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|---|---|---|
Given two real numbers $p>1$ and $q>1$ so that $\frac{1}{p} + \frac{1}{q} = 1$ and $pq = 4$, what is $q$?
|
2
| |
In triangle $ABC$, where $AB = 50$, $BC = 36$, and $AC = 42$. A line $CX$ from $C$ is perpendicular to $AB$ and intersects $AB$ at point $X$. Find the ratio of the area of $\triangle BCX$ to the area of $\triangle ACX$. Express your answer as a simplified common fraction.
|
\frac{6}{7}
| |
Let $ABC$ be a triangle with $AB = 5$ , $AC = 8$ , and $BC = 7$ . Let $D$ be on side $AC$ such that $AD = 5$ and $CD = 3$ . Let $I$ be the incenter of triangle $ABC$ and $E$ be the intersection of the perpendicular bisectors of $\overline{ID}$ and $\overline{BC}$ . Suppose $DE = \frac{a\sqrt{b}}{c}$ where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ .
*Proposed by Ray Li*
|
13
| |
A printer prints text pages at a rate of 17 pages per minute and graphic pages at a rate of 10 pages per minute. If a document consists of 250 text pages and 90 graphic pages, how many minutes will it take to print the entire document? Express your answer to the nearest whole number.
|
24
| |
Solve the following equations using appropriate methods:
$(1)\left(3x-1\right)^{2}=9$.
$(2)x\left(2x-4\right)=\left(2-x\right)^{2}$.
|
-2
| |
Medians $\overline{AD}$ and $\overline{BE}$ of $\triangle ABC$ intersect at an angle of $45^\circ$. If $AD = 12$ and $BE = 16$, then calculate the area of $\triangle ABC$.
|
64\sqrt{2}
| |
In Mr. Lee's classroom, there are six more boys than girls among a total of 36 students. What is the ratio of the number of boys to the number of girls?
|
\frac{7}{5}
| |
When $0.76\overline{204}$ is expressed as a fraction in the form $\frac{x}{999000}$, what is the value of $x$?
|
761280
| |
How many positive integers less than $201$ are multiples of either $6$ or $8$, but not both at once?
|
42
| |
Given that \( M \) is the midpoint of the height \( D D_{1} \) of a regular tetrahedron \( ABCD \), find the dihedral angle \( A-M B-C \) in radians.
|
\frac{\pi}{2}
| |
Given that
$\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}$
find the greatest integer that is less than $\frac N{100}$.
|
137
|
Let $f(x) = (1+x)^{19}.$ Applying the binomial theorem gives us $f(x) = \binom{19}{19} x^{19} + \binom{19}{18} x^{18} + \binom{19}{17} x^{17}+ \cdots + \binom{19}{0}.$ Since $\frac 1{2!17!}+\frac 1{3!16!}+\dots+\frac 1{8!11!}+\frac 1{9!10!} = \frac{\frac{f(1)}{2} - \binom{19}{19} - \binom{19}{18}}{19!},$ $N = \frac{2^{18}-20}{19}.$ After some fairly easy bashing, we get $\boxed{137}$ as the answer.
~peelybonehead
|
Let $N$ be a three-digit integer such that the difference between any two positive integer factors of $N$ is divisible by 3 . Let $d(N)$ denote the number of positive integers which divide $N$. Find the maximum possible value of $N \cdot d(N)$.
|
5586
|
We first note that all the prime factors of $n$ must be 1 modulo 3 (and thus 1 modulo 6 ). The smallest primes with this property are $7,13,19, \ldots$ Since $7^{4}=2401>1000$, the number can have at most 3 prime factors (including repeats). Since $7 \cdot 13 \cdot 19=1729>1000$, the most factors $N$ can have is 6 . Consider the number $7^{2} \cdot 19=931$, which has 6 factors. For this choice of $N, N \cdot d(N)=5586$. For another $N$ to do better, it must have at least 6 factors, for otherwise, $N \cdot d(N)<1000 \cdot 5=5000$. It is easy to verify that $7^{2} \cdot 19$ is the greatest number with 6 prime factors satisfying our conditions, so the answer must be 5586 .
|
Solve the equation $27 = 3(9)^{x-1}$ for $x.$
|
2
| |
In trapezoid \( KLMN \), the lengths of the bases are \( KN = 25 \), \( LM = 15 \), and the lengths of the legs are \( KL = 6 \), \( MN = 8 \). Find the length of the segment connecting the midpoints of the bases.
|
20
| |
Amy and Bob choose numbers from $0,1,2,\cdots,81$ in turn and Amy choose the number first. Every time the one who choose number chooses one number from the remaining numbers. When all $82$ numbers are chosen, let $A$ be the sum of all the numbers Amy chooses, and let $B$ be the sum of all the numbers Bob chooses. During the process, Amy tries to make $\gcd(A,B)$ as great as possible, and Bob tries to make $\gcd(A,B)$ as little as possible. Suppose Amy and Bob take the best strategy of each one, respectively, determine $\gcd(A,B)$ when all $82$ numbers are chosen.
|
41
| |
A truck travels $\frac{b}{6}$ feet every $t$ seconds. There are $3$ feet in a yard. How many yards does the truck travel in $3$ minutes?
|
\frac{10b}{t}
|
1. **Identify the rate of travel**: The truck travels $\frac{b}{6}$ feet every $t$ seconds.
2. **Convert time from minutes to seconds**: Since there are $60$ seconds in a minute, $3$ minutes is $3 \times 60 = 180$ seconds.
3. **Set up a proportion to find the distance in feet**: We know the truck travels $\frac{b}{6}$ feet in $t$ seconds. We need to find how many feet, $x$, the truck travels in $180$ seconds. Using the proportion:
\[
\frac{\frac{b}{6}}{t} = \frac{x}{180}
\]
Simplifying the left side:
\[
\frac{b}{6t} = \frac{x}{180}
\]
4. **Solve for $x$**: Cross-multiplying to solve for $x$ gives:
\[
180 \cdot \frac{b}{6t} = x \implies \frac{180b}{6t} = x \implies x = \frac{30b}{t}
\]
5. **Convert feet to yards**: Since $x$ is the distance in feet and there are $3$ feet in a yard, we convert $x$ to yards by dividing by $3$:
\[
\frac{x}{3} = \frac{\frac{30b}{t}}{3} = \frac{10b}{t}
\]
6. **Match the answer**: The expression $\frac{10b}{t}$ corresponds to choice $\boxed{\text{(E)}}$.
|
The solution to the equation \(\arcsin x + \arcsin 2x = \arccos x + \arccos 2x\) is
|
\frac{\sqrt{5}}{5}
| |
Eight numbers \( a_{1}, a_{2}, a_{3}, a_{4} \) and \( b_{1}, b_{2}, b_{3}, b_{4} \) satisfy the equations:
\[
\left\{
\begin{array}{l}
a_{1} b_{1} + a_{2} b_{3} = 1 \\
a_{1} b_{2} + a_{2} b_{4} = 0 \\
a_{3} b_{1} + a_{4} b_{3} = 0 \\
a_{3} b_{2} + a_{4} b_{4} = 1
\end{array}
\right.
\]
Given that \( a_{2} b_{3} = 7 \), find \( a_{4} b_{4} \).
|
-6
| |
In a square \(ABCD\), let \(P\) be a point on the side \(BC\) such that \(BP = 3PC\) and \(Q\) be the midpoint of \(CD\). If the area of the triangle \(PCQ\) is 5, what is the area of triangle \(QDA\)?
|
20
| |
Given that $a$ is a positive real number and $b$ is an integer between $1$ and $201$, inclusive, find the number of ordered pairs $(a,b)$ such that $(\log_b a)^{2023}=\log_b(a^{2023})$.
|
603
| |
Given a set of data $x_1, x_2, x_3, \ldots, x_n$ with a mean of 2 and a variance of 3, calculate the mean and variance of the data set $2x_1+5, 2x_2+5, 2x_3+5, \ldots, 2x_n+5$ respectively.
|
12
| |
There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$. What is the sum of these values of $a$?
|
-16
| |
In a circle, a chord of length 10 cm is drawn. A tangent to the circle is drawn through one end of the chord, and a secant parallel to the tangent is drawn through the other end. The internal segment of the secant is 12 cm. Find the radius of the circle.
|
13
| |
The point with coordinates $(6,-10)$ is the midpoint of the segment with one endpoint at $(8,0)$. Find the sum of the coordinates of the other endpoint.
|
-16
| |
Given a parabola $C: y^2 = 3x$ with focus $F$, find the length of segment $AB$ where the line passing through $F$ at a $30^\circ$ angle intersects the parabola at points $A$ and $B$.
|
12
| |
If set $A=\{-4, 2a-1, a^2\}$, $B=\{a-5, 1-a, 9\}$, and $A \cap B = \{9\}$, then the value of $a$ is.
|
-3
| |
If $\|\mathbf{v}\| = 4,$ then find $\mathbf{v} \cdot \mathbf{v}.$
|
16
| |
How many three-digit numbers are not divisible by $5$, have digits that sum to less than $20$, and have the first digit equal to the third digit?
|
60
|
We are tasked with finding the number of three-digit numbers that satisfy the following conditions:
1. The number is not divisible by $5$.
2. The sum of the digits is less than $20$.
3. The first and third digits are the same.
Let's denote such a number as $\overline{xyx}$, where $x$ and $y$ are its digits. The number can be expressed as $101x + 10y$.
#### Step 1: Analyze the divisibility by $5$
For the number not to be divisible by $5$, $x$ cannot be $0$ or $5$. If $x = 0$, the number is not a three-digit number. If $x = 5$, then $\overline{xyx} = 505 + 10y$ is divisible by $5$.
#### Step 2: Analyze the sum of the digits
The sum of the digits of $\overline{xyx}$ is $2x + y$. We need $2x + y < 20$.
#### Step 3: Case analysis based on the value of $x$
- **Case $x = 1, 2, 3, 4$:** Here, $2x < 10$, so $2x + y < 20$ for any $y = 0, 1, ..., 9$. Each value of $x$ gives $10$ possible values for $y$.
- Total numbers = $10 \times 4 = 40$.
- **Case $x = 6$:** Here, $2x = 12$, so $12 + y < 20$ implies $y < 8$. Possible values for $y$ are $0, 1, ..., 7$.
- Total numbers = $8$.
- **Case $x = 7$:** Here, $2x = 14$, so $14 + y < 20$ implies $y < 6$. Possible values for $y$ are $0, 1, ..., 5$.
- Total numbers = $6$.
- **Case $x = 8$:** Here, $2x = 16$, so $16 + y < 20$ implies $y < 4$. Possible values for $y$ are $0, 1, 2, 3$.
- Total numbers = $4$.
- **Case $x = 9$:** Here, $2x = 18$, so $18 + y < 20$ implies $y < 2$. Possible values for $y$ are $0, 1$.
- Total numbers = $2$.
#### Step 4: Summing up all the cases
Summing the numbers from all cases, we get:
$$40 + 8 + 6 + 4 + 2 = 60.$$
Thus, the total number of three-digit numbers satisfying all the given conditions is $\boxed{\textbf{(B)}\ 60}$.
|
Determine the smallest positive integer \(n\) for which there exists positive real numbers \(a\) and \(b\) such that
\[(a + 3bi)^n = (a - 3bi)^n,\]
and compute \(\frac{b}{a}\).
|
\frac{\sqrt{3}}{3}
| |
Given $f(\alpha) = \frac{\sin(\frac{\pi}{2} + \alpha) + 3\sin(-\pi - \alpha)}{2\cos(\frac{11\pi}{2} - \alpha) - \cos(5\pi - \alpha)}$.
(I) Simplify $f(\alpha)$;
(II) If $\tan \alpha = 3$, find the value of $f(\alpha)$.
|
-2
| |
Given the sequence $\left\{ a_n \right\}$ such that $a_{n+1}+a_n={(-1)}^n\cdot n$ ($n\in \mathbb{N}^*$), find the sum of the first 20 terms of $\left\{ a_n \right\}$.
|
-100
| |
Given the graph of the power function $y=f(x)$ passes through the point $\left(\frac{1}{3},\frac{\sqrt{3}}{3}\right)$, then the value of $\log_{2}f(2)$ is \_\_\_\_.
|
\frac{1}{2}
| |
Compute the range of $y=|x+7|-|x-2|$.
|
[-9, 9]
| |
Find the sum of all real solutions to the equation \[\frac{x-3}{x^2+3x+2} = \frac{x-7}{x^2-7x+12}.\]
|
\frac{26}{3}
| |
Seven students are standing in a row for a graduation photo. Among them, student A must stand in the middle, and students B and C must stand together. How many different arrangements are there?
|
192
| |
The three sides of a right triangle form a geometric sequence. Determine the ratio of the length of the hypotenuse to the length of the shorter leg.
|
\frac{1+\sqrt{5}}{2}
|
Let the shorter leg have length $\ell$, and the common ratio of the geometric sequence be $r>1$. Then the length of the other leg is $\ell r$, and the length of the hypotenuse is $\ell r^{2}$. Hence, $$\ell^{2}+(\ell r)^{2}=\left(\ell r^{2}\right)^{2} \Longrightarrow \ell^{2}\left(r^{2}+1\right)=\ell^{2} r^{4} \Longrightarrow r^{2}+1=r^{4}$$ Hence, $r^{4}-r^{2}-1=0$, and therefore $r^{2}=\frac{1 \pm \sqrt{5}}{2}$. As $r>1$, we have $r^{2}=\frac{1+\sqrt{5}}{2}$, completing the problem as the ratio of the hypotenuse to the shorter side is $\frac{\ell r^{2}}{\ell}=r^{2}$.
|
How many of the 512 smallest positive integers written in base 8 use 5 or 6 (or both) as a digit?
|
296
| |
The function $g$ is defined on the set of integers and satisfies \[g(n)= \begin{cases} n-4 & \mbox{if }n\ge 1010 \\ g(g(n+7)) & \mbox{if }n<1010. \end{cases}\] Find $g(77)$.
|
1011
| |
Given $\overrightarrow{a}=(\tan (\theta+ \frac {\pi}{12}),1)$ and $\overrightarrow{b}=(1,-2)$, where $\overrightarrow{a} \perp \overrightarrow{b}$, find the value of $\tan (2\theta+ \frac {5\pi}{12})$.
|
- \frac{1}{7}
| |
Point $(x,y)$ is randomly picked from the rectangular region with vertices at $(0,0),(2010,0),(2010,2011),$ and $(0,2011)$. What is the probability that $x > 3y$? Express your answer as a common fraction.
|
\frac{670}{2011}
| |
Two concentric circles have radii $r$ and $R>r$. Three new circles are drawn so that they are each tangent to the big two circles and tangent to the other two new circles. Find $\frac{R}{r}$.
|
3
|
The centers of the three new circles form a triangle. The diameter of the new circles is $R-r$, so the side length of the triangle is $R-r$. Call the center of the concentric circle $O$, two vertices of the triangle $A$ and $B$, and $A B$ 's midpoint $D$. $O A$ is the average $R$ and $r$, namely $\frac{R+r}{2}$. Using the law of sines on triangle $D A O$, we get $\frac{\sin (30)}{A D}=\frac{\sin (90)}{A O} \Rightarrow R=3 r$, so $\frac{R}{r}=3$.
|
Given the function $f(x)=ax^{3}+2bx^{2}+3cx+4d$, where $a,b,c,d$ are real numbers, $a < 0$, and $c > 0$, is an odd function, and when $x\in[0,1]$, the range of $f(x)$ is $[0,1]$. Find the maximum value of $c$.
|
\frac{\sqrt{3}}{2}
| |
It is known that there exists a natural number \( N \) such that \( (\sqrt{3}-1)^{N} = 4817152 - 2781184 \cdot \sqrt{3} \). Find \( N \).
|
16
| |
A line of soldiers 1 mile long is jogging. The drill sergeant, in a car, moving at twice their speed, repeatedly drives from the back of the line to the front of the line and back again. When each soldier has marched 15 miles, how much mileage has been added to the car, to the nearest mile?
|
30
|
30.
|
Jacqueline has 2 liters of soda. Liliane has 60% more soda than Jacqueline, and Alice has 40% more soda than Jacqueline. Calculate the percentage difference between the amount of soda Liliane has compared to Alice.
|
14.29\%
| |
Alvin rides on a flat road at $25$ kilometers per hour, downhill at $35$ kph, and uphill at $10$ kph. Benny rides on a flat road at $35$ kph, downhill at $45$ kph, and uphill at $15$ kph. Alvin goes from town $X$ to town $Y$, a distance of $15$ km all uphill, then from town $Y$ to town $Z$, a distance of $25$ km all downhill, and then back to town $X$, a distance of $30$ km on the flat. Benny goes the other way around using the same route. What is the time in minutes that it takes Alvin to complete the $70$-km ride, minus the time in minutes it takes Benny to complete the same ride?
|
33
| |
Compute $\arccos (\cos 3).$ All functions are in radians.
|
3 - 2\pi
| |
Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$-sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
|
147
|
1. **Understanding the Problem:**
We have three identical square sheets of paper, each with side length $6$. The middle sheet is rotated $30^\circ$ clockwise, and the top sheet is rotated $60^\circ$ clockwise. We need to find the area of the resulting $24$-sided polygon.
2. **Breaking Down the Polygon:**
The polygon can be divided into $12$ identical quadrilaterals, each formed by the intersection of the rotated squares. We focus on one such quadrilateral $OBAC$.
3. **Analyzing Triangle $OBC$:**
- The side $OC$ is half the diagonal of the square, so $OC = \frac{1}{2} \times \sqrt{6^2 + 6^2} = 3\sqrt{2}$.
- The angle $\angle OCB = 75^\circ$ (since the square is rotated $30^\circ$).
4. **Finding the Area of Triangle $OBC$:**
- We use the formula for the area of a triangle: $\text{Area} = \frac{1}{2}ab\sin(C)$.
- Here, $a = b = 3\sqrt{2}$ (sides of the triangle), and $C = 75^\circ$.
- $\text{Area}_{OBC} = \frac{1}{2} \times (3\sqrt{2})^2 \times \sin(75^\circ) = 9 \times \sin(75^\circ)$.
5. **Analyzing Triangle $ABC$:**
- $\angle BAC = 120^\circ$ (since $\angle OAC = 45^\circ$ and $\angle OAB = 75^\circ$).
- We split $ABC$ into two $30-60-90$ triangles by drawing altitude from $A$ to $BC$.
- The length of $BC = 3\sqrt{3} - 3$ (from the rotation and geometry of the square).
- The height of these triangles is $\frac{3\sqrt{3}-3}{2\sqrt{3}} = \frac{3-\sqrt{3}}{2}$.
- $\text{Area}_{ABC} = \frac{1}{2} \times (3\sqrt{3}-3) \times \frac{3-\sqrt{3}}{2}$.
6. **Calculating the Area of Quadrilateral $OBAC$:**
- $\text{Area}_{OBAC} = \text{Area}_{OBC} - \text{Area}_{ABC}$.
- $\text{Area}_{OBC} = 9 \times \sin(75^\circ)$.
- $\text{Area}_{ABC} = \frac{1}{2} \times (3\sqrt{3}-3) \times \frac{3-\sqrt{3}}{2}$.
- Simplifying, $\text{Area}_{OBAC} = 9 - 3\sqrt{3}$.
7. **Total Area of the Polygon:**
- Since there are $12$ such quadrilaterals, the total area is $12 \times (9 - 3\sqrt{3}) = 108 - 36\sqrt{3}$.
8. **Final Answer:**
- The area of the polygon is expressed as $108 - 36\sqrt{3}$.
- Thus, $a = 108$, $b = 36$, and $c = 3$.
- Therefore, $a + b + c = 108 + 36 + 3 = \boxed{147}$.
|
Equilateral triangle $ABC$ has side length $840$. Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$. The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$, respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$, $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$. Find $AF$.
Diagram
[asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy]
|
336
|
Since $\triangle AFG$ is isosceles, $AF = FG$, and since $\triangle AEF$ is equilateral, $AF = EF$. Thus, $EF = FG$, and since these triangles share an altitude, they must have the same area.
Drop perpendiculars from $E$ and $F$ to line $BC$; call the meeting points $P$ and $Q$, respectively. $\triangle BEP$ is clearly congruent to both $\triangle BED$ and $\triangle FQC$, and thus each of these new triangles has the same area as $\triangle BED$. But we can "slide" $\triangle BEP$ over to make it adjacent to $\triangle FQC$, thus creating an equilateral triangle whose area has a ratio of $18:8$ when compared to $\triangle AEF$ (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio $18:8$ reduces to $9:4$, the ratio of their sides must be $3:2$. So, because $FC$ and $AF$ represent sides of these triangles, and they add to $840$, $AF$ must equal two-fifths of $840$, or $\boxed{336}$.
~Interstigation
|
The symbol $|a|$ means $a$ is a positive number or zero, and $-a$ if $a$ is a negative number.
For all real values of $t$ the expression $\sqrt{t^4+t^2}$ is equal to?
|
|t|\sqrt{1+t^2}
|
1. **Identify the expression to simplify**: We start with the expression given in the problem:
\[
\sqrt{t^4 + t^2}
\]
2. **Factor out $t^2$ from the expression**: Recognizing that both terms inside the square root are powers of $t$, we can factor out the lowest power, $t^2$:
\[
\sqrt{t^4 + t^2} = \sqrt{t^2(t^2 + 1)}
\]
3. **Separate the square root into a product**: Using the property of square roots that $\sqrt{ab} = \sqrt{a}\sqrt{b}$, we separate the expression:
\[
\sqrt{t^2(t^2 + 1)} = \sqrt{t^2} \cdot \sqrt{t^2 + 1}
\]
4. **Simplify $\sqrt{t^2}$**: The square root of a square is the absolute value of the original number, which is a standard result from the properties of square roots and absolute values:
\[
\sqrt{t^2} = |t|
\]
5. **Combine the simplified terms**: Substituting back into the expression, we get:
\[
|t| \cdot \sqrt{t^2 + 1}
\]
6. **Conclusion**: The simplified expression is $|t|\sqrt{t^2 + 1}$, which matches option $\textbf{(E)}$ in the multiple-choice answers.
Thus, the final answer is $\boxed{\textbf{(E)}}$.
|
Say that an integer $A$ is delicious if there exist several consecutive integers, including $A$, that add up to 2024. What is the smallest delicious integer?
|
-2023
| |
In triangle $ABC$, $AB = BC$, and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $BE =
10$. The values of $\tan \angle CBE$, $\tan \angle DBE$, and $\tan \angle ABE$ form a geometric progression, and the values of $\cot \angle DBE$, $\cot \angle CBE$, $\cot \angle DBC$ form an arithmetic progression. What is the area of triangle $ABC$?
[asy]
pair A,B,C,D,E;
A=(0,0);
B=(4,8);
C=(8,0);
E=(10,0);
D=(4,0);
draw(A--B--E--cycle,linewidth(0.7));
draw(C--B--D,linewidth(0.7));
label("$B$",B,N);
label("$A$",A,S);
label("$D$",D,S);
label("$C$",C,S);
label("$E$",E,S);
[/asy]
|
\frac{50}{3}
| |
Consider a $3 \times 3$ block of squares as the center area in an array of unit squares. The first ring around this center block contains unit squares that directly touch the block. If the pattern continues as before, how many unit squares are in the $10^{th}$ ring?
|
88
| |
For which values of \(a\) does the equation \(|x-3| = a x - 1\) have two solutions? Enter the midpoint of the interval of parameter \(a\) in the provided field. Round the answer to three significant digits according to rounding rules and enter it in the provided field.
|
0.667
| |
Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$. A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$, which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}$. The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n}$, where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n$.
|
230
| |
Given that $\frac 1n - \frac{1}{n+1} < \frac{1}{10}$, what is the least possible positive integer value of $n$?
|
3
| |
Parallelogram $ABCD$ has $AB=CD=6$ and $BC=AD=10$ , where $\angle ABC$ is obtuse. The circumcircle of $\triangle ABD$ intersects $BC$ at $E$ such that $CE=4$ . Compute $BD$ .
|
4\sqrt{6}
| |
During the first year, ABC's stock price starts at $ \$100 $ and increases $ 100\% $. During the second year, its stock price goes down $ 25\% $ from its price at the end of the first year. What is the price of the stock, in dollars, at the end of the second year?
|
\$150
| |
The point $(1,1,1)$ is rotated $180^\circ$ about the $y$-axis, then reflected through the $yz$-plane, reflected through the $xz$-plane, rotated $180^\circ$ about the $y$-axis, and reflected through the $xz$-plane. Find the coordinates of the point now.
|
(-1,1,1)
| |
The arithmetic mean of several consecutive natural numbers is 5 times greater than the smallest of them. How many times is the arithmetic mean smaller than the largest of these numbers?
|
1.8
| |
The limiting sum of the infinite series, $\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots$ whose $n$th term is $\frac{n}{10^n}$ is:
|
\frac{10}{81}
|
To find the sum of the series $\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots$, where the $n$th term is $\frac{n}{10^n}$, we can use a technique that involves rewriting the series in a more manageable form.
1. **Rewrite the series**: Notice that each term in the series can be expressed as a sum of several terms from a geometric series:
\[
\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots = \left(\frac{1}{10}\right) + \left(\frac{1}{10^2} + \frac{1}{10^2}\right) + \left(\frac{1}{10^3} + \frac{1}{10^3} + \frac{1}{10^3}\right) + \dots
\]
This can be rearranged as:
\[
\left(\frac{1}{10}\right) + \left(\frac{1}{10^2} + \frac{1}{10^2}\right) + \left(\frac{1}{10^3} + \frac{1}{10^3} + \frac{1}{10^3}\right) + \dots
\]
Which simplifies to:
\[
\sum_{n=1}^{\infty} n \frac{1}{10^n}
\]
2. **Use the formula for the sum of an infinite series**: The sum of the series $\sum_{n=1}^{\infty} n x^n$ where $|x| < 1$ is given by the formula $\frac{x}{(1-x)^2}$. Here, $x = \frac{1}{10}$, so we substitute this into the formula:
\[
\sum_{n=1}^{\infty} n \left(\frac{1}{10}\right)^n = \frac{\frac{1}{10}}{\left(1-\frac{1}{10}\right)^2}
\]
Simplifying the denominator:
\[
1 - \frac{1}{10} = \frac{9}{10}
\]
So, the expression becomes:
\[
\frac{\frac{1}{10}}{\left(\frac{9}{10}\right)^2} = \frac{\frac{1}{10}}{\frac{81}{100}} = \frac{1}{10} \cdot \frac{100}{81} = \frac{10}{81}
\]
3. **Conclusion**: The sum of the series $\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots$ is $\boxed{\textbf{(B)}\ \frac{10}{81}}$.
|
Along a straight alley, there are 400 streetlights placed at equal intervals, numbered consecutively from 1 to 400. Alla and Boris start walking towards each other from opposite ends of the alley at the same time but with different constant speeds (Alla from the first streetlight and Boris from the four-hundredth streetlight). When Alla is at the 55th streetlight, Boris is at the 321st streetlight. At which streetlight will they meet? If the meeting occurs between two streetlights, indicate the smaller number of the two in the answer.
|
163
| |
Calculate:<br/>$(1)\frac{2}{5}-\frac{1}{5}\times \left(-5\right)+\frac{3}{5}$;<br/>$(2)-2^{2}-\left(-3\right)^{3}\div 3\times \frac{1}{3}$.
|
-1
| |
Find the number of positive integers $n,$ $1 \le n \le 2000,$ for which the polynomial $x^2 + 2x - n$ can be factored as the product of two linear factors with integer coefficients.
|
45
| |
Suppose \( x_{1}, x_{2}, \ldots, x_{2011} \) are positive integers satisfying
\[ x_{1} + x_{2} + \cdots + x_{2011} = x_{1} x_{2} \cdots x_{2011} \]
Find the maximum value of \( x_{1} + x_{2} + \cdots + x_{2011} \).
|
4022
| |
Given points P and Q are on a circle of radius 7 and PQ = 8, find the length of the line segment PR, where R is the midpoint of the minor arc PQ.
|
\sqrt{98 - 14\sqrt{33}}
| |
Compute the following expression:
\[ 4(1 + 4(1 + 4(1 + 4(1 + 4(1 + 4(1 + 4(1 + 4(1 + 4)))))))) \]
|
1398100
| |
Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
|
146
|
1. **Identify the problem and apply the Principle of Inclusion-Exclusion (PIE):**
We need to find the number of days Mrs. Sanders does not receive a call from any of her grandchildren. Each grandchild calls her every 3, 4, and 5 days respectively. We use PIE to count the days she receives at least one call and subtract from the total days in the year.
2. **Calculate the days she receives at least one call:**
- The first grandchild calls every 3 days, so the number of days she receives calls from this grandchild is $\left \lfloor \frac{365}{3} \right \rfloor = 121$.
- The second grandchild calls every 4 days, so the number of days she receives calls from this grandchild is $\left \lfloor \frac{365}{4} \right \rfloor = 91$.
- The third grandchild calls every 5 days, so the number of days she receives calls from this grandchild is $\left \lfloor \frac{365}{5} \right \rfloor = 73$.
3. **Calculate the days she receives calls from at least two grandchildren (overlapping days):**
- The first and second grandchildren have a least common multiple (LCM) of 12 days, so the number of overlapping days for these two is $\left \lfloor \frac{365}{12} \right \rfloor = 30$.
- The first and third grandchildren have an LCM of 15 days, so the number of overlapping days for these two is $\left \lfloor \frac{365}{15} \right \rfloor = 24$.
- The second and third grandchildren have an LCM of 20 days, so the number of overlapping days for these two is $\left \lfloor \frac{365}{20} \right \rfloor = 18$.
4. **Calculate the days she receives calls from all three grandchildren:**
- The LCM of 3, 4, and 5 is 60 days, so the number of days all three grandchildren call her is $\left \lfloor \frac{365}{60} \right \rfloor = 6$.
5. **Apply the PIE formula:**
\[
\text{Total days with at least one call} = (121 + 91 + 73) - (30 + 24 + 18) + 6 = 285 - 72 + 6 = 219
\]
6. **Calculate the days with no calls:**
\[
\text{Days with no calls} = 365 - 219 = 146
\]
7. **Conclusion:**
The number of days during the next year that Mrs. Sanders did not receive a phone call from any of her grandchildren is $\boxed{\textbf{(D) }146}$.
|
Three circles with radii 2, 3, and 3 touch each other. What is the area of the triangle formed by joining the centers of these circles?
|
12
| |
Given $3^{7} + 1$ and $3^{15} + 1$ inclusive, how many perfect cubes lie between these two values?
|
231
| |
There are seven students taking a graduation photo in a row. Student A must stand in the middle, and students B and C must stand together. How many different arrangements are possible?
|
192
| |
An experimenter selects 4 out of 8 different chemical substances to place in 4 distinct bottles. If substances A and B should not be placed in bottle 1, the number of different ways of arranging them is ____.
|
1260
| |
A frustum of a right circular cone is formed by cutting a small cone off of the top of a larger cone. If a particular frustum has a lower base radius of 6 inches, an upper base radius of 3 inches, and a height of 4 inches, what is its lateral surface area? (The lateral surface area of a cone or frustum is the curved surface excluding the base(s).)
[asy]size(200);
import three; defaultpen(linewidth(.8)); currentprojection = orthographic(0,-3,0.5); pen dots = linetype("0 3") + linewidth(1);
real h = 2.3, ratio = (91-24)/(171-24);
picture p1, p2; /* p1 is left-hand picture */
triple A = (0,0,0), B = (0,0,h); draw(p1,(-1,0,0)..(0,-1,0)..(1,0,0)); draw(p1,(-1,0,0)..(0,1,0)..(1,0,0),dots); draw(p1,(-1,0,0)--B--(1,0,0));
add(p1);
triple vlift = (0,0,0.5);
path3 toparc1 = shift((0,0,h*(1-ratio)))*scale3(ratio)*((-1,0,0)..(0,1,0)..(1,0,0)), toparc2 = shift((0,0,h*(1-ratio)))*scale3(ratio)*((1,0,0)..(0,-1,0)..(-1,0,0));
draw(p2,(-1,0,0)..(0,-1,0)..(1,0,0)); draw(p2,(-1,0,0)..(0,1,0)..(1,0,0),dots);
draw(p2,(-1,0,0)--ratio*(-1,0,0)+(1-ratio)*B^^ratio*(1,0,0)+(1-ratio)*B--(1,0,0));
draw(p2,shift(vlift)*(ratio*(-1,0,0)+(1-ratio)*B--B--ratio*(1,0,0)+(1-ratio)*B));
draw(p2,toparc1--toparc2); draw(p2,shift(vlift)*toparc1,dots); draw(p2,shift(vlift)*toparc2);
draw(p2,shift(vlift)*((1-ratio)*B--B),linewidth(0.7)); dot(p2,shift(vlift)*((1-ratio)*B),linewidth(1.5));
label(p2,"frustum",(0,0,h/4));
add(shift((3.4,0,0))*p2);
[/asy]
|
45\pi
| |
Given the function $f(x)=2\sin (\\omega x)$, where $\\omega > 0$.
(1) When $ \\omega =1$, determine the even-odd property of the function $F(x)=f(x)+f(x+\\dfrac{\\mathrm{ }\\!\\!\\pi\\!\\!{ }}{2})$ and explain the reason.
(2) When $ \\omega =2$, the graph of the function $y=f(x)$ is translated to the left by $ \\dfrac{\\mathrm{ }\\!\\!\\pi\\!\\!{ }}{6}$ unit, and then translated upward by 1 unit to obtain the graph of the function $y=g(x)$. Find all possible values of the number of zeros of $y=g(x)$ in the interval $[a,a+10π]$ for any $a∈R$.
|
20
| |
Find the number of eight-digit positive integers that are multiples of 9 and have all distinct digits.
|
181440
|
Note that $0+1+\cdots+9=45$. Consider the two unused digits, which must then add up to 9. If it's 0 and 9, there are $8 \cdot 7!$ ways to finish; otherwise, each of the other four pairs gives $7 \cdot 7!$ ways to finish, since 0 cannot be the first digit. This gives a total of $36 \cdot 7!=181440$.
|
Robert read a book for 10 days. He read an average of 25 pages per day for the first 5 days and an average of 40 pages per day for the next 4 days, and read 30 more pages on the last day. Calculate the total number of pages in the book.
|
315
| |
Let \( m \) and \( n \) (with \( m > n \)) be positive integers such that \( 70^2 \) divides \( 2023^m - 2023^n \). What is the smallest value of \( m+n \)?
|
24
| |
Determine all real values of $A$ for which there exist distinct complex numbers $x_{1}, x_{2}$ such that the following three equations hold: $$ x_{1}(x_{1}+1) =A $$ x_{2}(x_{2}+1) =A $$ x_{1}^{4}+3 x_{1}^{3}+5 x_{1} =x_{2}^{4}+3 x_{2}^{3}+5 x_{2} $$
|
\[ A = -7 \]
|
Applying polynomial division, $$ x_{1}^{4}+3 x_{1}^{3}+5 x_{1} =\left(x_{1}^{2}+x_{1}-A\right)\left(x_{1}^{2}+2 x_{1}+(A-2)\right)+(A+7) x_{1}+A(A-2) =(A+7) x_{1}+A(A-2) .$$ Thus, in order for the last equation to hold, we need $(A+7) x_{1}=(A+7) x_{2}$, from which it follows that $A=-7$. These steps are reversible, so $A=-7$ indeed satisfies the needed condition.
|
Replace the symbols $\$ last\$ with the same numbers so that the equation becomes true: $\$ \$ |\frac{20}{x} - \frac{x}{15} = \frac{20}{15} \$ \$$
|
10
| |
Let $a,$ $b,$ $c$ be positive real numbers. Find the smallest possible value of
\[6a^3 + 9b^3 + 32c^3 + \frac{1}{4abc}.\]
|
6
| |
In a $3 \times 3$ table, numbers are placed such that each number is 4 times smaller than the number in the adjacent cell to the right and 3 times smaller than the number in the adjacent cell above. The sum of all the numbers in the table is 546. Find the number in the central cell.
|
24
| |
A reflection takes $\begin{pmatrix} -1 \\ 7 \end{pmatrix}$ to $\begin{pmatrix} 5 \\ -5 \end{pmatrix}.$ Which vector does the reflection take $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$ to?
|
\begin{pmatrix} 0 \\ -5 \end{pmatrix}
| |
Isabella's house has $3$ bedrooms. Each bedroom is $12$ feet long, $10$ feet wide, and $8$ feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy $60$ square feet in each bedroom. How many square feet of walls must be painted?
|
876
|
1. **Calculate the area of the walls in one bedroom**:
Each bedroom has four walls, two pairs of opposite walls. Each pair consists of:
- Two walls of dimensions $12$ feet (length) by $8$ feet (height).
- Two walls of dimensions $10$ feet (width) by $8$ feet (height).
The area of the walls for each dimension pair is calculated as follows:
- For the $12$ feet by $8$ feet walls:
\[
2 \times (12 \times 8) = 2 \times 96 = 192 \text{ square feet}
\]
- For the $10$ feet by $8$ feet walls:
\[
2 \times (10 \times 8) = 2 \times 80 = 160 \text{ square feet}
\]
Adding these together gives the total wall area in one bedroom:
\[
192 + 160 = 352 \text{ square feet}
\]
2. **Subtract the area occupied by doorways and windows**:
Each bedroom has $60$ square feet of space occupied by doorways and windows which will not be painted. Thus, the paintable wall area in one bedroom is:
\[
352 - 60 = 292 \text{ square feet}
\]
3. **Calculate the total area to be painted for all bedrooms**:
Isabella has $3$ bedrooms, so the total area to be painted is:
\[
3 \times 292 = 876 \text{ square feet}
\]
Thus, the total number of square feet of walls that must be painted is $\boxed{\textbf{(E) }876}$.
|
The arithmetic mean of 12 scores is 82. When the highest and lowest scores are removed, the new mean becomes 84. If the highest of the 12 scores is 98, what is the lowest score?
|
46
| |
A 9x9 chessboard has its squares labeled such that the label of the square in the ith row and jth column is given by $\frac{1}{2 \times (i + j - 1)}$. We need to select one square from each row and each column. Find the minimum sum of the labels of the nine chosen squares.
|
\frac{1}{2}
| |
The sequence 1,000,000; 500,000; 250,000 and so on, is made by repeatedly dividing by 2. What is the last integer in this sequence?
|
15625
| |
A certain intelligence station has four different passwords A, B, C, and D. Each week, one of the passwords is used, and the password for each week is equally likely to be randomly selected from the three passwords not used in the previous week. If password A is used in the first week, what is the probability that password A is also used in the seventh week? (Express your answer as a simplest fraction.)
|
1/3
| |
A rectangular prism has dimensions of 1 by 1 by 2. Calculate the sum of the areas of all triangles whose vertices are also vertices of this rectangular prism, and express the sum in the form $m + \sqrt{n} + \sqrt{p}$, where $m, n,$ and $p$ are integers. Find $m + n + p$.
|
40
| |
A digital 12-hour clock has a malfunction such that every time it should display a "2", it instead shows a "5". For example, when it is 2:27 PM, the clock incorrectly shows 5:57 PM. What fraction of the day will the clock show the correct time?
|
\frac{5}{8}
| |
A subset $S$ of the set $\{1,2, \ldots, 10\}$ is chosen randomly, with all possible subsets being equally likely. Compute the expected number of positive integers which divide the product of the elements of $S$. (By convention, the product of the elements of the empty set is 1.)
|
\frac{375}{8}
|
For primes $p=2,3,5,7$, let the random variable $X_{p}$ denote the number of factors of $p$ in the product of the elements of $S$, plus 1 . Then we wish to find $\mathbb{E}\left(X_{2} X_{3} X_{5} X_{7}\right)$. If there were only prime powers between 1 and 10, then all $X_{p}$ would be independent. However, 6 and 10 are non-prime powers, so we will do casework on whether these elements are included: - Case 1: none included. Note that $\mathbb{E}\left(X_{2} \mid 6,10 \notin S\right)=1+\frac{1}{2}(1+2+3)=4$, since each of $\{2,4,8\}$ has a $1 / 2$ chance of being included in $S$. Similarly, $\mathbb{E}\left(X_{3} \mid 6,10 \notin S\right)=\frac{5}{2}$ and $\mathbb{E}\left(X_{5} \mid 6,10 \notin S\right)=\mathbb{E}\left(X_{7} \mid 6,10 \notin S\right)=\frac{3}{2}$. The values of $X_{2}, X_{3}, X_{5}$, and $X_{7}$ are independent given that $6,10 \notin S$, so $\mathbb{E}\left(X_{2} X_{3} X_{5} X_{7} \mid 6,10 \notin S\right)=4 \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{3}{2}=\frac{45}{2}$. - Case 2: 6 included. Now, we have $\mathbb{E}\left(X_{2} \mid 6 \in S, 10 \notin S\right)=5$ and $\mathbb{E}\left(X_{3} \mid 6 \in S, 10 \notin S\right)=\frac{7}{2}$, since we know $6 \in S$. We still have $\mathbb{E}\left(X_{5} \mid 6 \in S, 10 \notin S\right)=\mathbb{E}\left(X_{7} \mid 6 \in S, 10 \notin S\right)=\frac{3}{2}$. The values of $X_{2}, X_{3}, X_{5}$, and $X_{7}$ are independent given that $6 \in S$ but $10 \notin S$, so $\mathbb{E}\left(X_{2} X_{3} X_{5} X_{7}\right)$ $6 \in S, 10 \notin S)=5 \cdot \frac{7}{2} \cdot \frac{3}{2} \cdot \frac{3}{2}=\frac{315}{8}$. - Case 3: 10 included. We have $\mathbb{E}\left(X_{2} \mid 10 \in S, 6 \notin S\right)=5$ and $\mathbb{E}\left(X_{5} \mid 10 \in S, 6 \notin S\right)=\frac{5}{2}$, since we know $10 \in S$. We also have $\mathbb{E}\left(X_{3} \mid 10 \in S, 6 \notin S\right)=\frac{5}{2}$ and $\mathbb{E}\left(X_{7} \mid 10 \in S, 6 \notin S\right)=\frac{3}{2}$, hence $\mathbb{E}\left(X_{2} X_{3} X_{5} X_{7} \mid 10 \in S, 6 \notin S\right)=5 \cdot \frac{5}{2} \cdot \frac{5}{2} \cdot \frac{3}{2}=\frac{375}{8}$. - Case 4: 6 and 10 included. We have $\mathbb{E}\left(X_{2} \mid 6,10 \in S\right)=6, \mathbb{E}\left(X_{3} \mid 6,10 \in S\right)=\frac{7}{2}$, and $\mathbb{E}\left(X_{5} \mid 6,10 \in S\right)=\frac{5}{2}$. We still have $\mathbb{E}\left(X_{7} \mid 6,10 \in S\right)=\frac{3}{2}$, hence $\mathbb{E}\left(X_{2} X_{3} X_{5} X_{7} \mid 6,10 \in S\right)=$ $6 \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2}=\frac{315}{4}$ The average of these quantities is $\frac{1}{4}\left(\frac{45}{2}+\frac{315}{8}+\frac{375}{8}+\frac{315}{4}\right)=\frac{375}{8}$, as desired.
|
On a particular street in Waterloo, there are exactly 14 houses, each numbered with an integer between 500 and 599, inclusive. The 14 house numbers form an arithmetic sequence in which 7 terms are even and 7 terms are odd. One of the houses is numbered 555 and none of the remaining 13 numbers has two equal digits. What is the smallest of the 14 house numbers?
(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3, 5, 7, 9 is an arithmetic sequence with four terms.)
|
506
| |
In a certain country, there are exactly 2019 cities and between any two of them, there is exactly one direct flight operated by an airline company, that is, given cities $A$ and $B$, there is either a flight from $A$ to $B$ or a flight from $B$ to $A$. Find the minimum number of airline companies operating in the country, knowing that direct flights between any three distinct cities are operated by different companies.
|
2019
| |
In triangle $ABC$, we have $\angle C = 90^\circ$, $AB = 13$, and $BC = 5$. What is $\tan A$?
|
\frac{5}{12}
| |
Today is 17.02.2008. Natasha noticed that in this date, the sum of the first four digits is equal to the sum of the last four digits. When will this coincidence happen for the last time this year?
|
25.12.2008
| |
Given $f(x)= \frac{1}{4^{x}+2}$, use the method of deriving the sum formula for an arithmetic sequence to find the value of $f( \frac{1}{10})+f( \frac{2}{10})+…+f( \frac{9}{10})$.
|
\frac{9}{4}
| |
A circle passes through the vertex of a rectangle $ABCD$ and touches its sides $AB$ and $AD$ at $M$ and $N$ respectively. If the distance from $C$ to the line segment $MN$ is equal to $5$ units, find the area of rectangle $ABCD$ .
|
25
| |
What is the maximum number of rooks that can be placed in an $8 \times 8 \times 8$ cube so that no two rooks attack each other?
|
64
| |
In a trapezoid with bases 3 and 4, find the length of the segment parallel to the bases that divides the area of the trapezoid in the ratio $5:2$, counting from the shorter base.
|
\sqrt{14}
| |
There are $12$ small balls in a bag, which are red, black, and yellow respectively (these balls are the same in other aspects except for color). The probability of getting a red ball when randomly drawing one ball is $\frac{1}{3}$, and the probability of getting a black ball is $\frac{1}{6}$ more than getting a yellow ball. What are the probabilities of getting a black ball and a yellow ball respectively?
|
\frac{1}{4}
| |
A right triangle has area 5 and a hypotenuse of length 5. Find its perimeter.
|
5+3 \sqrt{5}
|
If $x, y$ denote the legs, then $x y=10$ and $x^{2}+y^{2}=25$, so $x+y+\sqrt{x^{2}+y^{2}}=\sqrt{\left(x^{2}+y^{2}\right)+2 x y}+5=\sqrt{45}+5=5+3 \sqrt{5}$.
|
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