarithm of x is the power to which the number 10 must be raised to obtain the value x. It lets them investigate how to regroup and solve problems with whole numbers and eventually fractions and decimals.
Ask students to detail the process they Base Ten used to figure out how to write the numbers in base-ten numerals. show evidence of a decimal system. Counting Base Ten Blocks Worksheets. This system uses 10 as its base number, so that is why it is called the base-10 system. Base conversion calculator with steps: binary,decimal,octal,hex conversion. Base-10 system • our everyday number system is a Base-10 system. More Base Ten images. Starter Set includes: • 100 units • 30 rods • 10 flats • 1 cube • Base Ten Activity Book (LER 4296) Units are made of durable, washable plastic for years of hands-on activities.
Create a word bank with the words expanded, base-ten, and number names. Print the Spring Base Ten Blocks Math Activity for 1-10 and extend the learning! Also, explore tools to convert base-5 or base-10 to other numbers units or learn more about numbers conversions. Includes composing and decomposing numbers, regrouping, counting and addition using base 10 blocks.
Base-10 blocks are used to help children to experiment with basic addition and subtraction within the realms of base-10. Suppose Base Ten we have a number in base 10 and want to find out how to represent that number in, say, base 2. We hope you liked it!
The natural logs and natural antilogs can be converted to base-10 counterparts as follows: Natural logs usually use the symbol Ln instead of Log. Once the students are in the world they will encounter Mark Bigglesworth game show host who will ask them if they want to start the level or watch the tutorial on how to read base 10 numbers in Minecraft. Pre- K GRADE K GRADE 1 GRADE 2 GRADE 3 GRADE 4 GRADE 5 GRADE 6+.
I&39;m very pleased with this kit. That is, the first digit tells you how many ones you have; the second. Number Pieces helps students develop a deeper understanding of place value while building their computation skills with multi-digit numbers.
Students will practice ans. Grade 1 math worksheet on base 10 blocks. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. Free pdf worksheets with no login required. , by using objects or drawings, and record each composition or decomposition by a drawing or equation (such as 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones. For example, in the number 345, there is a hundreds place, a tens place and a ones place. Base Ten Form - Displaying top 8 worksheets found for this concept. Inside the level they will find 4 Minecraft math models.
This is geared towards elementary students. Place Value: Compare Numbers Using Base-10 Blocks Assessment or Practice Sheets Use this Place Value: Comparing Numbers using Base-10 Blocks sheet as a pre-test, post test, or practice sheet. Only 10s and 1&39;s (no hundreds). Some of the worksheets for this concept are Math 5th grade base ten crossword 1 name, Name score place value blocks, Place value blocks, Math mammoth light blue grade 2 b, Number and operations in base ten 1 20number, Representing numbers a, Breaking a number into tens and ones, 1st grade base ten blocks.
doc Author: Administrator Created Date: 12:47:17 PM. It is a great tool for remediation! Base-10 is used in most modern civilizations and was the most common system for ancient civilizations, most likely because humans have 10 fingers. If you don’t have any base ten blocks, you could substitute them with unifix cubes or legos. In base-2 (binary), we only have 2 characters, i. Use a base ten rod and ten units for children to physically move to the correct place value column. Kids can choose a game mode depending on their desired level of difficulty, and also select place values options of ones, tens and hundreds.
Instant free online tool for base-5 to base-10 conversion or vice versa. In base 10, each digit in a position of a number can have an integer value ranging from 0 to 9 Base Ten (10 possibilities). Students use the pieces to represent multi-digit numbers, regroup, add, subtract, multiply, and divide. See how it is done in this little demonstration (press play): Also try Decimal, and try other bases like 3 or 4. In the converter, the input number base must have only digits 0-9 and letters A-Z. 0 and 1, until we start over again. Base Ten Fun helps kids understand place value, addition and subtraction through the use of virtual manipulatives. Does it make sense that a finite fraction ("decimal") is infinite.
Great for centers or individual. 1 Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e. This series of base ten blocks worksheets is designed to help students of grade 1, grade 2, and grade 3 practice composition and decomposition of place value of whole numbers. A block containing thousand units can also be referred as &39;1000-block&39;, hundred units as &39;flat&39; and ten units as &39;rod&39;. The numbering system that children learn and the one most of us are familiar with is the base ten system.
The Log Base 10 Calculator is used to calculate the log base 10 of a number x, which is generally written as lg(x) or log 10 (x). It will help you understand how all these different bases work. Instead of handing out place value worksheets to your child, ask them to use base ten blocks to show the place value of each base ten numeral in a number.
3424 (base 10) =basebase 10) = 314f (base 16) 3a4f (base 16) =base 2). Have students orally label each number and the form it&39;s in. Base ten is used in our counting system and our monetary system.
How do we do this? 100 base ten units 50 base ten rods 10 base ten flats 1 base ten cube 1 base ten revised edition grades 3-6 book (107pgs) The individual units are 1cm cubes, and weigh aprox. Egyptian hieroglyphs dating back to 3000 B. The decimal numeral system (also called base-ten positional numeral system, and occasionally called denary / ˈ d iː n ər i / or decanary) is the standard system for denoting integer and non-integer numbers.
That is, the first digit tells you how many ones you have; the second tells you how many 10s you have; the third tells you how many 10x10 you have; the fourth tells you how many 10x10x10 you have; and so on. 1g, both of which are helpful in getting my kids to become more familiar with those forms of measurement. Base ten numerals form the basis for our counting system and thus, our monetary system.
By building number combinations with Base Ten Blocks, students ease into the concept of regrouping, or trading, and can see the logical development of. This system was handed over to Greece, although the Greeks and Romans commonly used base-5 as well. It is the extension to non-integer numbers of the Hindu–Arabic numeral system. See more videos for Base Ten.
Great for helping students understand abstract base ten concepts, including place value, estimation, operations, and more. 2 Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Write down a few numbers in various forms (expanded form, base-ten numerals, and number names). Explore our complete library of ad-free math videos at Thank you for watching our Place Value Song. The base we usually use is base-10, because we have 10 (when including 0) digits until we start over again (8,9,10). Well, there is a simple and easy method to follow.
### Base Ten
email: [email protected] - phone:(810) 941-6154 x 6976
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# Key Idea 5 - Measurement
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``` Key Idea 5—Measurement:
Students use measurement in both metric and English measure to provide a
major link between the abstractions of mathematics and the real world in
order to describe and compare objects and data.
Overview:
Students need to be able to estimate, make, and use measurement in real-world situations. They
should be aware that measurement is approximate, never exact. Elementary students learn how to use
appropriate measurement tools and come to understand the various measurement attributes. The
students begin to use statistical methods such as graphs, tables, and charts to display and interpret
data. Intermediate students continue to develop measurement skills, explore measures of central
tendency, and informally derive and apply measurement formulas. Commencement-level students
build on their measurement skills, using dimensional analysis techniques and statistical methods.
They need to understand error in measurement and its consequence on subsequent calculations.
58
59
60
VISIBLY VERTICAL
Description:
Students will use measurement to experience real-world problems
involving their own growth statistics throughout the school year. They
will describe and compare data by collecting, recording, and displaying
data in various ways. Students will measure each other’s height on the
first Friday of each month. They will keep monthly records of their
height. Each time the heights are recorded, the teacher will have meas-
urement activities for the students to investigate.
Key Idea 5 - Measurement 81
82 Key Idea 5 - Measurement
Elementary Performance Indicators
Students will:
• Understand that measurement is approximate, never exact.
• Select appropriate standard and nonstandard measurement tools in measurement activities.
• Understand the attributes of length.
• Estimate and find measures such as length using both nonstandard and standard units.
• Collect and display data.
• Use statistical methods such as graphs, tables, and charts to interpret data.
PreK – K Grades 1 – 2
1. Have students discuss different types of 1. Have students measure heights using
nonstandard tools (e.g., blocks, books, hands) nonstandard units.
to measure length. 2. Explain meterstick, meter, and centimeter.
2. Mark off heights of the students against the wall. 3. Have students measure their heights in
3. Ask the class to decide on a nonstandard tool to centimeters using metersticks (see Activity
measure their heights. Sheet 1).
4. As you compare student heights, use expressions 4. Have students investigate similarities and
such as taller than, shorter than, or the same differences of data.
as. 5. Make a pictograph of students’ heights for
5. Ask students to find classmates who are taller, one month.
shorter, and/or the same height as they are. 6. Make a bar graph of students’ heights for
another month.
7. Compare heights over time.
1. Have students decide how to obtain the most accurate measurement of their height. Ask students what
unit they should use. Ask students how they can make sure their measurements are accurate.
2. Measure the height of each student monthly (see Activity Sheet 2).
3. Using tally marks, have students organize data into a frequency table. Make a pictograph to display
the number of students in each interval using fractional symbols. For example, if one smiley face
represents two students, half a smiley face represents one student.
4. Discuss how to get the average class height. Have students make predictions.
5. Have students find the mean and the range of the data each month.
6. After collecting data for six months, have students plot their height for each of the previous months as a
line graph.
7. At the end of the year, have students graph the class mean for each month. Discuss the concept of
growth over time.
Key Idea 5 - Measurement 83
84 Key Idea 5 - Measurement
Intermediate Performance Indicators
Students will:
• Estimate, make, and use measurements in real-world situations.
• Select appropriate standard and nonstandard measurement units and tools to measure to a
desired degree of accuracy.
• Develop measurement skills and informally derive and apply formulas in direct
measurement activities.
• Use statistical methods and measures of central tendencies to display, describe, and
compare data.
• Explore and produce graphic representations of data using calculators/computers.
• Develop critical judgment for the reasonableness of measurement.
1. Have students decide how to obtain the most accurate measurement of their height. Discuss the
concept of accuracy. Ask students what unit they want to use. Introduce millimeters as a more precise
unit of measure. How can they be certain they are accurate?
2. Working in pairs, students should measure each other’s height monthly. This yearlong project is
described in Activity Sheet 3.
3. Have students organize data into a frequency table and display their results as a graph. (The teacher
can also display bar graphs, line graphs, and histograms on a graphing calculator.)
4. Looking at the display of the different graphs, discuss how to approximate the average class height.
Have students write a short essay making predictions about the mean.
5. Have students find the mean, median, mode, and range of the data each month.
6. In the sixth month, have the students plot their height for each of the previous months on a
coordinate plane. Have students write a summary of their personal statistics for the previous months’
data.
7. At the end of the year, have students graph the mean class height for each month. Discuss the
concept of growth over time.
8. At the end of the year, have students make individual line graphs of their own growth. Have students
write a short essay making predictions of their height at the end of the next year.
Key Idea 5 - Measurement 85
1. Have students decide how to get the most accurate measurement of their height. Ask students what
units they should use and why. Students should discuss how to obtain the most accurate measurement
with the tools they have. (Students could use a carpenter’s level to help get accurate heights.)
2. Working in pairs, students should measure each other’s height monthly and record the heights in
centimeters, to the nearest tenth of a centimeter (see Activity Sheet 4, Part 1).
3. Have students organize class data into a frequency table and cumulative frequency table and
display their results as histograms (see Activity Sheet 4, Part 2).
4. Have students find the mean, median, mode, and range of the data each month.
5. Compare students’ heights in centimeters (the metric system) to their heights in feet and inches (the
English system). Compare heights to the heights of famous people. Data are available on the Internet
or in resource books (see Activity Sheet 5).
6. In the sixth month, have the students plot their height for each of the previous months on graph paper.
Have students use their data to write a summary of their personal statistics.
7. At the end of the year, have students graph the mean for each month. Discuss the concept of growth
over time.
8. At the end of the year, have students make individual line graphs of their own growth. Have students
write a short summary making predictions of their height at the end of the next year.
86 Key Idea 5 - Measurement
Commencement Performance Indicators
Students will:
• Choose the appropriate tools for measurement.
• Use statistical methods including measure of central tendency to describe and compare
data.
• Understand error in measurement and its consequence on subsequent calculations.
• Apply the normal curve and its properties to familiar contexts.
• Use statistical methods, including scatterplots and lines of best fit, to make predictions.
• Apply the conceptual foundation of rate of change.
• Determine optimization points on a graph.
Math A Math B
1. Have students measure their height. Let them 1. Have students obtain their height. Let them
decide the unit of measure and how to get the decide the unit of measure and how to get the
most accurate measurement possible (see most accurate measurement possible.
Activity Sheet 6, Part 1). 2. Have the students display the data graphically.
2. Have the students display the data graphically. 3. Find mean, median, mode, range, and standard
3. Have students find the mean, median, mode, deviation for the class. Compare the class data
and range for the class data. to a normal distribution curve. Discuss
4. Have students construct a cumulative percentages.
frequency distribution chart and a histogram 4. Give each student a Medical Association Growth
of the data (see Activity Sheet 6, Part 2). Chart. Have students find in which percentile
5. Have students construct a box and whisker plot their own height falls and predict what their
of the data. Have students write a short essay height will be in six months, one year, and two
about what the plot represents. years (see Activity Sheet 7).
6. Give each student a Medical Association Growth 5. Have students input their data into the statistical
Chart, found on pages 103 – 104 or at function of a graphing calculator.
www.cdc.gov/growthcharts. Have students find 6. Have students find the equation of best fit for
in which percentile their own height falls. Have their data. Plot the data and the equation of best
students predict what their height will be in six fit on graph paper. Write a summary explaining
months, one year, and two years (see Activity where the best-fit equation approximates the
Sheet 6, Part 3). data. Be specific about where it fits well and
7. Have students input their data into the statistical where it does not. Interpolate half-year points
function of a graphing calculator and find the and extrapolate points beyond the graph.
mean, median, mode, range, and upper and
lower quartiles of the data.
8. Extension: Have students find the line of best fit
for their data. Plot the data and the line of best
fit on graph paper. Write a summary explaining
where the best-fit line approximates the data. Be
specific about where it fits well and where it
does not. Interpolate half-year points and
extrapolate points beyond the graph.
Key Idea 5 - Measurement 87
88 Key Idea 5 - Measurement
Activity Sheet 1
MY MONTHLY HEIGHT RECORD
___________________________________’s Height Record
September October November December January
cm cm cm cm cm
I notice that
_____________________________________________________________________________
_____________________________________________________________________________
February March April May June
cm cm cm cm cm
The greatest difference was
_____________________________________________________________________________
_____________________________________________________________________________
Key Idea 5 - Measurement 89
90 Key Idea 5 - Measurement
Activity Sheet 2
MONTHLY HEIGHT RECORD
Month __________
Height (cm) Describe how to find the mean and the range for the data.
mean
range
Key Idea 5 - Measurement 91
92 Key Idea 5 - Measurement
Activity Sheet 3
VISIBLY VERTICAL JOURNAL
First page should include the following information:
• Title (Visibly Vertical)
• School Year
• Name
• Class/Section
• Teacher
Project:
Working in pairs, students will measure each other’s height on the first Friday of each month and
record the data.
Journal Entries:
• After a small group discussion concerning the concept of accuracy in measurement, write a
brief summary of the class data.
• Make a frequency table of the first month’s heights and display results as a graph.
• After observing different types of graphs displayed on a graphing calculator and discussing how
to calculate the class average, write a short summary about your prediction of the mean.
• Organize a chart to keep track of the mean, median, mode, and range of the data each month.
• In the sixth month, plot your height for each of the previous months on graph paper. Write a
summary of your personal statistics for the previous months’ data.
• At the end of the year, graph the class mean for each month and discuss growth over time.
• At the end of the year, make a line graph of your own growth. Predict your height at the end of
the next school year and justify your prediction.
• Write a summary of what you learned from doing this project.
Key Idea 5 - Measurement 93
94 Key Idea 5 - Measurement
Activity Sheet 4
PART 1: VISIBLY VERTICAL—MONTHLY CHART
Sept Oct Nov Dec Jan Feb Mar Apr May June
Mean
Median
Mode
Range
Key Idea 5 - Measurement 95
PART 2
Heights for ____________ (month)
Interval Tally Frequency Interval Cumulative
Frequency
1. Construct a frequency histogram.
2. Construct a cumulative frequency histogram.
3. What is the difference between the frequency histogram and the cumulative frequency
histogram?
May:
Make a graph of your own height for the previous months. Find the mean, median, mode, and
mean. Make predictions about what your height will be in six months, one year, and three years, if
the current trend continues.
June:
Make a line graph of the mean class height for each month over the last 10 months. What trend do
you see?
96 Key Idea 5 - Measurement
Activity Sheet 5
HEIGHTS OF SOME FAMOUS PEOPLE
Angelina Jolie 5’ 7”
Carmen Electra 5’ 4”
David Duchovny 6’ 0”
Gwyneth Paltrow 5’ 10”
Julia Roberts 5’ 9”
Leonardo DiCaprio 6’ 0”
Mel Gibson 6’ 0”
Reese Witherspoon 5’ 6”
Sandra Bullock 5’ 8”
Tyrone “Muggsy” Bogues 5’ 3”
Wilt Chamberlain 7’ 1”
Key Idea 5 - Measurement 97
98 Key Idea 5 - Measurement
Activity Sheet 6
PART 1: VISIBLY VERTICAL—HEIGHTS
HEIGHT HEIGHT
1. 15.
2. 16.
3. 17.
4. 18.
5. 19.
6. 20.
7. 21.
8. 22.
9. 23.
10. 24.
11. 25.
12. 26.
13. 27.
14. 28.
1. Find the:
a. Mean _______________ d. Range ____________________
b. Median _______________ e. Upper quartile ______________
c. Mode _______________ f. Lower quartile ______________
2. Construct a box and whisker plot that represents the heights of the class.
3. Input the class heights into a graphing calculator. Calculate the mean, median, mode, range,
upper quartile, and lower quartile. Compare your results.
4. Calculator results:
a. Mean _______________
b. Median _______________
c. Mode _______________
d. Range _______________
e. Upper quartile _______________
f. Lower quartile _______________
5. Using a graphing calculator, create a box and whisker plot from the data. How does this graph
compare to your box and whisker plot?
Key Idea 5 - Measurement 99
PART 2
Divide the range into 6 – 10 subdivisions of equal length.
Complete a frequency and a cumulative frequency distribution chart.
Interval Tally Frequency Interval Cumulative
Frequency
Construct a cumulative frequency histogram from the data above.
PART 3
Find your height on the growth chart.
1. In what percentile is your height found?
2. What does the percentile represent?
3. Using your percentile curve, enter the height at each age according to the growth chart.
AGE (YEARS) HEIGHT AGE (YEARS) HEIGHT
1 10
2 11
3 12
4 13
5 14
6 15
7 16
8 17
9 18
Enter the data into your graphing calculator.
Using the statistics function, find the line of best fit.
1. What is the equation of the line of best fit?
2. What is the slope of the line of best fit?
3. What does the slope of this line represent?
4. What is the y-intercept?
5. What does the y-intercept of this line represent?
6. Write a summary explaining why, or why not, your equation is a good representation of your
growth. Where does it fit best?
7. Interpolate. (How tall were you when you were 5 1/2 years old?)
8. Extrapolate. (According to the line, how tall will you be when you are 30 years old?)
9. Do these values make sense? Explain.
100 Key Idea 5 - Measurement
Activity Sheet 7
HEIGHT AND WEIGHT FOR GROWING CHILDREN
Background:
Doctors use charts to assess children’s growth. They compare the height and weight of the child to
that of other children of the same age. You have been given a copy of the chart they use to monitor a
child’s growth. In order to computerize this information, we need to translate the charts into functions.
Find your own height on the chart and determine in which percentile your height falls.
Comparing Models for a Child’s Growth:
Construct a table of values for your percentile for each age on your curve (2 – 18). Use your
calculator to determine the regression equations for your data. Check three different models: linear
(LINReg), logarithmic (LNReg), and exponential (EXPReg). Construct a graph that includes all
three functions and the actual data points. Compare each model (function) to the actual data. Write
a description of how well each model fits the given data. Be specific about where it fits well and
where it doesn’t.
Construct the Best Model:
Break the data into sections and choose which type of model is best for each section. For each new
section, find a new regression equation that fits. Construct a new chart that uses the different
equations for the different sections of the curve (thus creating a perfect fit for your data). Describe
how well your new model fits the data. Discuss the decisions you made about where and how to
break up the data. Describe alternate solutions you tried.
Interpolate and Extrapolate:
Choose a point between two of your actual data points (half-year data point) and compare the
values given by the different models. Which values make the most sense? Which would you throw
out? Choose a data point after the values on your curve and compare the values given by the
different models. Which values make the most sense? Which would you throw out?
Rate of Change:
Describe how the rate of change of a child’s growth varies over the years. Describe how this is
reflected in each of your models.
End Behavior of Models:
Describe what is happening to the growth of the child at the end of your curve. Compare this to the
behavior of each of your models if the curve were continued. Discuss why your model will or will
not give accurate values for an 80-year-old.
Summarize:
What function(s) would you use to model your percentile curve? Justify why this is the best model
for the data. What unique observations, applications to the real world, or general rule do you feel fit
this project? Elaborate.
Key Idea 5 - Measurement 101
102 Key Idea 5 - Measurement
Growth Chart I
2 to 20 years: Girls NAME
Stature-for-age and Weight-for-age percentiles RECORD #
12 13 14 15 16 17 18 19 20
Mother’s Stature Father’s Stature cm in
Date Age Weight Stature BMI*
AGE (YEARS) 76
190
74
185 S
72
180 T
70 A
95
175 T
90
68 U
170 R
75 66
165 E
in cm 3 4 5 6 7 8 9 10 11 50
64
160 25 160
62 62
155 10 155
60 5 60
150 150
58
145
56
140 105 230
54
S 135 100 220
T 52
A 130 95 210
50
T 125 90 200
U
48 190
R 120 85
E 95 180
46
115 80
44 170
110 90 75
42 160
105 70
150 W
40 75
100 65 140 E
38 I
95 60 130 G
50
36 90 H
55 120
25 T
34 85 50 110
10
32 80
5
45 100
30
40 90
80 35 35 80
W 70 70
30 30
E 60 60
I 25 25
G 50 50
H 20 20
40 40
T
15 15
30 30
10 10
lb kg AGE (YEARS) kg lb
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Published May 30, 2000 (modified 11/21/00).
SOURCE: Developed by the National Center for Health Statistics in collaboration with
the National Center for Chronic Disease Prevention and Health Promotion (2000).
http://www.cdc.gov/growthcharts
103 Key Idea 5 - Measurement
Growth Chart II
2 to 20 years: Boys NAME
Stature-for-age and Weight-for-age percentiles RECORD #
12 13 14 15 16 17 18 19 20
Mother’s Stature Father’s Stature cm in
Date Age Weight Stature BMI*
AGE (YEARS) 76
95
190
74
90
185 S
75
72
180 T
50 70 A
175 T
25 68 U
170 R
10 66
165 E
in cm 3 4 5 6 7 8 9 10 11 5
64
160 160
62 62
155 155
S 60 60
T 150 150
A 58
T 145
U 56
140 105 230
R
54
E 135 100 220
52
130 95 95 210
50
125 90 200
90
48 190
120 85
46 180
115 80
75
44 170
110 75
42 160
105 50 70
150 W
40
100 65 140 E
25
38 I
95 60 130 G
10
36 90 5 H
55 120
T
34 85 50 110
32 80 45 100
30
40 90
80 35 35 80
W 70 70
30 30
E 60 60
I 25 25
G 50 50
H 20 20
40 40
T
15 15
30 30
10 10
lb kg AGE (YEARS) kg lb
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Published May 30, 2000 (modified 11/21/00).
SOURCE: Developed by the National Center for Health Statistics in collaboration with
the National Center for Chronic Disease Prevention and Health Promotion (2000).
http://www.cdc.gov/growthcharts
104 Key Idea 5 - Measurement
```
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# 65% Of What Number Is 26
65% Of What Number Is 26?
Have you ever been faced with a math problem that seems to be too complex to solve? Perhaps you’ve encountered the question, “65% of what number is 26?” and found yourself puzzled. Fear not, for in this article, we will break down this problem step by step and provide you with a clear understanding of how to solve it. So, let’s dive in!
To find 65% of a number, we need to multiply that number by 0.65. In this case, we are trying to find 65% of an unknown number, which we will refer to as “x.” Therefore, our equation becomes:
0.65 * x = 26
To solve for x, we need to isolate it on one side of the equation. We can do this by dividing both sides of the equation by 0.65:
x = 26 / 0.65
Now, let’s calculate this:
x = 40
Hence, the unknown number is 40. Therefore, 65% of 40 is 26.
FAQs
Q: How do I calculate percentages?
A: To calculate a percentage, divide the given value by 100 and then multiply it by the percentage. For example, to find 65% of a number, multiply the number by 0.65.
Q: Can I use a calculator to solve this problem?
A: Absolutely! Using a calculator will make it easier and faster to find the solution. However, it is always beneficial to understand the underlying concept and steps involved in the calculation.
Q: Why do we divide by 0.65?
A: We divide by 0.65 because 0.65 represents 65% in decimal form. Dividing by 0.65 helps us solve for the unknown number, as it cancels out the multiplying factor.
See also How to Square Numbers in C++
Q: Can this method be applied to find other percentages of a number?
A: Yes, the same method can be applied to find any percentage of a number. Simply replace 0.65 with the desired percentage in decimal form in the equation.
Q: What if the percentage is given in a different form, such as a fraction?
A: If the percentage is given as a fraction, convert it to decimal form before using it in the equation. For example, if the percentage is given as 3/5, divide 3 by 5 to get 0.60 and then multiply it by the number to find the percentage.
Q: Is there an alternative method to find the solution?
A: Yes, there is another way to solve this problem. You can set up a proportion by using the fact that 65% is equivalent to 0.65. The equation will be: x/100 = 26/0.65. Cross-multiply and solve for x to find the answer.
Q: Are there any practical applications of this concept?
A: Calculating percentages is an essential skill in various real-life situations. It can be used in finance, sales, statistics, and many other fields. Understanding how to find percentages helps in analyzing data, determining discounts, and making informed decisions.
In conclusion, finding 65% of a number can be easily accomplished by multiplying the number by 0.65. By following the steps outlined in this article, you can solve the problem “65% of what number is 26?” and apply the same method to other percentage calculations. Remember, practice makes perfect, so keep honing your math skills, and you’ll soon become a master at solving these types of problems.
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# MATH 101: ALGEBRA I WORKSHEET, DAY #3. Fill in the blanks as we finish our first pass on prerequisites of group theory.
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1 MATH 101: ALGEBRA I WORKSHEET, DAY #3 Fill in the blanks as we finish our first pass on prerequisites of group theory 1 Subgroups, cosets Let G be a group Recall that a subgroup H G is a subset that is a group under the binary operation of G The subgroup criterion says that H G is a subgroup if and only if It shows that if φ : G G is a group homomorphism, then φ(g) G is a subgroup: Let H G be a subgroup Define an equivalence relation on G by a b if and only if a 1 b H This is an equivalence relation because The equivalence classes are ah = {ah : h H} for a G; note ah = bh if and only if H = a 1 bh Let G/H be the set of equivalence classes We similarly define equivalence on the other side, with classes Ha for a G, and write H\G for the set of equivalence classes Lagrange s theorem states: It implies Fermat s little theorem: because Date: Wednesday, 14 September
2 2 Centralizer, normalizer, stabilizer Let A G be a subset The centralizer of A in G is C A (G) =, consisting of the elements of G The centralizer C A (G) G is a subgroup: The center of G is Z(G) = C G (G) Similarly, the normalizer of A in G is N A (G) = Note that C G (A) N G (A) The normalizer N A (G) G is a subgroup Let G X act on a set X Let x X The stabilizer of x G is the set Stab G (x) = G x = {g G : gx = x} We verify that G x G is a subgroup Recall that the kernel of the action is the subgroup (1) Let G X = G act on itself by conjugation, so that Then the centralizer of A in terms of the group action is (2) Let X = P(G) be the power set of G, consisting of Then G X acts on X by conjugation via (g, A) gag 1 = {gag 1 : a A} Under this action, N G (A) is (3) Let the group N G (A) A act on A by conjugation Then C G (A) N G (A) is 2
3 3 Cyclic groups A group G is cyclic if, and we write In this case, we have an isomorphism Consequently, any two cyclic groups of the same order are isomorphic Lemma Let G be a cyclic group, generated by x, and let a Z (a) If x has infinite order, then G = x a if and only if (b) If x has order n <, then G = x a if and only if ; in particular, the number of generators of G is Every subgroup of a cyclic group is cyclic: In fact, the subgroups of a finite cyclic group correspond bijectively with via 3
4 4 Generating subgroups Let G be a group Given any nonempty collection {H i } i I of subgroups of G, the intersection i I H i G is a subgroup Let A G The subgroup generated by A, written A G, is A = A H H G The subgroup generated by A is the smallest subgroup of G containing A in the sense that A more concrete way of thinking about A is that it consists of all elements of G that can be written as words in elements of A A 1, where A 1 = {a 1 : a A}: this means A subgroup H G is finitely generated if H Example The subgroup of GL 2 (C) generated by is A = ( ) 0 1, B = 1 0 ( ) 0 i i 0 Example The subgroup of GL 2 (Q) generated by S = ( ) ( ) , T = is infinite even though S, T have orders because 4
5 5 Kernels, normal subgroups, quotient groups Let φ : G G be a group homomorphism The kernel of φ is the subgroup ker φ = G Lemma φ is injective if and only if ker φ = {1} Proof Lemma Let K G be a subgroup Then the following conditions are equivalent: (i) For all a G, we have ak = Ka; (ii) For all a G, we have aka 1 K; and (iii) For all a G, we have aka 1 = K Proof Proves itself A subgroup K G is normal if the equivalent conditions (i) (iii) hold, and we write K G Example Every subgroup of an abelian group is normal Proposition If K G, then G/K is a group under (ak)(bk) = abk if and only if K G Proof 5
6 If K G, then the group G/K is the quotient group Kernels of surjective group homomorphisms are the same as normal subgroups in the sense that In general, if φ : G G is a group homomorphism, then it factors: Theorem (First isomorphism theorem) There are also three more isomorphism theorems, concerning intersections, quotients, and lattices Example The group Q/Z looks like 6
7 6 Composition series A (finite or infinite) group is simple if #G > 1 and A composition series for G is a sequence of subgroups 1 = N 0 N 1 N k 1 N k = G such that N i N i+1 and N i+1 /N i is a simple group for all 0 i k 1; we then call the set of groups N i+1 /N i the composition factors Example A composition series for S 3 is: A composition series for D 8 is: A composition series for A 4 is: Example Let F be a field with #F = p A composition series for the Heisenberg group H(F ) is: Theorem (Jordan Hölder) Every nontrivial finite group has a composition series and the composition factors in any composition series are unique (up to reordering) To classify all finite groups, we classify all finite simple groups and then try to put them back together to get all finite groups 7
8 Theorem Every finite simple group is isomorphic to either one in a list of 18 infinite families of simple groups or one of 26 sporadic simple groups Examples of simple groups include: 8
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### 3.10 $$\int \sinh ^3(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx$$
Optimal. Leaf size=77 $\frac{(a+b)^2 \cosh ^3(c+d x)}{3 d}-\frac{(a+b) (a+3 b) \cosh (c+d x)}{d}-\frac{b (2 a+3 b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d}$
[Out]
-(((a + b)*(a + 3*b)*Cosh[c + d*x])/d) + ((a + b)^2*Cosh[c + d*x]^3)/(3*d) - (b*(2*a + 3*b)*Sech[c + d*x])/d +
(b^2*Sech[c + d*x]^3)/(3*d)
________________________________________________________________________________________
Rubi [A] time = 0.0978984, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.087, Rules used = {3664, 448} $\frac{(a+b)^2 \cosh ^3(c+d x)}{3 d}-\frac{(a+b) (a+3 b) \cosh (c+d x)}{d}-\frac{b (2 a+3 b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d}$
Antiderivative was successfully verified.
[In]
Int[Sinh[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^2,x]
[Out]
-(((a + b)*(a + 3*b)*Cosh[c + d*x])/d) + ((a + b)^2*Cosh[c + d*x]^3)/(3*d) - (b*(2*a + 3*b)*Sech[c + d*x])/d +
(b^2*Sech[c + d*x]^3)/(3*d)
Rule 3664
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Rule 448
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]
Rubi steps
\begin{align*} \int \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right ) \left (a+b-b x^2\right )^2}{x^4} \, dx,x,\text{sech}(c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-b (2 a+3 b)-\frac{(a+b)^2}{x^4}+\frac{(a+b) (a+3 b)}{x^2}+b^2 x^2\right ) \, dx,x,\text{sech}(c+d x)\right )}{d}\\ &=-\frac{(a+b) (a+3 b) \cosh (c+d x)}{d}+\frac{(a+b)^2 \cosh ^3(c+d x)}{3 d}-\frac{b (2 a+3 b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.516529, size = 71, normalized size = 0.92 $\frac{-3 \left (3 a^2+14 a b+11 b^2\right ) \cosh (c+d x)+(a+b)^2 \cosh (3 (c+d x))+4 b \text{sech}(c+d x) \left (-6 a+b \text{sech}^2(c+d x)-9 b\right )}{12 d}$
Antiderivative was successfully verified.
[In]
Integrate[Sinh[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^2,x]
[Out]
(-3*(3*a^2 + 14*a*b + 11*b^2)*Cosh[c + d*x] + (a + b)^2*Cosh[3*(c + d*x)] + 4*b*Sech[c + d*x]*(-6*a - 9*b + b*
Sech[c + d*x]^2))/(12*d)
________________________________________________________________________________________
Maple [B] time = 0.052, size = 162, normalized size = 2.1 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cosh \left ( dx+c \right ) +2\,ab \left ( 1/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{\cosh \left ( dx+c \right ) }}+4/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{\cosh \left ( dx+c \right ) }}-8/3\,\cosh \left ( dx+c \right ) \right ) +{b}^{2} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{6}}{3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-{\frac{8\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{16\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3\,\cosh \left ( dx+c \right ) }}-{\frac{16\,\cosh \left ( dx+c \right ) }{3}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x)
[Out]
1/d*(a^2*(-2/3+1/3*sinh(d*x+c)^2)*cosh(d*x+c)+2*a*b*(1/3*sinh(d*x+c)^4/cosh(d*x+c)+4/3*sinh(d*x+c)^2/cosh(d*x+
c)-8/3*cosh(d*x+c))+b^2*(1/3*sinh(d*x+c)^6/cosh(d*x+c)^3-2*sinh(d*x+c)^4/cosh(d*x+c)^3-8/3*sinh(d*x+c)^2/cosh(
d*x+c)^3+16/3*sinh(d*x+c)^2/cosh(d*x+c)-16/3*cosh(d*x+c)))
________________________________________________________________________________________
Maxima [B] time = 1.19186, size = 358, normalized size = 4.65 \begin{align*} -\frac{1}{24} \, b^{2}{\left (\frac{33 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{30 \, e^{\left (-2 \, d x - 2 \, c\right )} + 240 \, e^{\left (-4 \, d x - 4 \, c\right )} + 322 \, e^{\left (-6 \, d x - 6 \, c\right )} + 177 \, e^{\left (-8 \, d x - 8 \, c\right )} - 1}{d{\left (e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )} + 3 \, e^{\left (-7 \, d x - 7 \, c\right )} + e^{\left (-9 \, d x - 9 \, c\right )}\right )}}\right )} - \frac{1}{12} \, a b{\left (\frac{21 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{20 \, e^{\left (-2 \, d x - 2 \, c\right )} + 69 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d{\left (e^{\left (-3 \, d x - 3 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )}\right )}}\right )} + \frac{1}{24} \, a^{2}{\left (\frac{e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac{9 \, e^{\left (d x + c\right )}}{d} - \frac{9 \, e^{\left (-d x - c\right )}}{d} + \frac{e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
-1/24*b^2*((33*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + (30*e^(-2*d*x - 2*c) + 240*e^(-4*d*x - 4*c) + 322*e^(-6*d*
x - 6*c) + 177*e^(-8*d*x - 8*c) - 1)/(d*(e^(-3*d*x - 3*c) + 3*e^(-5*d*x - 5*c) + 3*e^(-7*d*x - 7*c) + e^(-9*d*
x - 9*c)))) - 1/12*a*b*((21*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + (20*e^(-2*d*x - 2*c) + 69*e^(-4*d*x - 4*c) -
1)/(d*(e^(-3*d*x - 3*c) + e^(-5*d*x - 5*c)))) + 1/24*a^2*(e^(3*d*x + 3*c)/d - 9*e^(d*x + c)/d - 9*e^(-d*x - c)
/d + e^(-3*d*x - 3*c)/d)
________________________________________________________________________________________
Fricas [B] time = 2.05485, size = 664, normalized size = 8.62 \begin{align*} \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{6} +{\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{6} - 6 \,{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 3 \,{\left (5 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} - 2 \, a^{2} - 12 \, a b - 10 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} - 3 \,{\left (11 \, a^{2} + 86 \, a b + 91 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 3 \,{\left (5 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} - 12 \,{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 11 \, a^{2} - 86 \, a b - 91 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} - 26 \, a^{2} - 220 \, a b - 210 \, b^{2}}{24 \,{\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
[Out]
1/24*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^6 - 6*(a^2 + 6*a*b + 5*b^2)*cosh
(d*x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 - 2*a^2 - 12*a*b - 10*b^2)*sinh(d*x + c)^4 - 3*(11*a^2
+ 86*a*b + 91*b^2)*cosh(d*x + c)^2 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 - 12*(a^2 + 6*a*b + 5*b^2)*cosh(
d*x + c)^2 - 11*a^2 - 86*a*b - 91*b^2)*sinh(d*x + c)^2 - 26*a^2 - 220*a*b - 210*b^2)/(d*cosh(d*x + c)^3 + 3*d*
cosh(d*x + c)*sinh(d*x + c)^2 + 3*d*cosh(d*x + c))
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)**3*(a+b*tanh(d*x+c)**2)**2,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 1.64637, size = 392, normalized size = 5.09 \begin{align*} \frac{{\left (a^{2} e^{\left (3 \, d x + 36 \, c\right )} + 2 \, a b e^{\left (3 \, d x + 36 \, c\right )} + b^{2} e^{\left (3 \, d x + 36 \, c\right )} - 9 \, a^{2} e^{\left (d x + 34 \, c\right )} - 42 \, a b e^{\left (d x + 34 \, c\right )} - 33 \, b^{2} e^{\left (d x + 34 \, c\right )}\right )} e^{\left (-33 \, c\right )} - \frac{{\left (9 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 138 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 177 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 26 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 316 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 322 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 24 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 216 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 240 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 36 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 30 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a^{2} - 2 \, a b - b^{2}\right )} e^{\left (-3 \, c\right )}}{{\left (e^{\left (3 \, d x + 2 \, c\right )} + e^{\left (d x\right )}\right )}^{3}}}{24 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
[Out]
1/24*((a^2*e^(3*d*x + 36*c) + 2*a*b*e^(3*d*x + 36*c) + b^2*e^(3*d*x + 36*c) - 9*a^2*e^(d*x + 34*c) - 42*a*b*e^
(d*x + 34*c) - 33*b^2*e^(d*x + 34*c))*e^(-33*c) - (9*a^2*e^(8*d*x + 8*c) + 138*a*b*e^(8*d*x + 8*c) + 177*b^2*e
^(8*d*x + 8*c) + 26*a^2*e^(6*d*x + 6*c) + 316*a*b*e^(6*d*x + 6*c) + 322*b^2*e^(6*d*x + 6*c) + 24*a^2*e^(4*d*x
+ 4*c) + 216*a*b*e^(4*d*x + 4*c) + 240*b^2*e^(4*d*x + 4*c) + 6*a^2*e^(2*d*x + 2*c) + 36*a*b*e^(2*d*x + 2*c) +
30*b^2*e^(2*d*x + 2*c) - a^2 - 2*a*b - b^2)*e^(-3*c)/(e^(3*d*x + 2*c) + e^(d*x))^3)/d
|
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|
Collision of Particles
1. Oct 10, 2005
shawpeez
Could someone point me in the right direction, im totally lost
particle A moves along the line y=30m with a constant velocity v of magnitude 3.0m/s and parallel to the x axis. At the instant particle A leaves the y axis, particle B leaves the origin with zero initial speed and constant acceleration a of magnitude 0.40m/s^2. What angle between a and the positive direction of the y axis would result in a collision?
2. Oct 10, 2005
mezarashi
A math exercise once you can visualize what's going on. You have a particle moving along y=30m. Now you are at the origin with zero velocity. You want to crash into the dude moving at y=30. The only thing is that you don't have a steering wheel. Meaning you have to calculate the angle you want to start off with knowing that you can accelerate at 0.40 m/s/s.
Kinematics equations. You have two equations, one in the x and the other in the y.
What are the conditions for collision?
1. The displacement in y is equal to 30m.
2. The displacement in x is equal to the displacement of particle A at collision time (which we don't yet know).
With those two conditions, we write out the variables we know and are looking for in the x and y (I gave you some hints, fill in the rest):
x-axis
a: asin(angle)
vinitial:
vfinal: ?
displacement:
time: ?
y-axis:
a: acos(angle)
vinitial:
vfinal: ?
displacement:
time: ?
3. Oct 10, 2005
shawpeez
Wouldn't x - axis be a: acos(angle) and y -axis be a: asin(angle)
4. Oct 10, 2005
shawpeez
Ok this is what i've tried so far
Particle B
Y-axis
displacement= 30m
30 = V0(t) + 1/2a(t^2)
t= 12.25s
v = v0 + at
v = (.40)12.25
v= 4.9m/s
In 12.25s particle A travels 3.0m/s * 12.25s = 36.75m
X-axis
displacement = 36.75m
36.75 = vo + 1/2a(t^2)
36.75 =(.2)(t^2)
t= 13.56s
v= vo + at
= (.4)(13.56)
= 5.42m/s
I don't know if im on the right track or not but this is as far as i can get
5. Oct 10, 2005
shawpeez
still pretty much stuck, could anyone help
6. Oct 11, 2005
shawpeez
from this i took the position r=(36.75m)i + (30m)j
then i found the angle relative to the x axis
Tan(theta)= 30/36.75
theta = arctan .82
= 39.4
90 - 39.4 = 50.6 which would be the angle from the y-axis
can someone tell me if im doing this right i would appreciate it.
7. Oct 11, 2005
mezarashi
Normally yes, except the question asks
so you've got to be more flexible and truly understand the geometry problem instead of thinking that cosine is always an x-axis "thing".
How did you get the value for t? The value for a is acos(angle). I thought we were just discussing how we don't know the angle and that's what we're finding? Same goes for your x-axis analysis. The conditions we DO know are:
$$d_b_y = \frac{1}{2}a_1t^2 = 30m$$
$$d_b_x = \frac{1}{2}a_2t^2 = d_a_x = vt$$
where a1 and a2 are acos(angle) and asin(angle). How many unknowns, how many variables do we have?
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# The Locker Problem. Purpose: Participants will determine attributes of numbers and their factors.
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1 The Locker Problem Purpose: Participants will determine attributes of numbers and their factors. Overview: Participants will determine which locker numbers have exactly two factors, an odd number of factors, an even number of factors, and a large number of factors. They will determine which lockers will be left open and which will be left closed. They will investigate which locker numbers are visited the most frequently and why. They will determine which lockers are visited exactly twice and why. TExES Mathematics 4-8 Competencies. The beginning teacher: B Analyzes and describes relationships between number properties, operations, and algorithms for the four basic operations involving integers, rational numbers, and real numbers. I.003.A Demonstrates an understanding of ideas from number theory (e.g., prime factorization, greatest common divisor) as they apply to whole numbers, integers, and rational numbers, and uses these ideas in problem situations. I.003.E Applies properties of the real numbers to solve a variety of theoretical and applied problems C Makes, tests, validates, and uses conjectures about patterns and relationships in data presented in tables, sequences, or graphs. TEKS Mathematics Objectives. The student is expected to: 4.4.A Model factors and products using arrays and area models. 4.4.B Represent multiplication and division situations in picture, word, and number form. 4.4.C Recall and apply multiplication facts through 12 x D Use multiplication to solve problems involving two-digit numbers. 5.3.B Use multiplication to solve problems involving whole numbers (no more than three digits times two digits without technology). 5.3.C Use division to solve problems involving whole numbers. 5.3.D Identify prime factors of a whole number and common factors of a set of whole numbers. 5.5.B Use lists, tables, charts, and diagrams to find patterns and make generalizations. 5.5.C Identify prime and composite numbers using concrete models and patterns in factor pairs. 6.1.E Identify factors and multiples including common factors and common multiples. 6.2.C Use multiplication and division of whole numbers to solve problems. 7.2.E Simplify numerical expressions involving order of operations and exponents. 7.2.F Select and use appropriate operations to solve problems and justify the selections. 8.2.A Select and use appropriate operations to solve problems and justify the selections.
2 Terms. Factor, divisor, multiple, exponent, perfect square number, prime factors Materials. Transparencies Activity Sheets Graph paper Transparencies. Transparency: The Locker Problem Transparency: Solution Activity Sheet(s). Activity Sheet: The Locker Problem Procedure: Steps 1. Place the transparency The Locker Problem on the overhead projector and have the participants read the problem. Questions/Math Notes State the rules of the game in your own words? 2. With the help of the participants, act out the first few steps of The Locker Problem. Have the participants form a circle and explain that each person represents a locker. Number the participants (they will each represent a locker). The locker is closed when a person s back faces the inside of the circle; the locker is open when the person s front faces the inside of the circle. Ask Student #1 to role-play the part of the 1 st student in the problem. That person is to walk around the circle and open every locker. (It is effective to have Student #1 hand each person in the circle a 3x5 card with the number 1 written on it.) Ask Student #2 to role-play the part of the 2 nd student in the problem. That person is to walk around the circle and change the position of every 2 nd locker (beginning with locker #2). (It is effective to have Student #2 hand each even-numbered person in the circle a 3x5 card with the number 2 written on it. What number locker do you represent? What is the initial position of the lockers? (All lockers are closed so all persons should face the outside of the circle.) What does the 1 st student in this problem do? (The first student opens every locker.) Have Student #1 (starting at locker #1) tap each person on the shoulder to open the locker. As participants are tapped, they should rotate so that they are facing the inside of the circle. What does the 2 nd student do? Are his/her actions different from the 1 st student s? (The 2 nd student closes every other locker beginning with locker #2.) Ask Student #2 to begin with locker #2 and tap every other person on the shoulder. Those who are tapped should change the position of their locker. Half the circle should be facing inward and half the circle (those who were just
3 tapped) should be facing outward. Ask Student #3 to role-play the part of the 3 rd student in the problem. That person is to walk around the circle and change the position of every 3 rd locker (beginning with locker #3). (It is effective to have Student #3 hand each person whose position is changed a 3x5 card with the number 3 written on it. How do the actions of the 3 rd student differ from the 2 nd student s? (The 3 rd student changes the state of every 3 rd locker, beginning with locker #3.) Have Student #3 begin with locker #3 and tap every 3 rd person on the shoulder. Those who are tapped should change the state of their locker. Continue the pattern through at least Student #8. 3. Have participants work with a partner on the Activity Sheet: The Locker Problem using graph paper to build a table that shows the actions of the first students. Monitor their work and ask scaffolding questions as needed to clarify and extend their thinking about the actions of the students in the problem and the results of their actions. 4. Debrief the activity by having several groups share how they solved different pieces of the problem. What information do you need to collect in order to answer the questions? How are you going to record the actions of each student? Will this approach provide you with the information you need to answer the questions? How did you determine how many lockers will be open? How did the method you used for recording the data assist you in answering this question? How did you determine which lockers will be open? Why are these lockers open? How did you determine which lockers switched positions the most times? What are these lockers? Why did their position change so many times? 5. Have participants summarize in writing three key ideas that can be gleaned from this activity. Which lockers switched positions exactly twice? How did you determine the numbers of the lockers? How did you count these lockers?
4 Possible Solution: 1. The lockers that will be open have numbers that are perfect squares and are less than These lockers are: (1) 2, (2) 2, (3) 2, (4) 2, (5) 2, (31) 2. Hence, there are 31 lockers of the 1000 total that will be open. Open: 31 lockers Closed: 969 lockers 2. Open lockers are numbered: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961. These lockers are open because they have an odd number of factors. They are open because they have a double factor (e.g., 4 x 4). 16 = 1 x = 2 x 8 16 = 4 x 4 16 has 5 factors. It requires an odd number of factors for a locker to end in an open position. An even number of factors leaves a locker in a closed position. 3. Lockers which change their position the most are the ones whose numbers have the most factors. 840 has 32 factors. 840 = 2 3 x 3 1 x 5 1 x = 1 x = 10 x = 2 x = 12 x = 3 x = 14 x = 4 x = 15 x = 5 x = 20 x = 6 x = 21 x = 7 x = 24 x = 8 x = 28 x 30 Can you find one that has more factors? 4. Lockers that will change their state exactly twice are numbered with prime numbers. The prime numbers between 1 and 1000 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997 Hence, there are 158 lockers that will change their position exactly twice. Visit to see a listing of the first 1000 primes.
5 Supporting Websites: The Locker Problem Use Joe O'Reilly's applet to simulate The Locker Problem for the first 100 lockers: Visit the Math Forum for several interesting ideas on how to use this problem with Middle School students. Check out the Locker Boards and the clear instructions on how to use Claris Works Spreadsheets for recording the data. See Middle School student responses to The Locker Problem when it was posted as a Math Forum Problem of the Week (POW). Listings of Prime Numbers For a listing of the first 1000 prime numbers, visit: Sieve of Eratosthenes To find the prime numbers less than 400, use the Eratosthenes Sieve: The classroom version is excellent also Reference: House, P. A. (1980). Making a Problem of Junior High School Mathematics. The Arithmetic Teacher, 28(2),
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+0
-1
126
9
+397
deleted.
Nov 13, 2019
edited by sinclairdragon428 Nov 13, 2019
edited by sinclairdragon428 Nov 13, 2019
edited by sinclairdragon428 Nov 20, 2019
#1
+106981
+1
This question was answered not that long ago.
I expect that you mean the numbers 1,2,3,4,5,6,7,8,and 9
Nov 13, 2019
#2
+106981
+1
What have you done already towards solving this ?
Have you worked out which triples add to a multiple of 3?
Nov 13, 2019
#3
+397
-1
i worked out that if i have the numbers in ascending/descending order, it will be divisible by three (because x-1 + x + x+1 would be equal to 3x)
im not sure how to use that information to work out the total number though (or if there are other ways to place them).
I would love some guidance :)
sinclairdragon428 Nov 13, 2019
#4
+106981
0
First I will state that i haven ot attempted to work this out.
So you have worked out that these cannot be together
1,2,3
2,3,4
3,4,5
4,5,6
5,6,7
6,7,8
7,8,9,
8,9,1
9,1,2
Mmm
There are lots of others too. like 3,6,9 etc
This is really tricky.
Nov 13, 2019
#5
+106981
0
Here is where it was asked last time.
Seems you will be accused of cheating....
https://web2.0calc.com/questions/help-quick_12
Nov 13, 2019
#6
+106981
0
I'd give you some guidence if I had any but I really don't
I'd play around some more with when trippled can or cannot go together but I do not have any brainwaves.
Nov 13, 2019
#7
+397
0
ok :) thanks for the help so far
sinclairdragon428 Nov 13, 2019
#8
+19913
0
https://web2.0calc.com/questions/help-pls-fast_2
Nov 13, 2019
#9
+117
-2
Please do not post solutions to this problem! This is an active homework problem.
To the original poster: I realize that homework may be challenging. If you wish to receive some help from the staff or other students, I encourage you to use the resources that the online classes provide, such as the Message Board. Thanks.
Nov 13, 2019
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# Statistics
posted by .
When finding a confidence interval for a population mean based on a sample of size 11, which assumption is made?
The standard deviation is fixed.
The sampling distribution of z is normal.
The sampling distribution of the sample means is approximately normal.
There is no special assumption made.
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# Identify and write the like terms in the following group of terms.$7p,8pq, - 5pq, - 2p,3p$
Last updated date: 10th Aug 2024
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Hint:The terms having the same variable with same exponents are called like terms. Check the variables and exponents of all the variables. Then the terms with the similar exponents and variables will come under like terms.
Here, the given terms are $7p,8pq, - 5pq, - 2p,3p$
Among the given terms we should find the like terms. Initially for that we should consider the terms having same variables as a group
In the given group we have two different variables they are $p{\text{ & }}pq$
Let us group the terms that have $p$ as its variable.
$7p, - 2p$ and $3p$ are the terms which have variable $p$.
These three terms are known as terms of variable $p$, since they contain the same variable p to the same power. (The power of p is 1 in all the three terms)
Now let us group all the terms that contain the variable $pq$.
In the given group of variables there are two terms with variable $pq$.
$8pq$ and $- 5pq$ are the two terms which have the variable $pq$
These two terms are known as terms of variable $pq$, since they contain the same variable p and q to the same power. (The powers of p and q is 1 in both terms)
Hence,
We have found that the like terms in the given group are $\left( {7p, - 2p,3p} \right)$ and $\left( {8pq, - 5pq} \right)$
Note:We know that like terms are terms that contain the same variables raised to the same exponent (power). Only the numerical coefficients are different.
Constants are always said to be like terms because in every constant term there may be any number of variables which have the exponent zero. Unlike terms are the terms which have different variables and exponents.
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Posted by: atri | February 27, 2009
## Homework is out!
I’ll hand out copies in class today but you can also find a copy on the webpage. The writeup contains all the relevant instructions. The homework is due March 23 in class. Please use the comments section if you have any questions on the homework. (I’ll make this post sticky so it is easy to access.)
UPDATE (March 1): The hint for 4(b) was not quite right. I have fixed it in the current version of the homework on the course webpage (follow the link above).
## Responses
1. I have some questions for the homework. For the benefit of others I will write them here.
Q1:
****removed by Atri****
Also: What does it mean to have a non-linear code? Say we have a q-ary (n,k,d) code, and k need not be an integer. So k can be e.g. 2½. What does this code look like?
Q3:
Once again I think the understanding of non-linear code will help me. That being said, here’s my question:
That thing to be proved is the lower bound for the distance. The H^-1 function, is that the function we called Hq in class? Or is it 1/Hq? (q being 2 here, I suppose)
—–
Thank you. There is more to come, I am sure.
2. Hmm, my first post go messed up. I guess I used some wrong characters or something. So here’s my second try, please disregard the first one:
—————-
I have some questions for the homework. For the benefit of others I will write them here.
Q1:
***removed by Atri****
Q2:
I assume that the tricks lies in using a proof that states that a q-ary (n,k) code has the distance d leads to it is a q-ary (n,k,d) code. I can’t find that we have a proof for this anywhere. I can only find a proof that states that a code with a given distance leads to the fact that a minimum number of errors can be corrected. Where did we have a proof for the distance of a code?
Also: What does it mean to have a non-linear code? Say we have a q-ary (n,k,d) code, and k need not be an integer. So k can be e.g. 2½. What does this code look like?
Q3:
Once again I think the understanding of non-linear code will help me. That being said, here’s my question:
That thing to be proved is the lower bound for the distance. The H^-1 function, is that the function we called Hq in class? Or is it 1/Hq? (q being 2 here, I suppose)
—–
Thank you. There is more to come, I am sure.
3. JD,
Q1) I removed your comment as I think it gives away the real essence of the problem. However, as “hint” to you and others, let me mention a fact that I should have done earlier in the class. For any linear code of dimension $k$ and block length $n$, its $k\times n$ generator matrix has rank $k$. (Convince yourself that this follows from the fact that the code has dimension $k$.)
Q2) I’m not sure I get your question but I think your confusion is with the notation. The notation $(n,k,d)_q$ denotes a $q$-ary code with dimension $k$, block length $n$ and distance $d$.
Regarding your second point, in Q2 you cannot assume anything about the code other than what is given (i.e. block length, dimension, distance, alphabet size and whether the code is linear or not). Hence, what you have been asked to prove are “general purpose” conversion methods.
If your question regarding non-linear code is you want an example of how such a code might look like, here is an example of a $(4,1/2)_4$ code (I used $1/2$ instead of your $5/2$ to make the example easier). Consider the code over $\mathbb{Z}_4$ which has the codewords $(0,0,0,1)$ and $(1,1,1,1)$. Note that this code has the claimed parameters but it is not a linear code, e.g., as the vector $(0,0,0,1)+(1,1,1,1)=(1,1,1,2)$ is not in the code.
However, I repeat again: in Q2 you cannot use any specific property of the codes other than the parameters of the code that are given.
Q3) Sorry for using the $H^{-1}(\cdot)$ notation without defining it. I have added a definition in the homework: see the updated version on the course webpage. (And yes, I did mean $q=2$.)
4. A few more questions:
Q5b:
What does this mean delta = delta(gamma, p)? I think the proof is very similar to Shannon’s Capacity Theorem proof, but do I replace (p + gamma) with delta?
Q6a:
How many matrices G do we have? What are the properties of these matrices? And what do they do/how are they used?
Why do we have more than one matrix? I thought a code was generated by using a single matrix.
Q6b:
In the original proof of Shannon’s Capacity Theorem we find the expectation of the probability that y is received when E(x) was sent. We do this by splitting the vector space up into those y’s that lie within the hamming ball centered around E(x) with distance (p+gamma)n and those y’s that lie outside of this ball. Now we have an erasure channel with a linear code, so how do we split up the vector space in this case?
Q7:
First, alpha is a single element from Fq, right?
What does the alpha^i mean? I suppose it would be a function applied something like a^2 = a¤a, but what kind of function is “¤” then?
I just have a hard time using the first hint, since my trials with small q’s and n’s does not sum to zero (I’m trying addition and multiplication as functions). So there is obviously something I’m missing.
Thank you.
5. JD,
Below are my responses to your questions. Please let me know if the explanations do not make sense.
Q5b) $\delta=\delta(\gamma, p)$ means that $\delta$ is a number that depends only on $\gamma$ and $p$. In particular, if $\gamma$ and $p$ are constants, then so is $\delta$.
Also I don’t think the proof of 5b is similar to that of Shannon’s capacity theorem (at least not for the proof idea I have in mind).
Q6a) This part of the question has nothing to do with codes per se. As the question says $G$ is a $k\times n$ matrix over $\mathbb{F}_q$ and that is all you have to “play” with. Also the sentence in the parenthesis was stated in a slight confusing manner: I have fixed it in the writeup now. Maybe the new statement in the parenthesis will make the question clearer.
Even though this part of Q6 has nothing to do with codes per se, you will use the statement in part (a) to prove the claimed statement in part (b).
Q6b) The proof here is somewhat different from the proof that we saw in the class. In particular, in part (b) of this question note that you are asked to show the probability of a decoding error conditioned on the erasures taking place in the locations indexed by $J$.
The reason problem 6 is divided into four parts is partly because the proof for the erasure channel is a bit different from that for the $BSC_p$ that we did in the class. The four parts in some sense are the four main steps in the proof and in this problem you are supposed to prove each of these steps in sequence.
Q7) Yes, $\alpha$ is an element in $\mathbb{F}_q$ (but is not the zero element). Also $\alpha^i$ is $\alpha$ “multiplied” with itself $i$ times. The multiplication here is not just “normal” multiplication (say on reals or integers). We defined how multiplication is defined for finite fields in lecture 5 earlier in the semester. Unfortunately, the final polished notes are not up yet.
In short when $q=p$ is a prime, then the elements in the field are $\{0,1,\dots,p-1\}$ and the “addition” and “multiplication” are the “usual” addition and multiplication but modulo $p$. When $q=p^s$ is a prime power, then the elements of $\mathbb{F}_q$ are polynomials of degree $s-1$ over $\mathbb{F}_p$ and the addition and multiplication are polynomial addition and multiplication modulo an irreducible polynomial of degree $s$.
If you are not that comfortable with general finite fields, to get some intuition going I encourage you to think about the case when $q$ is a prime number.
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# Rank these three quantities from least to greatest.
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Rank these three quantities from least to greatest. [#permalink]
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23 Oct 2018, 04:02
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Ⅰ. $$\frac{49!}{(7^{49})^2}$$
Ⅱ. $$\frac{49!}{(7!)^2}$$
Ⅲ. $$\frac{49!}{(42!)(7!)}$$
Rank these three quantities from least to greatest.
A. I, II, III
B. I, III, II
C. II, I, III
D. II, III, I
E. III, I, II
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Re: Rank these three quantities from least to greatest. [#permalink]
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Bunuel wrote:
Ⅰ. $$\frac{49!}{(7^{49})^2}$$
Ⅱ. $$\frac{49!}{(7!)^2}$$
Ⅲ. $$\frac{49!}{(42!)(7!)}$$
Rank these three quantities from least to greatest.
A. I, II, III
B. I, III, II
C. II, I, III
D. II, III, I
E. III, I, II
Ⅰ. $$\frac{49!}{(7^{49})^2}$$ Since Denominator is too big so this will be in decimals and a very small number in comparison to others
Ⅱ. $$\frac{49!}{(7!)^2} = 7*8*47*23*3*11*43*42!/7!$$
Ⅲ. $$\frac{49!}{(42!)(7!)} = \frac{49*48*47*46*45*44*43*42!}{42!7!}= 7*8*47*23*3*11*43$$
I is least and II is largest
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Re: Rank these three quantities from least to greatest. [#permalink] 23 Oct 2018, 04:37
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# Relative Velocity with plane and wind
1. Mar 21, 2013
### Woolyabyss
1. The problem statement, all variables and given/known data
An aircraft flew due east fro P to Q at u1 km/h.Wind speed from south west was v km/h.On the return journey from Q to P,due west, the aircraft's speed was u2 km/h, the wind velocity being unchanged.If the speed of the aircraft in still air was x km/h, x>v,show by resolving the perpendicular to PQ.or otherwise,that
u1 - u2 = v√2
2. Relevant equations
3. The attempt at a solution[/b/
Can somebody help me with this?I think I made a simple mistake.
vpw = (xcosa)i - (xsina)j
vw = (v√2/2)i + (v√2/2)j
vp = (xcosa + v√2/2)i +(xsina + v√2/2)
since the j component must be zero sina = 2v/2x ............. cosa =√(4x^2 -2v^2)/2x
vp =x√(4x^2 -2v^2)/2x + v√2/2 = (v√2 + √(4x^2 -2v^2))/2 = u1
return journey
since the plane is travelling in the opposite direction
vpw = (-xcosa)i - (xsina)j
vp = (-xcosa + v√2/2)i +(-xsina + v√2/2)j
since j is zero cosa =√(4x^2-2v^2)/2x
u2=vp= (v√2 - √4x^2 -2v^2)/2 i
u1 - u2 = (√(4x^2 - 2v^2) +√(2v^2))/2 - (√(4x^2 -2v^2) - √(2v^2))/2
If I simplify it doesn't work out.Should I have multiplied u2 by minus 1 since its going in the minus i direction?
2. Mar 21, 2013
### Simon Bridge
Probably the easiest way to check your working is to draw out the vector triangles and do the geometry.
3. Mar 21, 2013
### Woolyabyss
Sorry im not exactly sure what you mean.I get the correct answer if I use the vector as opposed to the magnitude of the vector.
Last edited: Mar 21, 2013
4. Mar 21, 2013
### Simon Bridge
The solution involves adding and subtracting vectors - there are two main ways of doing this. You have been resolving each vector into components N-S and E-W.
Have you tried sketching the vectors out using head-to-tail?
5. Mar 22, 2013
### Woolyabyss
I just figured where I went wrong thanks anyway.
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# Efficient way to find $a^b \bmod {n}$
I am not sure whether or not this is a duplicate question. I'm wondering what is an efficient way to compute $$x \equiv a^b \bmod{n}$$ where $a,b,n \in \mathbb{Z}$ and $a,b < n$?
For example say I need to compute $13^{59} \bmod{77}$, how do I do it manually?
• Manually? Why? Commented Mar 6, 2011 at 7:29
• @TonyK : Because in exams we are not allowed to use a computer. :) BTW this is a homework question. Commented Mar 6, 2011 at 7:30
• By the way, there's no need for any restriction on $a$ or $b$, though you can take $a$ modulo $n$ and $b$ modulo $\phi(n)$. Commented Mar 6, 2011 at 7:32
• If this is for exam purpose, you may want to use Chinese Remainder Theorem to split your modulus. In your example, considering mod 7 or 11 + Euler's theorem is not a lot of work.
– user325
Commented Mar 6, 2011 at 7:35
This case is rather simple, since $(13,77)=1$ and $\phi(77) = 60$. So by Euler's theorem $13^{60} \equiv 1 \pmod{77}$. Now $13^{59}\cdot 13 \equiv 1 \pmod{77}$, Hence you can find it using the generalized Euclidean algorithm.
In general if $(a,n)=1$, then I'd take $b \pmod{\phi(n)}$ and proceed with taking iterated squares.
The usual trick is repeated squaring. If you want to raise a number $x$ to the 59th power, you follow this algorithm:
x1 = x
x2 = x1 * x1
x4 = x2 * x2
x8 = x4 * x4
x16 = x8 * x8
x32 = x16 * x16
x59 = x32 * x16 * x8 * x2 * x1
Usually the algorithm is given using the following recurrence equations: $$x^{2n} = (x^n)^2, x^{2n+1} = (x^n)^2\cdot x.$$ Using these equations, you can write a function that raises a number to a power.
Since you're computing everything modulo $n$, don't forget to reduce modulo $n$ each time. This will keep the numbers small and so the effort manageable.
• Knuth (in TAOCP) discusses this at length - repeated squaring is not optimal, though it uses at most twice as many multiplications as necessary, and a lot of squarings, which may be a bit faster. Commented Mar 6, 2011 at 7:34
$13^{59} \pmod 7 = (-1)^{59} \pmod 7 = 6$.
$13^{59} \pmod {11} = 2^{59} \pmod {11} = 2^9 \pmod {11} = 6$.
So the answer is $6$.
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# Adding Fractions - PowerPoint PPT Presentation
Adding Fractions. Adding Like Fractions. 1. Add numerators 2. keep denominators the same 3. Reduce or make mixed number. Try Some!. = 6 7. +. 1 5. ÷ 2 ÷ 2. 2. 9 3 10 10. = 12 10. = 6 5. 1. +. 1 2. = 12 8. ÷ 4 ÷ 4. = 3 2.
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### Adding Fractions
Adding Like Fractions
1. Add numerators
2. keep denominators the same
3. Reduce or make mixed number
= 6
7
+
1
5
÷ 2
• ÷ 2
• 2. 93
10 10
= 12
10
= 6
5
1
+
1
2
= 12
8
÷ 4
• ÷ 4
= 3
2
1
• 3. 66
8 8
+
1. 42
7 7
Adding unlike fractions
1. find common denominator (stack)
2. make new fractions
3. add new numerators
4. keep denominator the same
5. Reduce or make into mixed number
2. 23
3 4
+
+
4
7
2
14
2
3
3
4
8
14
2
14
8
12
9
12
x 2 x 2
x 4 x 4
=
=
x 3 x 3
+
=
+
=
5
7
5
12
10
14
17
12
1
÷ 2 ÷ 2
1. 42
7 14
Adding fractions & mixed numbers
1. find common denominator (stack)
2. make new fractions
3. add new numerators
4. keep denominator the same
5. add whole number to fraction
6. Reduce or make into mixed number
4
2. 31
4 6
3
+
+
3
4
1
6
3
2
3
1
6
9
12
2
12
3
4
6
1
6
x 3x 3
x 2 x 2
=
=
4
+
4
x 2x 2
+
=
=
11
12
5
6
4
3
1. 21
3 6
Adding Mixed Numbers
1. Stack fractions and Add whole numbers
2. find common denominator if necessary
3. make new fractions
4. add fractions
5. Reduce or make into mixed number
1
4
2. 31
4 6
3
2
+
+
1
2
3
1
5
3
2
3
1
6
10
15
3
15
3
4
6
1
6
x 5x 5
1
x 2 x 2
=
=
4
2
2
4
x 3x 3
=
+
=
+
13
15
5
5
5
6
1. 21
3 6
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## Lesson 4 – How Does Light Travel? Part II: Refraction
Posted on: Friday, February 4th, 2022 In: Learning Optics with Austin
Why does a pencil look like it is snapped in half when partially placed in a container of water? The short answer is refraction.
Pencil Demonstration of Refraction
A more detailed explanation is that the rays of light bends when they pass from one medium to another. This bending of light from air to water produces a virtual image of the pencil in water that is different from the pencil’s actual position.
Refraction Diagram
The way we measure refraction is through the refractive index (n). We define ⍺ to be the angle of incidence and Ɣ to be the angle of refraction. By the relationship sin(⍺) / sin(Ɣ) = n0 / n1, the refractive index (n) = n0 / n1. Thus, the larger n is, the smaller the angle of refraction which also means a greater bending of light.
Equation for Refractive Index
While it is the case when light travels from air to water that the angle of refraction is smaller than the angle of incidence, this is not always true. In fact, if light were to travel from water to air, then the angle of incidence would be greater than the angle of refraction. What is always true though, is that the angle of incidence is equivalent to the angle of reflection. This brings us back to the topic of reflectivity and transmission covered last week.
Reflection, Refraction, and Transmission Combined
Not all light that hits a surface goes through refraction. Depending on the conditions of the surface, light is also reflected. This is why the equation for reflectivity and transmission adds up to 100%. Thus, a material with high reflectivity will have low transmission and vice versa. Reflectivity and transmission are two crucially related variables that are considered in the designing of optical components.
Next: Lesson 5 – Intro to Optical Components
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# NEWTON’S SECOND LAW.
## Presentation on theme: "NEWTON’S SECOND LAW."— Presentation transcript:
NEWTON’S SECOND LAW
The effect of an applied force is to cause the body to accelerate in the direction of the force.
The acceleration is in direct proportion to the force and in inverse proportion to the mass of the body.
F = ma m, mass in kg a, acceleration in m/s2 F, force in Newtons
1 N = 1 kg•m/s2
To find the weight of an object in newtons, use “g” for the acceleration. Fw = mg
NEWTON’S THIRD LAW
When one body exerts a force on another, the second body exerts on the first a force of equal magnitude in the opposite direction.
For every action there is an equal and opposite reaction.
Action and reaction iinvolves two different forces acting on two different objects.
Mass is a measure of inertia in that the larger the mass, the more an object tends to resist a change in motion.
Mass and weight are not equivalent.
Equilibrium is the condition in which all forces on an object add up to zero.
If a person is standing in a room, his weight must be equal to the force with which the floor is pushing upward.
This upward pushing force is called “ the normal force”.
If a person is moving with constant velocity, all forces must be in equilibrium.
Forces are vectors. Forces can be added using vector addition.
Resolution of forces procedure for finding component forces from an original force vector.
A lawn mower has a mass of 130 kg
A lawn mower has a mass of kg. A person tries to push the mower up a hill that is inclined 15° to the horizontal. How much force does the person have to exert along the handle just to keep the mower from the rolling down the hill, assuming that the handle is parallel to the ground?
FRICTION is a force that resists motion
FRICTION is a force that resists motion. It involves objects that are in contact with each other.
CAUSES OF FRICTION Deformation of a surface
CAUSES OF FRICTION Deformation of a surface. Interlocking of irregularities of surfaces. How can friction be decreased?
Polishing surfaces decreases friction
Polishing surfaces decreases friction. However, making surfaces very smooth actually increases friction due to the forces of attraction between the molecules of substances.
Friction is often desirable
Friction is often desirable. Without friction you couldn’t walk across the room, you couldn’t get your car to start moving, nails would spring out of pieces of wood.
Starting friction - Maximum frictional force between stationary objects. (force needed to make an object start moving)
Sliding Friction - frictional force between objects that are sliding with respect to each other.
1. Friction acts parallel to the surfaces that are in contact and in the direction opposite to the motion of the object or to the net force tending to produce such motion.
2. Friction depends on the nature of the materials in contact and the smoothness of their surfaces.
3. Sliding friction is less than or equal to starting friction.
4. Friction is practically independent of the area of contact.
5. Starting or sliding friction is directly proportional to the force pressing the two surfaces together.
Coefficient of Sliding Friction - The ratio of the force of sliding friction to the normal (perpendicular) force pressing the surfaces together.
µ = Ff / FN
Increasing friction sand, tire chains,snow tires, rosin.
Decreasing friction - wax, oil, self-lubricating alloys, rolling friction, Teflon©.
Decreasing friction - wax, oil, self-lubricating alloys, rolling friction, Teflon©. (What makes Teflon© stick to the pan?)
A box is pushed across a level floor at a constant velocity by a force of 100 N. If the box weighs 200 N, what is the coefficient of friction between the box and the floor?
A box is pushed across a level floor at a constant velocity by a force of 100 N. If the box has a mass of kg, what is the coefficient of friction between the box and the floor?
A box with a mass of 175 kg is pulled along a level floor with a constant velocity. If the coefficient of friction between the box and the floor is 0.34, what horizontal force is exerted in pulling the box?
A force of 50 N is directed 30º above horizontal
A force of 50 N is directed 30º above horizontal. What are the vertical and horizontal components of that force?
A 120 N box is placed on a 25º incline
A 120 N box is placed on a 25º incline. What are the perpendicular and parallel components?
A wooden block weighing 130 N rests on an inclined plane
A wooden block weighing 130 N rests on an inclined plane. The coefficient of sliding friction between the block and the plane is Find the angle of the inclined plane at which the block will slide down the plane at constant speed once it has started moving.
A crate is pulled with a constant velocity up an inclined floor that makes an angle of 12° with the horizontal. The crate weighs 950 N and the pulling force parallel to the floor is 460 N. Find the coefficient of friction between the crate and the floor.
A box weighing 200 N is pulled along a level floor at constant speed by a rope that makes an angle of 40.0° with the floor. If the force on the rope is 100 N, (a) what is the horizontal component (Fh) of this force? (b) What is the normal force (FN)? (c) What is the coefficient of sliding friction (µ)?
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# An elementary course of practical mathematics, Part 1
### Popular passages
Page 187 - ... fourth ; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth...
Page 220 - Iff a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.
Page 184 - When there is a series of quantities, such that the ratios of the first to the second, of the second to the third, of the third to the fourth, &c., are all equal ; the quantities are said to be in continued proportion.
Page 26 - Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient, and subtract the product from the dividend.
Page 179 - Ratios tnat are equal to the same ratio are equal to one another.
Page 185 - If three quantities are proportional, the first is to the third, as the square of the first to the square of the second ; or as the square of the second, to the square of the third.
Page 184 - IF there be any number of magnitudes, and as many others, which, taken two and two, in a cross order, have the same ratio; the first shall have to the last of the first magnitudes the same ratio which the first of the others has to the last.
Page 93 - The first and fourth terms of a proportion are called the extremes, and the second and third terms, the means. Thus, in the foregoing proportion, 8 and 3 are the extremes and 4 and 6 are the means.
Page 180 - Division, when the difference of the first and second is to the second as the difference of the third and fourth is to the fourth...
Page 181 - If four magnitudes are in proportion, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference.
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Prof. Elvezio Ronchetti Research Center for Statistics and GSEM
Ass : Linda Mhalla University of Geneva
S203031 Probability and Statistics II Fall 2017
EXAM January 22, 2018
– The exam lasts 3h.
– You may answer in English or in French.
– You are not allowed to leave the room during the exam.
– Open book exam.
– Justify your answers and do not forget to write your name on your copy.
Good luck !
Exercise 1 (6 points)
A gambling game works as follows. A random variable X is produced ; you win 1 Fr if X > 0 and
you lose 1 Fr if X < 0. Suppose first that X has a normal (0, 1) distribution. Then the game is clearly
fair. Now suppose the casino gives you the following option. You can make X have a normal (b, 1)
distribution, but to do so you have to pay c · b Fr which is not returned to you even if you win. Here c > 0
is set by the casino, but you can choose any b > 0.
1. Defines the random variable G, your gain, as a function of X .
2. For what values of c is it advantageous for you to use this option ?
Exercise 2 (8 points)
Let X and Y have joint density
fX,Y (x, y) =
{
e−y if 0 < x < y <∞,
0 otherwise.
1. Compute Pr(Y > X + 1).
2. Give the marginal density of Y .
3. Give the conditional density of X|Y and compute the conditional expectation E(X|Y ).
4. Give the distribution of E(X|Y ).
Exercise 3 (16 points)
Let X1, . . . , Xn be an i.i.d. sample from a continuous distribution with cumulative distribution func-
tion
FX(x|θ) =
{
a + be−x
2/θ if x > 0,
0 otherwise,
where θ ∈ Θ = (0,∞) is an unknown parameter, a and b are some constants.
1. Show that a = 1 and b = −1.
2. Find fX(.|θ) the probability density function of X .
3. Show that Eθ(Xi) =
piθ/2 and use this fact to derive a method of moments estimator θˆMM of θ.
Hint : You can use the fact that
∫ +∞
−∞
1√
2piσ
x2e−x
2/(2σ2)dx = σ2, ∀σ2 > 0.
4. It can be shown that Eθ(X2i ) = θ. Using this fact, find Eθ(θˆMM).
Is θˆMM an unbiased estimator of θ ?
Is θˆMM a consistent estimator of θ ?
5. Let θ˜ = n[min(X1, . . . , Xn)]2. Find the cumulative distribution function of θ˜.
Hint : {min(X1, . . . , Xn) > u} if and only if {X1 > u} ∩ {X2 > u} ∩ · · · ∩ {Xn > u}.
6. Is θ˜ an unbiased estimator of θ ?
7. Find the log-likelihood function and show that the maximum likelihood estimator of θ is given by
θˆML =
1
n
n∑
i=1
X2i .
8. Compute the median of the distribution of Xi and give its maximum likelihood estimator.
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# Stuck on a complex number question dealing with the rotation of complex numbers in polar form .
The Question :
Two opposite vertices of a square are represented by complex numbers $9+12i$ and $-5+10i$. Find the complex number representing the other two vertices of the square.
My attempt :
Considering $ABCD$ as the square and $O$ as the origin the known points $A$ and $C$
$OA= 9+12i$
$OC= -5+10i$
By vector addition : $OA=OC+CA$ therefore , $CA=OA-OC$ we get, $CA=9+12i+5-10i=14+2i$
rotating $AC$ using polar for onto $AB$ (Clockwise) $AB=\sqrt{2} CA (\cos 45+ i \sin 45 )$ (45 degrees)
$AB=(14+2i)(1+i)$
$AB=14+14i+2i-2$
$AB=12+16i$
$OA+AB=OB$
$9+10i+12+16i=OB$
$OB=21+26i$ however the answer in the book is $1+18i$ and $3+4i$
The center of the square is $\frac{A+C}2=2+11i$. So, $MA=7+i$ and you get another vertex $B$ of the square rotating $A$ around $M$ by $\frac\pi2$ radians. That is$$B=M+iMA=1+18i.$$And you'll get the fourts vertex $D$ doing the smae thing, but with an angle of $-\frac\pi2$ radians:$$D=M-iMA=3+4i.$$Your error was to rotate around the origin.
• I see.. Because the square could have been inclined in any direction to the origin , we have to rotate it with respect to a stationary point . – Harshit Pandey Aug 29 '17 at 16:02
Here is another way to do this. Let
$$z_1=9+12i\\ z_3=-5+10i\\ s=\frac{|z3-z2|}{\sqrt{2}},\quad \text{side of the square}\\ \theta=\arg (z3-z2),\quad \text{its angle on the plane}$$
Then
$$z_2=z_1+se^{i(\theta-\pi/4)}=1+18i\\ z_4=z_1+se^{i(\theta+\pi/4)}=3+4i$$
(I've verified this solution.)
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# Maths - Matrix algebra - 4D arithmatic
The following calculator allows you to calculate 4x4 matrix arithmetic. Enter the values into the matrix into the top two matries and then press "+ - or * " to display the result in the bottom matrix:
=
To add matrices just add the corresponding elements, for 4x4 matrices then,
a00 + b00 a01 + b01 a02 + b02 a03 + b03 a10 + b10 a11 + b11 a12 + b12 a13 + b13 a20 + b20 a21 + b21 a22 + b22 a23 + b23 a30 + b30 a31 + b31 a32 + b32 a33 + b33
=
a00 a01 a02 a03 a10 a11 a12 a13 a20 a21 a22 a23 a30 a31 a32 a33
+
b00 b01 b02 b03 b10 b11 b12 b13 b20 b21 b22 b23 b30 b31 b32 b33
### Matrix Subtraction
To subtract matrices just subtract the corresponding elements, for 4x4 matrices then,
a00 - b00 a01 - b01 a02 - b02 a03 - b03 a10 - b10 a11 - b11 a12 - b12 a13 - b13 a20 - b20 a21 - b21 a22 - b22 a23 - b23 a30 - b30 a31 - b31 a32 - b32 a33 - b33
=
a00 a01 a02 a03 a10 a11 a12 a13 a20 a21 a22 a23 a30 a31 a32 a33
-
b00 b01 b02 b03 b10 b11 b12 b13 b20 b21 b22 b23 b30 b31 b32 b33
### Matrix Multiplication
The multiplication of two 4x4 matrices is:
ma00*mb00 + ma01*mb10 + ma02*mb20 + ma03*mb30 ma00*mb01 + ma01*mb11 + ma02*mb21 + ma03*mb31 ma00*mb02 + ma01*mb12 + ma02*mb22 + ma03*mb32 ma00*mb03 + ma01*mb13 + ma02*mb23 + ma03*mb33 ma10*mb00 + ma11*mb10 + ma12*mb20 + ma13*mb30 ma10*mb01 + ma11*mb11 + ma12*mb21 + ma13*mb31 ma10*mb02 + ma11*mb12 + ma12*mb22 + ma13*mb32 ma10*mb03 + ma11*mb13 + ma12*mb23 + ma13*mb33 ma20*mb00 + ma21*mb10 + ma22*mb20 + ma23*mb30 ma20*mb01 + ma21*mb11 + ma22*mb21 + ma23*mb31 ma20*mb02 + ma21*mb12 + ma22*mb22 + ma23*mb32 ma20*mb03 + ma21*mb13 + ma22*mb23 + ma23*mb33 ma30*mb00 + ma31*mb10 + ma32*mb20 + ma33*mb30 ma30*mb01 + ma31*mb11 + ma32*mb21 + ma33*mb31 ma30*mb02 + ma31*mb12 + ma32*mb22 + ma33*mb32 ma30*mb03 + ma31*mb13 + ma32*mb23 + ma33*mb33
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A162203 The mountain path of the primes (see comment lines for definition). 12
2, 2, 2, 3, 1, -1, 1, 3, 1, -1, 1, 3, 1, -3, 1, 4, 1, -2, 1, 5, 1, -1, 1, 3, 1, -3, 1, 6, 1, -2, 1, 4, 1, -3, 1, 3, 1, -2, 1, 5, 1, -3, 1, 7, 1, -4, 1, 3, 1, -1, 1, 3, 1, -1, 1, 9, 1, -7, 1, 5, 1, -2, 1, 6, 1, -4, 1, 4, 1, -4, 1, 5, 1, -3, 1, 6, 1, -2, 1, 6 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS On the infinite square grid we draw an infinite straight line from the point (1,0) in direction (2,1). We start at stage 1 from the point (0,0) drawing an edge ((0,0),(2,0)) in a horizontal direction. At stage 2 we draw an edge ((2,0),(2,2)) in a vertical direction. We can see that the straight line intercepts at the number 3 (the first odd prime). At stage 3 we draw an edge ((2,2),(4,2)) in a horizontal direction. We can see that the straight line intercepts at the number 5 (the second odd prime). And so on (see illustrations). The absolute value of a(n) is equal to the length of the n-th edge of a path, or infinite square polyedge, such that the mentioned straight line intercepts, on the path, at the number 1 and the odd primes. In other words, the straight line intercepts the odd noncomposite numbers (A006005). The position of the x-th odd noncomposite number A006005(x) is represented by the point P(x,x-1). So the position of the first prime number is represented by the point P(2,0) and position of the x-th prime A000040(x), for x>1, is represented by the point P(x,x-1); for example, 31, the 11th prime, is represented by the point P(11,10). See also A162200, A162201 and A162202 for more information. LINKS Omar E. Pol, Graph of the mountain path function for prime numbers Omar E. Pol, Illustration: The mountain path of the primes FORMULA From Nathaniel Johnston, May 10 2011: (Start) a(2n+1) = 1 for n >= 2. a(2n) = (-1)^n*(A162341(n+2) - 1) = (-1)^n*(A052288(n) - 1) + 1 for n >= 2. (End) EXAMPLE Array begins: ===== X..Y ===== 2, 2; 2, 3; 1,-1; 1, 3; 1,-1; 1, 3; 1,-3; 1, 4; 1,-2; 1, 5; CROSSREFS Cf. A000040, A006005, A008578, A162200, A162201, A162202, A162340, A162341, A162342, A162343, A162344. Sequence in context: A307298 A216674 A106795 * A071455 A288724 A198862 Adjacent sequences: A162200 A162201 A162202 * A162204 A162205 A162206 KEYWORD easy,sign AUTHOR Omar E. Pol, Jun 27 2009 EXTENSIONS Edited by Omar E. Pol, Jul 02 2009 More terms from Nathaniel Johnston, May 10 2011 STATUS approved
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# Likelihood function of a gamma distributed sample
I missed the day of class where we went over likelihood functions, and I had a quick question. If $X_1,...X_n$ are i.i.d. ${\Gamma}(\alpha,\beta)$ r.v.s, I'm trying to find the likelihood function for $\alpha$ and $\beta$. From what I can tell, the likelihood function is defined as $f(x_1;\alpha,\beta)*...*f(x_1;\alpha,\beta)$. If i'm not mistaking, $f(x_1;\alpha,\beta)$ should be the same as the pdf for the gamma distribution (although not a pdf, on account of x being a fixed value here?), so would the likelihood function in this case be ${(\text{pdf-of-}\Gamma)}^n$? Further, I'm a bit confused on what the support for this function should be; in this case x is fixed, am I correct in supposing that the support is then $\alpha>0, \beta>0$? I hope this makes sense - I get the feeling I might be very confused about the whole thing. Thank you for any help!
-
If $X$ follows a gamma distribution with shape $\alpha$ and scale $\beta$, then its probability density is
$$p_{\alpha, \beta}(x) = \frac{ x^{\alpha-1} e^{-x/\beta}}{\Gamma(\alpha) \beta^\alpha }$$
Sometimes this is re-parameterized with $\beta^{\star} = 1/\beta$, in which case you will need to change this accordingly.
The likelihood function is just the density viewed as a function of the parameters. So, the log-likelihood function for an IID sample $X_1, ..., X_n$ from this distribution with realized values $x_1, ..., x_n$ is
$$L(\alpha, \beta) = \sum_{i=1}^{n} \log \big( p_{\alpha, \beta}(x_i) \big) = (\alpha-1) \sum_{i=1}^n \log(x_i) - \frac{1}{\beta} \sum_{i=1}^{n}x_i - n\alpha \log(\beta) - n\log( \Gamma(\alpha) )$$
which can be maximized jointly as a function of $\alpha, \beta$ to get the MLE.
-
How do we differentiate gamma$\alpha$ to get the value of $\alpha$ ? – idpd15 Sep 3 '15 at 12:27
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## Exercise 3.1 Part 2
Question 6: Find the angle measures x in the following figures.
Solution: We know that, angle sum of a quadrilateral = 360⁰
So, 50° + 130° + 120°+ x = 360°
Or, 300°+ x = 360°
Or, x = 360°- 300°= 60°
Question 6 (b)
Solution:
We know that, angle sum of a quadrilateral = 360⁰
So, 90° + 60°+ 70°+ x = 360°
Or, 220° + x = 360°
Or, x = 360°- 220°= 140°
Question 6 (c)
Solution:Solution
We know that angle sum of a pentagon = 540o
110° + 120° + 30° + x + x = 540°
Or, 260° + 2x = 540°
Or, 2x = 540° - 260° = 280°
Or, x = 280°÷2 = 140°
Question 6 (d)
Solution:Solution: Angle sum of a pentagon = (5 – 2) xx 180⁰ = 3 xx 180⁰ = 540⁰
Since, it is a regular pentagon, thus, its angles are equal
So, x + x + x + x + x = 540°
Or, 5x = 540°
Or, x = 540°÷5 = 108°
Question 7:
Solution:
We know that angle sum of a triangle = 180⁰
Thus, 30⁰ + 90⁰ + C = 180⁰
Or, 120⁰ + C = 180⁰
Or, C = 180⁰ – 120⁰
Or, C = 60⁰
Now, y = 180° - C
Or, y = 180° - 60° = 120°
Similarly, z = 180°- 30° = 150°
Similarly, x = 180° - 90° = 90°
Hence, x + y + z = 90° + 120° + 150°= 360°
Alternate method: We know that sum of external angles of a polygon = 360⁰
Hence, x + y + z = 180°
Solution:
We know that angle sum of a quadrilateral = 360⁰
A + 60° + 80° + 120° = 360°
Or, A + 260° = 360°
Or, A = 360° - 260° = 100°
Hence, w = 180° - 100° = 80°
Similarly, x = 180° - 120° = 60°
Similarly, y = 180° - 80° = 100°
Similarly, z = 180° - 60° = 120°
Hence, x + y + z + w = 60° + 100° + 120° + 80° = 360°
Alternate method: We know that sum of external angles of a polygon = 360⁰
Hence, x + y + z + w = 360°
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14 September, 08:47
# Answer for lady stole 100 from owner she bought 70 dollars worth of products and got back 30 how much did the owner lose
+3
1. 14 September, 09:09
0
Answer: The total loss incurred by the store owner is 100\$.
At first look, the answer might seem to be 200\$ or 130\$, but that's not the correct one. The explanation is below:
The lady stole 100\$ from the owner therefore loss incurred is 100\$.
The lady went back to the store and:
Step 1 - The lady hands back the 100\$ to the store owner. Loss incurred is back to 0\$.
Step 2 - The store owner hands over merchandise worth 70\$. Loss Incurred is now 70\$.
Step 3 - The store owner hands over 30\$ worth of change. Loss Incurred is (70 + 30) \$ = 100\$
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# (a) (b) (c) Figure 1 : Graphs, multigraphs and digraphs. If the vertices of the leftmost figure are labelled {1, 2, 3, 4} in clockwise order from
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1 4 Graph Theory Throughout these notes, a graph G is a pair (V, E) where V is a set and E is a set of unordered pairs of elements of V. The elements of V are called vertices and the elements of E are called edges. We typically denoted by V (G) = V the vertex set of G and E(G) = E the edge set of G. If u, v V (G), then u and v are adjacent if {u, v} E(G). We refer to u and v as the endpoints of the edge {u, v}. It is convenient to draw V (G) as a set of points in the plane and each edge as a curve between its endpoints in the plane, as in Figure 1(a). A multigraph is a pair (V, E) where V is a set and E is a multiset of unordered pairs of elements of V we allow E to contain repetitions of the same unordered pair. In other words, more than one edge can join two vertices. Sometimes we even allow loops on a vertex see Figure 1(b). A directed graph digraph for short is a pair (V, E) where V is a set and E is a set of ordered pairs of G. The elements of E are called arcs, and we draw them in the plane as curves with an arrow indicating the direction of the arc see Figure 1(c). (a) (b) (c) Figure 1 : Graphs, multigraphs and digraphs. If the vertices of the leftmost figure are labelled {1, 2, 3, 4} in clockwise order from the topmost vertex, then the edge set of that graph is {{1, 2}, {2, 3}, {3, 4}, {4, 2}}. Under the same labelling, the multigraph in Figure 1(b) would have edge multiset {{1, 2}, {2, 3}, {3, 4}, {4, 2}, {4, 2}, {3, 3}}. Finally, the digraph in Figure 1(c) would have arc set {(2, 1), (2, 3), (3, 4), (4, 2)}. 1
2 4.1 Graph theoretic terminology In the next few short sections, we describe the notation that will be used in the remainder of the notes to refer to various properties of graphs Degrees and Neighbourhoods The neighborhood of v in G is denoted Γ G (v), and is the set of vertices of G which are adjacent to v. The degree of a vertex v in a graph G is the number of edges incident with v, and is denoted d G (v), or just d(v) for short. We write δ(g) and (G) for the smallest and largest degree of a vertex in G, respectively. For example, if G is the graph in Figure 1(a), then δ(g) = 1 and (G) = 3. The neighborhood of a set X V (G), which we denote by Γ(X), is the set of vertices of V (G)\X with at least one neighbor in X. For sets X, Y V (G), we write e(x, Y ) for the set of edges of G with one endpoint in X and the other endpoint in Y. An important fact involving the degrees of a graph G, which we will use on numerous occasions, is the handshaking lemma: Lemma 1 For any graph G = (V, E), d G (v) = 2 E. v V Proof When we add up the degrees of vertices of G, every edge of G is counted twice, once for each of its endpoints. An important consequence of the handshaking lemma is that the number of vertices of odd degree in any graph must be even otherwise the sum on the left above would be odd. So, for example, if you try to draw a graph on five vertices in which every vertex has degree three, you would fail. We refer to graphs where all the degrees are the same, say all degrees are equal to r, as r-regular graphs. The handshaking lemma gives an easy way to count the number of edges in a graph: it is just half the sum of the degrees of the vertices Basic classes of graphs Let K n denote the complete graph on n vertices: this is the graph consisting of all ( ) n 2 possible edges on n vertices. A bipartite graph is a graph G such that V (G) can be partitioned into two sets A and B such that each edge of G has one endpoint in A and one endpoint in B. The sets A and B are called the parts of G. Let K s,t denote 2
3 the complete bipartite graph with s vertices in one part and t vertices in the other: this graph consists of all st possible edges between a set of size s and a set of size t. We refer to K 1,t as a star. The n-dimensional cube, or n-cube, is the graph whose vertex set is the set of binary strings of length n, and whose edge set consists of pairs of strings differing in one position. A tree is a connected graph without cycles, and a forest is a vertex-disjoint union of trees and isolated vertices. The vertices of degree one in a tree or forest are often called leaves. A cycle is a connected graph all of whose vertices have degree two, and a path is a tree with exactly two vertices of degree one. Those vertices are sometimes called the endpoints of the paths, and the other vertices of the path are called internal vertices. Two paths are internally disjoint if they share no internal vertices. The number of edges in K n is ( n 2), and by the handshaking lemma, we can verify this: every vertex in K n is joined to all n 1 other vertices, so every vertex has degree n 1. By the handshaking lemma E(K n ) = 1 d(v) = 1 (n 1) = 1 ( ) n n(n 1) =. 2 v V v V For the n-dimensional cube, Q n, there are 2 n vertices. Each vertex v is joined to n other vertices namely flip one position in the string v to get strings adjacent to v, and there are n possible positions to do a flip. So every vertex of the n-cube has degree n, and E(Q n ) = 1 d(v) = 1 n = n n = n2 n 1. v V v V Subgraphs If H and G are graphs and V (H) V (G) and E(H) E(G), then H is called a subgraph of G. To denote that H is a subgraph of G, we write H G. If in addition V (H) = V (G) then H is called a spanning subgraph of G. If G = (V, E) is a graph and X V and L E, then G X denotes the graph whose vertex set is V \X and whose edge set consists of all edges disjoint from X. Also G L denotes the graph (V, E\L). In the case that X = {x}, we write G x instead of G X and if L = {e} we write G e instead of G {e}. For example, if G is the graph in Figure 1(a) and x is the vertex of degree three in Figure 1(a), and y is the vertex of degree one, then G x consists of a disjoint union of K 1 and K 2, and G y is K 3. 3
4 4.1.4 Walks and Paths A walk in a graph G = (V, E) is an alternating sequence of vertices and edges, whose first and last elements are vertices, and such that each edge joins the vertices immediately preceding it and succeeding it in the sequence. For example, a{a, d}d{d, e}e{e, a}a{a, d}d is a walk in the graph in Figure 3. If the vertices of a walk are all distinct, then it is referred to as a path. The length of a walk is the number of steps taken in the walk. The first and last vertices of a walk are called end vertices and the remaining vertices are called internal vertices. For example, the walk given above is not a path, but a{a, d}d{d, e}e is a path. The walk above has length four. For convenience, we often omit the edges and list the vertices in order along the path, such as ade instead of a{a, d}d{d, e}e. If the endpoints of a walk are u and v, then we say the walk is a uv-walk. If the first and last vertex of the walk are the same, then the walk is referred to as a closed walk. A cycle is a closed walk with the same number of edges as vertices every vertex of a cycle has degree two. For example, in Figure 3, the different cycles are adea, abea, bcdeb, adeba, adcba, aedcba, and aebcda. d c e a b Figure 3 : Walks, paths and cycles. 4.2 Connected graphs A graph is connected if any pair of vertices in the graph are joined by at least one path. If a graph is not connected, we say it is disconnected. The components of a graph G = (V, E) are the maximal (i.e. with as many edges as possible) connected subgraphs of G. The graphs in Figure 1(a) and Figure 3 are connected, whereas the graph below has three components. 4
5 Figure 4 : Components. The simplest type of connected graph is a tree: by definition, a tree is a connected graph without cycles a connected acyclic graph. To describe the structure of trees, we define the notion of a bridge. A bridge of a graph G is an edge e E(G) such that G e has more components than G. For example, in Figure 1(a), the top edge e is a bridge since G has one component but G e has two components. It is easy to spot the bridges of a graph, using the following lemma: Lemma 2 An edge e E(G) is a bridge if and only if e is not contained in any cycle in G. Proof If e is contained in a cycle C of G, then C e is a path joining the ends of e. But that means G e is connected, so e could not have been a bridge. Since a tree has no cycles, every edge of a tree must be a bridge. We can now characterize which graphs are trees in a few ways. Proposition 3 Each of the following is equivalent to a graph G being a tree: 1. The graph G is connected and acyclic. 2. The graph G is connected and every edge of G is a bridge. 3. The graph G is connected and has V (G) 1 edges. Proof Clearly Proposition 3.1 is the definition of G being a tree. Since a connected graph is acyclic if and only if every edge of the graph is a bridge, by the last lemma, Proposition 3.1 and Proposition 3.2 are equivalent. We proved Proposition 3.1 implies Proposition 3.3 by strong induction on the number of vertices in the tree, so it remains to show that Proposition 3.3 implies Proposition 3.1. To see that, if G is connected with V (G) 1 edges, we remove an edge of any cycle and that does not disconnect G, by Lemma 2. We continue removing edges of G in cycles until all the cycles are gone. But then the remaining graph T is connected and acyclic, so must be a tree. 5
6 Since it has V (G) vertices, we know it must have V (G) 1 edges. But G itself has V (G) 1 edges, so G = T. The last part of the proof of this proposition is important. It says that in any connected graph G, while there is a cycle, pick an edge of the cycle and remove it. By Lemma 2, we did not disconnect the graph, so if we repeat this procedure we eventually obtain a spanning subgraph of G which is acyclic and connected a tree. We call this a spanning tree of the graph. In general, a graph has many spanning trees. Proposition 4 Any connected graph contains a spanning tree. The proof gives a fairly quick way to find a spanning tree of a graph: search for a cycle and remove an edge of the cycle, and repeat until there are no cycles left. 4.3 Block Decomposition We just gave three equivalent characterizations of trees. In general, we would like to describe how to build connected graphs. The main result will be the block decomposition theorem. We require some definitions. A cut vertex of a graph G is a vertex G such that G x is disconnected. A block of a graph is a maximal connected subgraph with no cut vertex a subgraph with as many edges as possible and no cut vertex. So a block is either K 2 or is a graph which contains a cycle. For example in a tree, every block is K 2. The block decomposition of a graph is just the set of all the blocks of the graph. An example of a block decomposition is shown below. Figure 5 : Blocks. 6
7 In the picture, there are fourteen blocks. Seven of the blocks are K 2, four of the blocks are triangles, one of the blocks is K 5, and there are two other blocks. The block decomposition theorem says that block decompositions of graphs have a treelike structure. To make this precise, given a graph G, we form a new graph B where the vertices of B consist of all cut vertices of G and also all blocks of G, and where a block is joined to all cut vertices of G contained in it. An example of this graph is shown below for the figure above: Figure 6 : The graph B. In the figure, the black vertices represent cut vertices of G, and the grey vertices represent blocks of G. Here is the block decomposition theorem: Theorem 5 Let G be a connected graph. Then B is a tree. Proof By adding edges inside the blocks of G, we do not change B, so we can assume every block of G is a complete graph. Since G is connected, clearly B is connected too. Now we show B has no cycles. The vertices of a cycle C B are alternately blocks of G and cut vertices of G, by definition of B. This is shown in the figure below: 7
8 B v B B v Figure 7 : Cycle in B. In the figure, the blocks are shown as grey dots and labelled B and the cut vertices are black dots labelled v. Let the cut vertices of G in order along C be v 0, v 1,..., v k, v 0. Then v 0 v 1 v 2... v k v 0 is a cycle C G. If B C, then B C is a subgraph of G consisting of the complete graph B together with the cycle C containing an edge of B and at least one edge not in B. Therefore B C has no cut vertex, and must be a block of G. However, this contradicts the definition that B is block. Using this lemma, we give a first example of a structure theorem in graph theory. Define a theta graph to be any graph consisting of the union of three internally disjoint paths between two points. Proposition 6 Let G be a connected graph containing no theta graph. Then every block of G is a cycle or K 2 and G is a tree of cycles and K 2 s, as shown in Figure 8 below. Proof Let B be a block of G. If B K 2, then B contains a cycle, C. If B C, then there is a path P in B such that P C is a theta graph. Therefore B = K 2 or B is a cycle. We know by the last result that G is then a tree of cycles and K 2 s. 8
9 Figure 8 : Tree of cycles and K 2 s 4.4 k-connected graphs A cut of a connected graph G is a set of vertices or edges whose removal from G gives a disconnected graph. A graph G is k-connected if every cut of G has size at least k, and k-edge-connected if every edge cut of G has size at least k. So far we have characterized trees and connected graphs using block decompositions. The situation for k-connected graphs is more complicated when k 2. Fortunately, Menger s Theorems give a necessary and sufficient condition for a graph to be k-connected or k-edge-connected. Let u and v be vertices in a graph G, and let P and Q be uv-paths. Then P and Q are internally disjoint if the only vertices they have in common are u and v. A uv-separator is a set W V (G)\{u, v} such that u and v are in distinct components of G W. Finally, let κ(u, v) denote the minimum size of a uv-separator in G. Theorem 7 (Menger s Theorem Vertex Form) Let u and v be non-adjacent vertices in a graph G. Then the maximum number of pairwise vertex disjoint uv-paths in G is κ(u, v). In particular, a graph is k-connected if and only if each pair of its vertices is connected by at least k pairwise internally disjoint paths. It is important to note that condition that u and v are not adjacent in this theorem. Clearly, there is no uv-separator if u and v are adjacent. The vertex-form of Menger s Theorem has an edge-form. A set L E(G) is a uv-edge-separator if u and v are in different components of G L. Let λ(u, v) denote the minimum size of a uv-edgeseparator in G. 9
10 Theorem 8 (Menger s Theorem Edge Form) Let u and v be any vertices in a graph G. Then λ(u, v) equals the maximum number of pairwise edge-disjoint uv-paths in G. In particular, a graph is l-edge-connected if and only if each pair of its vertices is connected by at least l pairwise edge-disjoint paths. We will see how these theorems follow readily from the max-flow min-cut theorem. Note that it is possible to prove these theorems directly by structural arguments, but this is beyond the scope of this course. It is an interesting consequence of Menger s Theorem that if G is a graph with minimum degree δ(g), and κ(g) and λ(g) are the sizes of the smallest vertex cut and edge cut of G respectively, then κ(g) λ(g) δ(g). This is intuitively true in the sense that we can do more damage to a graph by removing vertices than edges, so we expect κ(g) λ(g). We prove this result as follows Corollary 9 Let G be any graph which is not a complete graph. Then κ(g) λ(g) δ(g). Proof To see that λ(g) δ(g), note that if we remove all the edges around a vertex of minimum degree, we disconnected G and therefore that set of δ(g) edges is an edge cut of G. Since λ(g) is the smallest possible size of an edge cut of G, we get λ(g) δ(g). To prove κ(g) λ(g), we use the two versions of Menger s Theorem given above. By definition, κ(g) = min{κ(u, v) : u, v V (G) are not adjacent}. Similarly λ(g) = min{λ(u, v) : u, v V (G)}. Now by Menger s Theorems, if u and v are not adjacent, then κ(u, v) λ(u, v) since a set of internally disjoint uv-paths is also a set of edge-disjoint uv-paths. If u and v are adjacent e = {u, v} E(G) then we remove the edge between them. Clearly that reduces λ(u, v) by exactly one, and also reduces the maximum number of edge-disjoint uv-paths by one. So we get κ(g e) λ(g) 1. However, κ(g e) κ(g) 1 so we get κ(g) 1 λ(g) 1 and the proof is complete: κ(g) λ(g). 10
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# Pedal Triangles
November 23, 2010
This investigation looks at some properties of pedal triangles and makes use of Geometer SketchPad.
## 1. The Pedal Triangle
A pedal triangle is defined as the projection of some point onto a triangle.
Here is the definition on Wikipedia.
Here is the definition on Wolfram Mathworld.
This is what it looks like:
Figure 1: Pedal Tiangle of point P on triangle ABC
We find the pedal triangle by projecting point P onto each side of the triangle. Projection is done by constructing a perpendicular line onto some side, passing through P:
This process can be repeated for this pedal triangle again in the same way. Our problem look at repeating this process until we have a pedal triangle of a pedal triangle of a pedal triangle, known as the 3rd pedal triangle of ABC.
## 2. The Conjecture and its Construction
We want to show:
The third pedal triangle is similar to the original triangle ABC
The proof is in showing that each respective angle of the two triangles are equal eg.
Figure 1: Two similar triangles
The construction is given in the video below, it makes use of a tool that draws a pedal triangle once its been given a point and the three vertices of the triangle. The tool can be downloaded here.
So we need to show that triangle LJK is similar to ABC.
## 3. Building Blocks of the Proof
Here are some important building blocks (B) we will be using for the proof:
• B1: The hypotenuse of right-angled triangle is the diameter of its circumcircle
• B2: All angles subtended by the same arc, are equal.
## 4. Proof of the Conjecture
Figure 4.1: The problem
To show that triangle LJK is similar to ABC.
This is how we go about it:
Figure 4.2: ABC and EFG
Here in Figure 4.2 we simplify the problem by only showing the construction of the first two triangles. Additionally, we construct PC, PF and PD. Recall that PF is perpendicular to BC and PD to AC by construction (because that is the definition of the projection of P onto BC and AC).
Figure 4.3 Using B1
We use the building block B1 in constructing this circle: PC is the diagonal of the circle though points P, F and C. D also lies on this line by sharing PC as a base with PFC and having right angle PDC.
Figure 4.4: Using B2
By using the building block B2 we know that angle PFD equals angle PCD by the subtended arc of PD.
Figure 4.5: Using B2 again
Using B2 again, we also show angle FCP equal to angle FDP.
Figure 4.6: Naming the angles
We split angle ACB up into two parts and call them lowercase a and b respectively.
Figure 4.7: Naming all angles
In the same way as figure 4.6, we split angle ABC and angle CAB into two parts and apply the same reasoning to get angles a though f into the first pedal triangle.
Figure 4.8 The first pedal triangle
Let's have a closer look at figure the first pedal triangle.
Figure 4.9: Only the first pedal triangle.
We will go to work on this triangle using the same reasoning as with the main triangle.
Figure 4.10: The second pedal triangle
This is the second pedal triangle constructed inside the first.
Figure 4.11: Using B1 and B2
Figure 4.12: B1 and B2 applied to all angles.
Figure 4.13: The second pedal triangle
4.14: The third pedal triangle inside the first
Figure 4.15 Using B1 and B2
Figure 4.16: Using B1 and B2 for all angles
Figure 4.17: The third pedal triangle
Finally, by comparing with Figure 4.7, we have
• angle JKL = a + b = angle ACB
• angle KJL = c + d = angle CBA
• angle JLK = e + f = angle BAC
Therefore:
Triangle JKL is similar to triangle CBA.
In closing
Remember that when we work with proofs and conjectures, we try to find all applicable building blocks that can be applied to the problem and then, systematically, sift out as much information as possible using the given facts and also the comparisons that can be made using what we know.
There are usually many ways to solve a given problem and reaching dead-ends are sometimes vital in reaching a solution.
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# Interactive Voting - 1995 maths paper b
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### Interactive Voting - 1995 maths paper b
1. 1. 1995 MATHS PAPER B
2. 2. 1995 Mathematics Paper B Input your name and press send. Next Page
3. 3. 1a. Here is a table of the number of stamps used each day in school. Next Page 1995 Mathematics Paper B How many 25p stamps did the school use on Wednesday?
4. 4. 1. Here is a table of the number of stamps used each day in school. Next Page 1995 Mathematics Paper B This is a graph of one kind of stamp they used. 1b. What was the value of the stamp In pence?
5. 5. Q2. In this block, the pairs of numbers joined together have a difference of 7. Choose one more pair of numbers which has a difference of 7 . 1995 Mathematics Paper B Next Page Answer key
6. 6. Q2. In this block, the pairs of numbers joined together have a difference of 7. Choose one more pair of numbers which has a difference of 7 . 1995 Mathematics Paper B Next Page
7. 7. Q3. Numbers are missing on four of these calculator buttons. Write in numbers to make the answer 47. 1995 Mathematics Paper B Next Page
8. 8. Q4a. On sports day children get points for how far they jump. 1995 Mathematics Paper B Next Page Teaching Slide Joe jumped 138cm. How many points does he get? Sam said, “I jumped 1.5 metres. I get 4 points”. Give a reason why Sam is correct.
9. 9. Q4. On sports day children get points for how far they jump. 1995 Mathematics Paper B Next Page Joe jumped 138cm. How many points does he get? 80cm 100cm 120cm 140cm 160cm 180cm 138 cm 1 point 2 points 3 points 4 points 5 points
10. 10. Q4b. On sports day children get points for how far they jump. 1995 Mathematics Paper B Next Page Each child puts a cross on a line to show how far they jumped. Sam puts her cross at 1.5 metres. Lynn jumps 1.14 metres. Put a cross on the line for Lynn’s jump.
11. 11. 1995 Mathematics Paper B Q5a. Beth sells comics. Ali buys one Magna and one Mindy. He gives Beth 50p. How many pennies change will he get? 25p + 19p = 25p +19p = 44p 50p – 44p 06p Next Page
12. 12. 1995 Mathematics Paper B Q5b 215 people go to the sale. It costs 15p each to go in. Work out how much they pay altogether – answer in pence. 2 marks for the correct answer. 1 mark for working out. Next Page XL File
13. 13. <ul><li>Q6 </li></ul><ul><li>Raj thinks of a number. He says </li></ul><ul><li>“ If I subtract 14 from it, I get 29.” </li></ul><ul><li>What is Raj’s number? </li></ul>1995 Mathematics Paper B Next Page - Teaching slide
14. 14. <ul><li>Q6 </li></ul><ul><li>Raj thinks of a number. He says </li></ul><ul><li>“ If I subtract 14 from it, I get 29.” </li></ul><ul><li>What is Raj’s number? </li></ul>1995 Mathematics Paper B Next Page K eep I t S imple S tupid! ? – 14 = 29 ? – 3 = 2 What is ? What did you do with 3 and 2 to know that ? = 5 You went 3 + 2 = 5 so… 14 + 29 = ? 14 + 29 = 43
15. 15. <ul><li>Q7 Which symbol in the circle +, –, *, or ÷ would mak e the calculation correct? </li></ul>1995 Mathematics Paper B Next Page - Teaching slide <ul><ul><li>+ </li></ul></ul><ul><ul><li>- </li></ul></ul><ul><ul><li>* </li></ul></ul><ul><ul><li>÷ </li></ul></ul>
16. 16. <ul><li>Q7 </li></ul>1995 Mathematics Paper B Next Page K eep I t S imple S tupid! ? * 5 = 30 What is ? (18 0 3) It must be 6. So 18 0 3 = 6 18 + 3 = 21 18 – 3 = 15 18 * 3 = plenty 18 ÷ 3 = 6 Missing operation = ÷
17. 17. Q8a. Here are two shapes made with centimetre squares. Each shape has 5 squares. Write ONE other thing which is the same about the two shapes. 1995 Mathematics Paper B Next Page Answer Key
18. 18. Q8. Here are two shapes made with centimetre squares. Each shape has 5 squares. Write ONE other thing which is the same about the two shapes. 1995 Mathematics Paper B Next Page <ul><li>(a) A mathematical criterion such as: </li></ul><ul><ul><li>· “They are symmetrical.” </li></ul></ul><ul><ul><li>· “Each has three squares across.” </li></ul></ul><ul><ul><li>· “Both have the same perimeter.” </li></ul></ul><ul><ul><li>· “They have 4 joined-on lines.” </li></ul></ul><ul><ul><li>· “They have the same area.” </li></ul></ul><ul><ul><ul><ul><li>Do not accept criteria which are implicit in the fact that the shapes are made from cm squares, eg: </li></ul></ul></ul></ul><ul><ul><ul><ul><ul><li>· “They have straight edges.” </li></ul></ul></ul></ul></ul>
19. 19. Q8b. Here are more shapes made with centimetre squares. 1995 Mathematics Paper B Next Page Which shape has a perimeter of 10cm?
20. 20. 1995 Mathematics Paper B Next Page Q9a. Children in Year 6 make number patterns. This group uses the rule ‘2 less than’. What is the missing number?
21. 21. 1995 Mathematics Paper B Next Page Teaching slide Q9b. This group uses the rule ‘divide by 3’. What is the missing number?
22. 22. <ul><li>Q9 </li></ul>1995 Mathematics Paper B Next Page K eep I t S imple S tupid! ? ÷ 3 = 54 Simple version: 6 ÷ 3 = 2 What would you do with 3 and 2 to make 6? 3 * 2 = 6 Switching things around then 3 * 54 = ? 3 * 54 = 162 Check 162 ÷ 3 = 54
23. 23. Q10 On the grid, draw the reflection of the shape in the mirror line. 1995 Mathematics Paper B Next Page Teaching slide
24. 24. Q10 On the grid, draw the reflection of the shape in the mirror line. 1995 Mathematics Paper B Next Page
25. 25. Q11 What are the missing numbers? 1995 Mathematics Paper B Next Page What is this number?
26. 26. Q11 What are the missing numbers? 1995 Mathematics Paper B Next Page What is this number? XL File
27. 27. Q12a Here is a map of an island. Estimate the area of the wood. 1995 Mathematics Paper B Next Page Write on the board
28. 28. Q12b A boat can safely carry 145 kilograms. 1995 Mathematics Paper B Next Page Answer key. Work out if the boat can safely carry Mary and Bob
29. 29. Q12b A boat can safely carry 145 kilograms. 1995 Mathematics Paper B Next Page <ul><li>(b) Correct addition of 59.5 and 80.3 to make 139.8 </li></ul><ul><li>OR </li></ul><ul><li>Estimation such as 60 + 81 <145. </li></ul><ul><ul><li>No mark is awarded for drawing a conclusion. The mark is awarded for correct calculation even if conclusion is wrong. </li></ul></ul>
30. 30. Q13a 427 children visit a castle. They go in groups of 15. One group has less than 15. Every group of children has one adult with them. How many adults will need to go? 1995 Mathematics Paper B Next Page
31. 31. Q13b Mr Todd buys 7 drinks at 48p each and 8 drinks at 52p each. What is the total cost of the 15 drinks? You must show your working. Answer like this: 3.21 (2 Marks) 1995 Mathematics Paper B Next Page XL File
32. 32. Q14 Rob runs 100 metres ten times. These are his times in seconds. What is his mean (average) time? 1995 Mathematics Paper B Next Page XL File
33. 33. Q15. Write in the missing digit. The answer does not have a remainder. 1995 Mathematics Paper B Next Page
34. 34. Q16. Here is a triangular box. 1995 Mathematics Paper B Next Page Answer Key Here is part of the net of the box, but two of its faces are missing. Draw accurately on the board , full size, ONE of the missing faces on the diagram below. You can use a ruler and protractor (angle measurer).
35. 35. Q16. Here is a triangular box. 1995 Mathematics Paper B Next Page (a) Award ONE mark for correct position of triangle as shown in one of the diagrams below. (b) Award ONE mark for accurate drawing of one triangle with right angle (90° ± 2.5°) AND length of lines as indicated ± 2mm. No marks awarded for triangles not attached to main stem.
36. 36. Q17a. Ann makes a pattern of L shapes with sticks. Ann says : “I find the number of sticks for a shape by first multiplying the shape-number by 4, then adding 3”. Work out the number of sticks for the shape that has shape-number 10. 1995 Mathematics Paper B Next Page
37. 37. Q17b. Ann makes a pattern of L shapes with sticks. Ann says : “I find the number of sticks for a shape by first multiplying the shape-number by 4, then adding 3”. 1995 Mathematics Paper B Next Page Ann uses 59 sticks to make another L shape in this pattern. What is its shape-number? XL File
38. 38. Q17c. Here is Ann’s rule again: “ I find the number of sticks for a shape by first multiplying the shape-number by 4, then adding 3”. Write a formula to work out the number of sticks for any L shape. Use S for the number of sticks and N for the shape-number. Ann makes a pattern of L shapes with sticks. 1995 Mathematics Paper B Next Page Answer Key S =
39. 39. Q17c. Here is Ann’s rule again: “ I find the number of sticks for a shape by first multiplying the shape-number by 4, then adding 3”. Write a formula to work out the number of sticks for any L shape. Use S for the number of sticks and N for the shape-number. Ann makes a pattern of L shapes with sticks. 1995 Mathematics Paper B Next Page <ul><li>(c) Award TWO marks for expressions such as: up to 2 </li></ul><ul><ul><li>· S = 4N + 3 </li></ul></ul><ul><ul><li>· S = 3 + 4N </li></ul></ul><ul><ul><li>· S = N + N + N + N + 3 </li></ul></ul><ul><li>If the answer is incorrect, award ONE mark for evidence of multiplying N by 4 in the expression, eg: </li></ul><ul><ul><li>· 4N </li></ul></ul><ul><ul><li>· 4 × N </li></ul></ul><ul><ul><li>· N.4 </li></ul></ul><ul><ul><li>· N + N + N + N </li></ul></ul><ul><li>OR award ONE mark for evidence of adding 3 in the expression, eg: </li></ul><ul><ul><li>· N + 3 </li></ul></ul><ul><ul><ul><ul><li>Do not accept S = × 4 + 3 = N up to 2 </li></ul></ul></ul></ul>
40. 40. 1995 Mathematics Paper B Next Page Answer Key Q18a. Some children work out how much money two shopkeepers get from selling fruit. They use pie charts to show this. Mrs Binns gets £350 selling bananas. Estimate how much she gets selling oranges. XL File
41. 41. 1995 Mathematics Paper B Next Page Q18a. Some children work out how much money two shopkeepers get from selling fruit. They use pie charts to show this. Mrs Binns gets £350 selling bananas. Estimate how much she gets selling oranges. (a) Award ONE mark for an answer in the range £85 to £125, inclusive. XL File
42. 42. 1995 Mathematics Paper B End of test Q18b. Mrs Binns gets a total of £1000 and Mr Adams gets a total of £800. How much more does Mrs Binns get than Mr Adams for selling peaches ? XL File
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# Bernstein–Vazirani problem in book as exercise
I´ve solved the Exercise 7.1.1 (Bernstein–Vazirani problem) of the book "An introduction to quantum computing" (Mosca et altri). The problem is the following:
Show how to find $a \in Z_2^n$ given one application of a black box that maps $|x\rangle|b\rangle \to |x\rangle |b \oplus x · a\rangle$ for some $b\in \{0, 1\}$.
I´d say we can do it like this:
• First i go from $|0|0\rangle \to \sum_{i \in \{0,1\}^n}|i\rangle| + \rangle$ using QFT and Hadamard
• Then I apply the oracle: $$\sum_{i \in \{0,1\}^n}(-1)^{(i,a)} |i\rangle| + \rangle$$
• Then I read the pase with an Hadamard (since we are in $Z_2^n$ our QFT is an Hadamard) $$|a\rangle |+ \rangle$$
I think is correct. Do you agree?
This is not correct: you need to use the state $|-\rangle=(|0\rangle-|1\rangle)/\sqrt{2}$ instead of $|+\rangle$.
The important thing is that you've missed showing how the black box map that you've stated gives the oracle output that you've stated. To see this, apply the map on $$|x\rangle|+\rangle\mapsto|x\rangle(|0\oplus x\cdot a\rangle+|1\oplus x\cdot a\rangle)/\sqrt{2}=|x\rangle(|0\rangle+|1\rangle)/\sqrt{2}.$$ When the $|+\rangle$ state is there, you get no phase. Meanwhile, with the $|-\rangle$ state, $$|x\rangle|-\rangle\mapsto|x\rangle(|0\oplus x\cdot a\rangle-|1\oplus x\cdot a\rangle)/\sqrt{2}=\left\{\begin{array}{cc} |x\rangle(|0\rangle-|1\rangle)/\sqrt{2} & x\cdot a=0 \\ |x\rangle(|1\rangle-|0\rangle)/\sqrt{2} & x\cdot a=1\end{array}\right..$$ This can simply be written as $(-1)^{x\cdot a}|x\rangle|-\rangle$.
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# Regression Wisdom - PowerPoint PPT Presentation
Regression Wisdom. Getting to Know Your Scatterplot and Residuals. Important Terms. Extrapolation (203) Outlier (205) Leverage (206) Influential Point (206) Lurking Variable (208). Residuals.
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## PowerPoint Slideshow about 'Regression Wisdom' - kalkin
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### Regression Wisdom
Getting to Know Your Scatterplot and Residuals
• Extrapolation (203)
• Outlier (205)
• Leverage (206)
• Influential Point (206)
• Lurking Variable (208)
• Recall – Residuals are the difference between data values and the corresponding values predicted by the regression model
• Residual = Observed Value – Predicted Value
e =
(page 172)
• We want our plot of residuals to be boring
• It should have no structure, direction, shape, none of that stuff.
• When it does, there is something else going on in the data that explains the variation of the two variables.
- We can form subsets of the same population to try and achieve a better analysis of the data.
- Sometimes the easiest way to achieve this is to examine a plot or histogram of residuals
• You can perform regression analysis on each subset of the larger population, noting correlation and all appropriate summary statistics for each subset.
• Extrapolations require the very questionable assumption that nothing about the relationship between x and y changes even at extreme values of x and beyond
• Our Linear Model:
• Plug in a new x, it gives you a predicted
• But the farther the new x-value is from , the less trust we can place in the predicted y value.
• Once we venture into new x territory such a prediction is called an extrapolation
• If your x variable is Time, extrapolation becomes a prediction about the future!
• Example: Mid-1970s, oil cost \$17 a barrel in 2005 dollars
• But suddenly, within a few years, the price skyrocketed to over \$40 a barrel
• If you used this data for your model, you might be predicting oil prices today in the hundreds upon hundreds of dollars per barrel while if you had done your analysis before the spike in prices, you might still be predicting around 17\$ a barrel.
• Outliers can have big impacts on your fitted regression line.
• Points with large residuals always deserve special attention.
• A data point with an unusually large x-value from the mean is said to have high leverage
• High Leverage doesn’t mean the point changes the overall picture.
• If the point lines up with the pattern of other points, including it doesn’t change our estimate of the line
• But by sitting so far from it may strengthen the relationship, inflate the correlation and R-Squared
• A point is influentialif omitting it from the analysis gives a very different model
• Influence depends on both leverage and residual
• A case with high leverage whose y-value sits right on the line is not influential.
• Removing this point may not change the slope but may change R-Squared
• A point is influentialif omitting it from the analysis gives a very different model
• Influence depends on both leverage and residual
• A case with modest leverage but a very large residual can be influential.
• With enough leverage, the regression line can be pulled right to it. Then its highly influential but will have a small residual
• A point is influentialif omitting it from the analysis gives a very different model
• Influence depends on both leverage and residual
• The only thing to do is to do your analysis twice:
• Once with the point
• Once omitting the point
Does the unusual point have high-leverage, a large residual, and is it influential?
High Leverage
Not Influential
Small Residual
Not high leverage
Not influential
Large Residual
High Leverage
Influential
Not Large Residual
Lurking Variables, Causation and is it influential?
• No matter how strong the association
• No matter how large the value
• No matter how straight the line
• There is NO way to conclude from regression alone that one variable causes the other.
• There may always be a lurking variable that causes the apparent association
Lurking Variable Example and is it influential?
• The scatterplot shows the Life Expectancy of men and women in 41 different countries
• These values are plotted against the square root of Doctors per person in that country.
Lurking Variable Example and is it influential?
• There is a strong positive correlation,
• This confirms our expectation that more doctors per person improves healthcare, leading to longer lifetimes and greater life expectancy.
Lurking Variable Example and is it influential?
• Can we conclude though that doctors cause greater life expectancy? Perhaps, but increasing numbers of doctors and greater life expectancy may both be results of a larger change.
Lurking Variable Example and is it influential?
• Here is a similar looking scatterplot now comparing life expectancy to the square root of TVs per person.
• This is an even stronger association!
A Final Note and is it influential?
• Beware of scatterplots of statistics of summarized data.
• For example,
### Homework and is it influential?
Pg 214, #1, 3, 4, 8, 10
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# Resources tagged with: Compound transformations
Filter by: Content type:
Age range:
Challenge level:
### There are 35 results
Broad Topics > Transformations and constructions > Compound transformations
### Cutting Corners
##### Age 7 to 11 Challenge Level:
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Square Tangram
##### Age 7 to 11 Challenge Level:
This was a problem for our birthday website. Can you use four of these pieces to form a square? How about making a square with all five pieces?
### Midpoint Triangle
##### Age 7 to 11 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Cut and Make
##### Age 7 to 11 Challenge Level:
Cut a square of paper into three pieces as shown. Now,can you use the 3 pieces to make a large triangle, a parallelogram and the square again?
### Sorting Symmetries
##### Age 7 to 11 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
##### Age 7 to 11 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### Twice as Big?
##### Age 7 to 11 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### Flight of the Flibbins
##### Age 11 to 14 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### Counting Triangles
##### Age 11 to 14 Challenge Level:
Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether?
### Square to L
##### Age 7 to 11 Challenge Level:
Find a way to cut a 4 by 4 square into only two pieces, then rejoin the two pieces to make an L shape 6 units high.
### Transforming the Letters
##### Age 7 to 11 Challenge Level:
What happens to these capital letters when they are rotated through one half turn, or flipped sideways and from top to bottom?
### Screwed-up
##### Age 11 to 14 Challenge Level:
A cylindrical helix is just a spiral on a cylinder, like an ordinary spring or the thread on a bolt. If I turn a left-handed helix over (top to bottom) does it become a right handed helix?
### Squares, Squares and More Squares
##### Age 11 to 14 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Triangular Tantaliser
##### Age 11 to 14 Challenge Level:
Draw all the possible distinct triangles on a 4 x 4 dotty grid. Convince me that you have all possible triangles.
### Simplifying Transformations
##### Age 11 to 14 Challenge Level:
How many different transformations can you find made up from combinations of R, S and their inverses? Can you be sure that you have found them all?
### Paint Rollers for Frieze Patterns.
##### Age 11 to 16
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach.
### Combining Transformations
##### Age 11 to 14 Challenge Level:
Does changing the order of transformations always/sometimes/never produce the same transformation?
### Making Rectangles, Making Squares
##### Age 11 to 14 Challenge Level:
How many differently shaped rectangles can you build using these equilateral and isosceles triangles? Can you make a square?
### Transformation Game
##### Age 11 to 14 Challenge Level:
Why not challenge a friend to play this transformation game?
### Decoding Transformations
##### Age 11 to 14 Challenge Level:
See the effects of some combined transformations on a shape. Can you describe what the individual transformations do?
### Matching Frieze Patterns
##### Age 11 to 14 Challenge Level:
Sort the frieze patterns into seven pairs according to the way in which the motif is repeated.
### Who Is the Fairest of Them All ?
##### Age 11 to 14 Challenge Level:
Explore the effect of combining enlargements.
### Transformations Tables
##### Age 7 to 11 Challenge Level:
These grids are filled according to some rules - can you complete them?
### Going Places with Mathematicians
##### Age 7 to 14
This article looks at the importance in mathematics of representing places and spaces mathematics. Many famous mathematicians have spent time working on problems that involve moving and mapping. . . .
### ...on the Wall
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Mathematical Patchwork
##### Age 7 to 14
Jenny Murray describes the mathematical processes behind making patchwork in this article for students.
### Bow Tie
##### Age 11 to 14 Challenge Level:
Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.
### Mirror, Mirror...
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
### 2001 Spatial Oddity
##### Age 11 to 14 Challenge Level:
With one cut a piece of card 16 cm by 9 cm can be made into two pieces which can be rearranged to form a square 12 cm by 12 cm. Explain how this can be done.
### The Frieze Tree
##### Age 11 to 16
Patterns that repeat in a line are strangely interesting. How many types are there and how do you tell one type from another?
### Frieze Patterns in Cast Iron
##### Age 11 to 16
A gallery of beautiful photos of cast ironwork friezes in Australia with a mathematical discussion of the classification of frieze patterns.
### Maurits Cornelius Escher
##### Age 7 to 14
Have you ever noticed how mathematical ideas are often used in patterns that we see all around us? This article describes the life of Escher who was a passionate believer that maths and art can be. . . .
### Trees and Friezes
##### Age 11 to 14 Challenge Level:
This problem is based on the idea of building patterns using transformations.
### Friezes Using Logo
##### Age 11 to 14 Challenge Level:
Experimenting with variables and friezes.
### Grouping Transformations
##### Age 11 to 18
An introduction to groups using transformations, following on from the October 2006 Stage 3 problems.
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Question Video: Analysis of the Equilibrium of a Uniform Rod Placed Horizontally on Two Supports | Nagwa Question Video: Analysis of the Equilibrium of a Uniform Rod Placed Horizontally on Two Supports | Nagwa
# Question Video: Analysis of the Equilibrium of a Uniform Rod Placed Horizontally on Two Supports Mathematics • Third Year of Secondary School
## Join Nagwa Classes
A uniform rod π΄π΅ having a length of 1.3 m and weighing 147 N is resting in a horizontal position on two supports, where the support πΆ is at the end π΄ and π· is at a distance π₯ from the end π΅. Find the reaction of the support π
_(πΆ) and the distance π₯, given that π
_(πΆ) = (2/5)π
_(π·).
05:40
### Video Transcript
A uniform rod π΄π΅ having a length of 1.3 meters and weighing 147 newtons is resting in a horizontal position on two supports, where the support πΆ is at the end π΄ and π· is at a distance π₯ from the end π΅. Find the reaction of the support π
πΆ and the distance π₯, given that π
πΆ is equal to two-fifths π
π·.
Because the rod is resting, we know that it is in equilibrium, which means two conditions are met. The net force on the rod is zero, and the net moment of force about any point on the rod is also zero. Moments of force are always calculated about some arbitrary reference point. Letβs call the magnitude of the force of interest πΉ. And weβll call the distance between where the force is acting and the reference point π. If the line connecting where the force is acting and the reference point is perpendicular to the line of action of the force itself, then the moment of this force about the reference point takes on a specially simple form.
The magnitude of the moment of force in this situation is given by the magnitude of the force times the distance to the reference point. We specify that this is the magnitude because moments of force also have a sign. If we imagine that this orange line were actually a solid rod pivoting about the reference point, the force as drawn would be pulling it in a clockwise direction. If the force pointed in the opposite direction, then it would be pulling the rod in a counterclockwise direction. The sign of the moment of force is a convention that we choose, which tells us whether the force is pointing clockwise or counterclockwise. Usually we arbitrarily pick the counterclockwise direction to be positive, but as long as weβre consistent, it really doesnβt matter.
Now that we have several equations and conditions, letβs draw a diagram to organize the information that we are given. Here we have a rod with a length of 1.3 meters resting in a horizontal position on the supports πΆ and π·. The support πΆ is located at the end π΄ as specified in the statement, and π· is located somewhere in the middle a distance of π₯ from the end π΅. There are also several forces acting on the rod. The weight of the rod is 147 newtons. And because the rod is uniform, the weight acts at its midpoint. Each support also provides a reaction force pushing up on the rod to counteract its weight.
Using the variables defined in the question, weβll call the reaction force at the support π· π
π·. And weβll call the reaction force at the support πΆ π
πΆ, where π
πΆ is two-fifths of π
π·. Finally, to complete this diagram, weβll just include that the distance from point π΄ to the support π· is 1.3 meters minus π₯ and that the weight 147 newtons is acting half of 1.3 meters, which is 0.65 meters or 65 centimeters, from either end. All right, letβs now use our two equilibrium conditions to solve for the two quantities that weβre looking for, the force π
πΆ and the distance π₯.
To make sure the net force is zero, we add all of the forces pointing in one direction and subtract all of the forces pointing in the opposite direction. This works out especially well here because all of the forces are parallel. So we have two-fifths times π
π·, which is just π
πΆ expressed in terms of π
π·, plus π
π· minus 147 equals zero. Two-fifths π
π· plus π
π· is seven-fifths π
π·. Adding 147 to both sides gives us seven-fifths π
π· equals 147. We can solve for π
π· by multiplying both sides by five-sevenths. We find that π
π· is 105 newtons. Then either by using the formula given to us in the statement or the fact that π
π· plus π
πΆ is the weight of 147 newtons, we find that π
πΆ is 42 newtons.
Now that we have found π
πΆ, one of the quantities that weβre looking for, we can use our condition for the net moment of force to find π₯, the other quantity that weβre looking for. Letβs choose our reference point to be the end of the rod at π΅. The reference point is on the rod. And all of the forces are acting on the rod. So the line connecting the reference point to where the forces are acting is just the rod itself. Since π
π·, the weight, and π
πΆ are all perpendicular to the rod, we can use our equation for the magnitude of the moments, force times distance. For the signs of the moments, observe that π
π· and π
πΆ point in the same orientation, while the weight points in the opposite orientation. So whatever sign we choose, the moment from π
π· and from π
πΆ will have the same sign and the moment from the weight will have the opposite sign.
Letβs arbitrarily choose the sign for the moments of π
π· and π
πΆ to be positive. The distance from π
π· to π΅ is π₯, so its moment is 105 newtons times π₯. The distance from π
πΆ to π΅ is 1.3 meters, so its moment is 42 newtons times 1.3 meters. Finally, the distance from the weight of the rod to π΅ is the same as the distance from the weight of the rod to π΄, which is 0.65 meters. So the moment of the weight is negative 147 newtons times 0.65 meters about π΅.
And as with the net force, this whole expression must evaluate to zero. If we plug all of the numerical parts into a calculator, we get negative 40.95. If we add 40.95 to both sides, we get 105π₯ equals 40.95, which we can then solve for π₯ by dividing both sides by 105. This gives us that π₯ equals 0.39 meters, or equivalently π₯ is 39 centimeters. And with that, weβve found both π
πΆ and π₯, the true quantities that we were looking for.
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# How To Find The Height Of A Rhombus?
A rhombus is a quadrilateral whose four sides are all equal. Its diagonals are perpendicular bisectors, and its opposite angles are equal. The method used to calculate the height of a rhombus depends on the information we have about it. For instance, given the length of its side and its area, we can calculate its height using the formula (area)/(side).
To determine how to calculate the height of a rhombus, we must first define it.
A rhombus is a four-sided plane figure (quadrilateral) that has the following properties:
-All its sides are equal;
-Its opposite sides are parallel to each other;
-Its opposite angles are equal;
-Its diagonals bisect each other at right angles. These diagonals are also angle bisectors;
-Its diagonals divide it into 4 congruent (same shape and size) triangles.
The method used to determine the height of a rhombus depends on the dimensions given.
1) Given the side(s) of a rhombus and its area:
The formula for finding the area of a rhombus is side \times perpendicular height.
Thus, the height of the rhombus in this case is \frac{area}{side} .
Example: Find the height of a rhombus whose area is 20 cm^{2} and side is 4 cm.
Height = \frac{area}{side} = \frac{20}{4} = 5 cm.
2) Given the diagonals of a rhombus:
If the diagonals d_{1} and d_{2} of a rhombus are given, then using the following two properties of rhombi, namely, the diagonals bisect each other at right angles and the diagonals divide the rhombus into 4 congruent triangles, we can find the area of one triangle as follows:
Area of 1 triangle = \frac{1}{2} \times bh = \frac{1}{2} \times \frac{d_{1}}{2} \times \frac{d_{2}}{2} =\frac{d_{1}d_{2}}{8}
And area of the 4 triangles in the rhombus is \frac{d_{1}d_{2}}{8} \times 4 = \frac{d_{1}d_{2}}{2} .
Having found the area of the rhombus, we can next determine the side of the rhombus using Pythagoras Theorem.
The base of one of the 4 triangles formed by the diagonals is \frac{d_{1}}{2} and the height is \frac{d_{2}}{2} . Therefore, the hypotenuse, which forms the side of the rhombus, is \sqrt{(\frac{d_{1}}{2})^{2} +(\frac{d_{2}}{2})^{2}} .
Finally, we know that the area of the rhombus can also be calculated using the formula side x height. Here, we have already found the area and the side. Therefore, the height of the rhombus is \frac{area}{side} , where area = \frac{d_{1}d_{2}}{2} and side = \sqrt{(\frac{d_{1}}{2})^{2} +(\frac{d_{2}}{2})^{2}} .
Example: Find the height of a rhombus whose diagonals are 90 mm and 400 mm.
The area of the rhombus is \frac{d_{1}d_{2}}{2} = \frac{90 \times 400}{2} = 18000 mm^{2} .
One of the four congruent right-angled triangles in the rhombus has a height \frac{d_{1}}{2} = 45 mm and a base \frac{d_{2}}{2} = 200 mm. Therefore, the height of this triangle, which is also the side of the rhombus, is \sqrt{45^{2} + 200^{2}} = 205 mm.
Thus, using the formula area of rhombus = side x height, we obtain height = \frac{area}{side} = \frac{18000}{205} = 87.805 mm (to 3 decimal places).
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# Math IA
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### Math IA
1. 1. Michael Pollock IB Math SL Period 7 March 31st 2011Lacsaps Fractions 1 1 3 1 1 2 6 6 1 1 4 4 10 10 10 1 1 7 6 7 15 15 15 15 1 1 11 9 9 11Finding the Numerator EquationThere are many patterns to be found in Lascaps Fractions. In order to meet our first objective offinding the numerator of the sixth row, I only viewed the numerators. This helped to single out patternswhich would eventually lead to an equation. _ _ _ 3 _ _ 6 6 _ _ 10 10 10 _ _ 15 15 15 15 _I took out the sides because they are simply ones, their numerators could be edited to match the rest oftheir respective rows, however their denominators would have to be changed as well.I put the numerators on a chart next to the row number just to get a clearer look at the possible patternsthat could be occurring:Row Number Numerator Value2 33 64 105 15
2. 2. There is a clear pattern here. Firstly, all numerators in a single row are identical. This helps simplify ourequation as we only need the row number and not the position within said row.To get from the numerator in row two, which is three, to the numerator in row three, which is six, weadd three to the numerator of row two. To get from the numerator in row three (6) to row four (10), weadd four, and on it goes. We can observe that the amount we add is increased by one each row. To findthe numerator of row n, we multiply n by half of n plus one, or as follows: n1 N n=n 2This was derived with a bit of logic and a bit of guess and check. I knew that I would have to multiplyn by some version of itself. To find this value I simply experimented with different values andeventually found one that worked with all the rows I was given. I used this to solve the next row for anumerator of 21, and when I checked by adding 6 to 15 (as in the previously found pattern), I found thesame value.When this equation is expanded we get this: n2 +n N ( n)= 2The expansion is used in order to graph. If we graph this we would be comparing the row number n tothe numerator number N. When we compare these we get this graph depicted below, with the rownumber on the x-axis and the numerator on the y-axis:As the row number increases, the numerator increases in greater increments. We observe exactly this inour triangle. For each subsequent row, the numerator increases by one more than it did for the previousrow. This shows that our equation matches the relationship accurately, even when it comes to highernumbers, without having to test those higher numbers.
3. 3. Using our equations we can easily find the numerators for the sixth and seventh rows:Numerator of Row Six 21Numerator of Row Seven 28And we can check this work easily by using the same pattern we found to begin with. The numerator ofthe fifth row (15) plus six is in fact equal to 21, just as our equation told us. And the numerator of thesixth plus seven is equal to 28, which is what our equation told us row number seven would be.
4. 4. Finding the Denominator EquationNow that we can find the value for the numerator, we need to find the value of the denominator. I willonce again draw the triangle, this time only showing denominators: _ _ _ 2 _ _ 4 4 _ _ 7 6 7 _ _ 11 9 9 11 _In this case we must take into account the row number and the position within the row to find ourvalues. This means that we need to add another variable to our equation.Once again though, there was a pattern that I observed. I will show the numerator and denominator ofthe first fraction in each row, which is located at what well call position one from now on:Row Number Numerator Value Denominator Value2 3 23 6 44 10 75 15 11We can view the difference between the numerator and denominator of each row in position one. Inrow two, they have a difference of one. In row three the difference is two. In row four, the difference isthree and so on. This shows me that there is some sort of relationship between the numerator anddenominator. I can then assume that the equation for the numerator can be used in the equation for thedenominator.Now in order to find the relationship between the position and denominator value I made another chart:Position Number Denominator Value0 151 112 93 94 115 15After a careful blend of logic and luck I finally came up with the second piece of my denominatorequation, which you can see on the next page:
5. 5. n+1 E n (r )=n ( )−r (n−r ) 2The first piece of the equation should be recognizable as the equation for the numerator. The secondpart includes a variable, r, which represents the position within the row of the value being searched for.The variable n did and still does represent the row number.Also, it is important to note that for this equation to work the first position must be treated as positionzero. This allows us to solve for one, which I will explain in more detail when looking at the entireequation.
6. 6. Finding the General EquationOnce we can find the numerator and denominator it is as simple as combining the two equations to findthe entire thing. Of course the numerator equation will be on top and the denominator on bottom, as so: n+1 n( ) 2 E n (r )= n+1 n( )−r ( n−r ) 2The top portion is our numerator equation, and the bottom is simply our denominator equation. Put oneover the other and we have the proper fraction which is present within the triangle.We can find entire rows using this equation, including the initial 1 in each first and last position ineach row. Well find the next two rows, six and seven, using this: 21 21 21 21 21 1 1 16 13 12 13 16 28 28 28 28 28 28 1 1 22 18 16 16 18 22The equation works for both the left and the right side of the triangle.It also works for the ones which begin and end each row, because when r is equal to zero you simplyend up with the numerator over itself, which is of course one.When attempting to find a value past the ending position one ends up with a value lower than 1. This isan easy mistake to catch, therefore we could use this for large rows without much confusion. Simplycalculate until a value under one is found and youve found the end of the row.Unfortunately plugging in each position gets tedious, as even at row seven there are eight differentpositions.Attempting to find negative row numbers simply doesnt work:Row Number Numerator Value-3 3-4 6-5 10-6 15This simply gives us the numerator values for the negative rows turned positive minus one. Rownumber -3 is the same as 2, number -4 is the same as 3, etc.
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Short-Question
Practical answers to complex questions
# What is the formula of circular measure?
## What is the formula of circular measure?
The length of an arc of a circle is given by s=r⋅θ s = r ⋅ θ , where r is the radius and θ is the angle subtended in radians.
How do you find the angle measure of a circle?
The unit of measure is the radian, the angle subtended at the centre of a circle by an arc of equal length to the radius. Since an arc of length r subtends an angle of 1 radian, the whole circumference, length 2πr, will subtend an angle of 2πr/r = 2π radians. Thus, 360° = 2π radians; 1 radian = 57.296°.
What is meant by circular measure?
Definition of circular measure : the measure of an angle in radians.
### How do you find the length of a circular measure?
A circle is 360° all the way around; therefore, if you divide an arc’s degree measure by 360°, you find the fraction of the circle’s circumference that the arc makes up. Then, if you multiply the length all the way around the circle (the circle’s circumference) by that fraction, you get the length along the arc.
How do you find the length of a chord in a circle?
Where, r is the radius of the circle. c is the angle subtended at the center by the chord. d is the perpendicular distance from the chord to the circle center….Chord Length Formula.
Formula to Calculate Length of a Chord
Chord Length Using Trigonometry Chord Length = 2 × r × sin(c/2)
What are the angles of a circle?
A circle has a total of 360 degrees all the way around the center, so if that central angle determining a sector has an angle measure of 60 degrees, then the sector takes up 60/360 or 1/6, of the degrees all the way around.
#### How long is a radian?
It follows that the magnitude in radians of one complete revolution (360 degrees) is the length of the entire circumference divided by the radius, or 2πr / r, or 2π. Thus 2π radians is equal to 360 degrees, meaning that one radian is equal to 180/π ≈ 57.295779513082320876 degrees.
How is the inscribed angle and the intercepted arc related?
An inscribed angle is an angle with its vertex on the circle and whose sides are chords. The intercepted arc is the arc that is inside the inscribed angle and whose endpoints are on the angle. The Inscribed Angle Theorem states that the measure of an inscribed angle is half the measure of its intercepted arc.
What do you need to know about circular measure?
Circular measure is one of the easiest chapters when it comes to understanding the basic concepts. Let’s first look at these concepts. The first thing that you need to know is the relation between degrees and radians. The smaller part of the circle is called the minor sector the larger part is the major sector.
## How is arc length represented in circular measure?
Circular Measure – Additional Mathematics. The arc length is the length of arc along AB. It is represented by s. r is the radius of circle and θ is subtended by the sector (of which the area is to be found).
How are radians used in a circular measure?
The radian is defined by a unit of measurement which is used in many parts of the mathematics area which include circular measure and more often, the branch of angle measurements where pi (Π) is used. There is a relation between radians and degrees of measuring angles. The relation is: Π radians equals 180 degrees.
How to calculate the area of a circle?
Well, there are two formulas used in this chapter. They are The area of any sector of a circle is 1 / 2 (r 2 Θ), where r is the radius of a circle and Θ is the angle inside the sector. The length of any arc of a circle is rΘ.
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# 9.05 Area
Lesson
Can you use an array to help solve a multiplication question?
### Examples
#### Example 1
Which of these number sentences describe the array?
There may be more than one correct answer.
A
16 \times 8 =2
B
8 \times 2 = 16
C
16 \times 2 =8
D
2 \times 8 =16
Worked Solution
Create a strategy
If we multiply the number of rows by the number of columns, we find the total number of squares.
Apply the idea
The number sentences which describe the array are options B and D.
Idea summary
We get the same answer whichever way we look at our array.
## Area of shapes
If we have a two-dimensional shape (2D), we can work out how much space it takes up (the area), by seeing how many unit squares fit inside the shape. In the video, you'll see how we do this, as well as how we can add up parts of unit squares.
### Examples
#### Example 2
Find the area of the shape below.
Worked Solution
Create a strategy
Count the total number of squares.
Apply the idea
There are 5 squares inside the shape.
\text{ Area }= 5 \text{ units}^2
Idea summary
To find the amount of space a shape takes up, we can use a unit square.
## Area of rectangles with arrays
Let's look at how we can use arrays and multiplication to find the area, or how much space is inside, a two-dimensional (2D) shape, by counting how many unit squares fit inside it.
### Examples
#### Example 3
What is the area of the rectangle?
Worked Solution
Create a strategy
Count the number of squares.
Apply the idea
There are 12 squares.
The area of the shape is 12 square units.
Idea summary
We can use arrays and multiplication to find the area of a rectangle by multiplying the number of rows by the number of unit squares in each row.
## Area of rectangles using length and width
In this video, we use arrays to work out the area of our rectangle but start naming the dimensions as length and width.
### Examples
#### Example 4
Find the area of the rectangle shown.
Worked Solution
Create a strategy
The length and width tell us how many unit squares each side is broken up into.
Use the area of a rectangle formula: \text{Area}=\text{Length} \times \text{Width}
Apply the idea
We can see that length is 12 \text{ cm} and the width is 2 \text{ cm}.
Idea summary
We don't need to know the unit of measurement when we use unit squares. We can also add up parts of unit squares, if they are not complete squares.
When we calculate area, the unit of measurement is squared. If we have sides measured in centimetres (cm), for example, area will be \text{cm}^2.
### Outcomes
#### MA3-10MG
selects and uses the appropriate unit to calculate areas, including areas of squares, rectangles and triangles
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# Daily Archives: January 16, 2016
## Summaries for Game Theory
I came across some problem regarding game theory. I studied for a little bit. Here are some of my summaries. Even though it is still superficial.
Impartial Game: In combinatorial game theory, an impartial game is a game in which the allowable moves depend only on the position and not on which of the two players is currently moving, and where the payoffs are symmetric. In other words, the only difference between player 1 and player 2 is that player 1 goes first.
Normal Play: The player who takes the last move wins.
Misere Play: The player who takes the last move loose.
In a normal play impartial game, we define P are the positions where previous player can win; N is the position where the next player can win. From P, it can only reach N state. From N, it can reach either N or P state. P is like a balanced position:
For most of a problem, we are asked if there is a MUST-WIN strategy for a given state.
Assume A is a set of integers. mex(A) is minimum integer number(greater than or equal to 0) which doesn’t appear in set A. g(x) = mex{g(y) : y ∈ F(x)}
For example, mex({2,4,5,6}) = 0, mex({0,1,2,6}) = 3. Consider a state graph like below:
We can get the g(x) for each point:
We can know that the point with value 0 are the P points.
For example, in Subtraction Game, we have a pile of stones. And each time, a player can only remove 1, 3 or 4 stones. The player who get the rest is the winner. In this game, it has N, P position like below:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 1 0 1 2 3 2 0 1 0 1 2 3 2 0 <— mex values
P N P N N N N P N P N N N N P
Nim game is that there are n piles of stone. Each player can take any number of stones from each pile. When n = 2, a set of state graph is like below:
We can see the P position are marked red. We see it is a P position if it complies formula
Wythoff’s Game. There are 2 piles of stones. A player either take same amount of stones from 2 piles, or any number of stones from one pile. It has below solution:
Sample points are like: (1,2), (3,5), (4,7), (6,10), (8,13), (9,15), …
Besides, golden cut rate has a beautiful formula:
Here are some nice articles for game theory:
http://web.mit.edu/sp.268/www/nim.pdf
https://math.mit.edu/research/highschool/primes/materials/2014/conf/12-2-Xiong.pdf
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# MAT275 Modeling with First Order
A body of mass 6 kg is projected vertically upward with an initial velocity 25 meters per second.We assume that the forces acting on the body are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude c|v(t)| where c=0.35kgs and v(t) is the velocity of the ball at time t. The gravitational constant is g=9.8m/s2.a) Find a differential equation for the velocity v:dvdt=b) Solve the differential equation in part a) and find a formula for the velocity at any time t:v(t)= Find a formula for the position function at any time t, if the initial position is s(0)=0:s(t)= How does this compare with the solution to the equation for velocity when there is no air resistance?If c=0, then v(t)=25−9.8t, and if s(0)=0, then s(t)=25t−4.9t2.We then have that v(t)=0 when t≈2.551, and s(2.551)≈31.888,and that the positive t solution to s(t)=0 is t≈5.102, which leads to v(5.102)=−25 meters per second.
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