instruction_seed stringlengths 2 10.7k | solution stringlengths 0 10.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
values | question_type stringclasses 5
values | problem_is_valid stringclasses 5
values | solution_is_valid stringclasses 4
values | source stringclasses 11
values | synthetic bool 2
classes |
|---|---|---|---|---|---|---|---|---|
8. Draw the midlines of each face of a regular tetrahedron, resulting in 12 line segments. Among these line segments, the number of "line segment pairs" that are skew lines is $\qquad$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result dire... | 8. 24 Detailed Explanation: 24 "skew segment pairs". Consider any median line $AB$, the face containing $AB$ has no skew segments with $AB$; the other face containing point $A$ has exactly one median line skew to $AB$; the other face containing point $B$ has exactly one median line skew to $AB$; the face not containing... | Other | math-word-problem | Yes | Yes | olympiads | false | |
10.3 On the board, an acute-angled $\triangle A B C$ was drawn, and the feet of the altitudes $A_{1}, B_{1}$ and $C_{1}$ were marked. Then the entire drawing was erased, except for the points $A_{1}, B_{1}$ and $C_{1}$. Can the original $\triangle A B C$ be reconstructed using a compass and a straightedge? | Solution: Angles $\angle A A_{1} B$ and $\angle B B_{1} A$ are equal, so the quadrilateral $A B_{1} A_{1} B$ is cyclic, $\angle B_{1} A_{1} C=\angle A$. Similarly, $\angle C_{1} A_{1} B=\angle A$. Therefore, the sides of triangle $A B C$ are the external angle bisectors of triangle $A_{1} B_{1} C_{1}$. This constructio... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find the integer $n$, $0 \le n \le 13$, such that \[n \equiv 1729 \pmod{13}.\] | To solve the problem, we need to compute $1729 \mod 13$. We start by finding how many times 13 fits into 1729.
\[
1729 \div 13 \approx 133.00
\]
Multiplying back to check:
\[
133 \times 13 = 1729
\]
Since there's no remainder, $1729 \equiv 0 \pmod{13}$. Hence, the integer $n$ satisfying $0 \leq n \leq 13$ and $n \equ... | 0 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Given the function f(x) = $$\begin{cases} 2^{x},x\leq 0 \\ log_{2}x,x>0 \end{cases}$$, find f(f(-2)) = _____. | Since the function f(x) = $$\begin{cases} 2^{x},x\leq 0 \\ log_{2}x,x>0 \end{cases}$$,
We first find f(-2). According to the function definition, for x ≤ 0, we use f(x) = 2^x.
So, f(-2) = 2^(-2) = $\frac {1}{4}$.
Next, we find f(f(-2)), which is equivalent to f($\frac {1}{4}$). According to the function definition, ... | -2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
The expression $z^2 - 6z + 17$ can be rewritten as the sum of a squared binomial and an integer. Find the integer. | We will complete the square for $z^2 - 6z + 17$.
1. The binomial to be squared will be of the form $z + b$. Squaring this binomial gives $z^2 + 2bz + b^2$. We want $2bz$ to equal the linear term of the original expression, $-6z$, thus $b = -3$.
2. Substituting $b = -3$ into the squared binomial, we have $(z - 3)^2... | 8 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Naeun, Leejun, Yoonseo, Seohyun, and Soyul are interviewed in order. When Yoonseo and Seohyeon switched, who will be the 4th person being interviewed? | The original order of interviews is:
1. Naeun
2. Leejun
3. Yoonseo
4. Seohyun
5. Soyul
If Yoonseo and Seohyun switch places, the new order will be:
1. Naeun
2. Leejun
3. Seohyun
4. Yoonseo
5. Soyul
So, the $\boxed{4th}$ person being interviewed after the switch will be Yoonseo. | 4th | Logic and Puzzles | math-word-problem | Yes | Yes | orca_math | true |
Let $E$ be the determinant of the matrix whose column vectors are $\mathbf{a}, \mathbf{b}, \mathbf{c}$. Find the determinant of the matrix whose column vectors are $2\mathbf{a} + \mathbf{b}, \mathbf{b} + 3\mathbf{c}, \mathbf{c} + 4\mathbf{a}$, in terms of $E$. | The determinant $E$ is given by $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$.
Then, the determinant of the matrix whose column vectors are $2\mathbf{a} + \mathbf{b}, \mathbf{b} + 3\mathbf{c}, \mathbf{c} + 4\mathbf{a}$ is given by
\[
(2\mathbf{a} + \mathbf{b}) \cdot ((\mathbf{b} + 3\mathbf{c}) \times (\mathbf{c} +... | 14E | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
"$x > 1$" is the condition for "$\log_{\frac{1}{2}}(x+2) < 0$" to be ( ).
A: A sufficient but not necessary condition
B: A necessary but not sufficient condition
C: A necessary and sufficient condition
D: Neither sufficient nor necessary condition | From $\log_{\frac{1}{2}}(x+2) 1$, which simplifies to $x > -1$.
Therefore, "$x > 1$" is a sufficient but not necessary condition for "$\log_{\frac{1}{2}}(x+2) < 0$".
Hence, the correct option is $\boxed{A}$.
This problem primarily tests the understanding of sufficient and necessary conditions. The key to solving th... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false |
4. Solve the equation $6(x-1)(x+2)-4(x-3)(x+4)=2(x-5)(x-6)$. | 4. $6\left(x^{2}+2 x-x-2\right)-4\left(x^{2}+4 x-3 x-12\right)=2\left(x^{2}-6 x-5 x+30\right) \quad 1$ BOD
$6\left(x^{2}+x-2\right)-4\left(x^{2}+x-12\right)=2\left(x^{2}-11 x+30\right) \quad 1$ BOD
$6 x^{2}+6 x-12-4 x^{2}-4 x+48=2 x^{2}-22 x+60 \quad$ 1 BOD
$6 x^{2}-4 x^{2}-2 x^{2}+6 x-4 x+22 x=60-48+12 \quad 1$ BOD... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2870 korona is to be divided into 4 parts, so that the ratio of the parts is as follows:
$$
A: C=2: 3, \quad D: B=3: 4, \quad C: E=4: 5, \quad D: A=6: 5
$$ | From the given ratios, it follows that
$$
D=\frac{6}{5} A, \quad B=\frac{8}{5} A, \quad C=\frac{3}{2} A, \quad E=\frac{15}{8} A
$$
thus
$$
A+\frac{8}{5} A+\frac{3}{2} A+\frac{6}{5} A+\frac{15}{8} A=2870
$$
from which
$$
A=400 K, \quad \text{so} \ B=640 K, \quad C=600 K, \quad D=480 K, \quad E=750 K
$$
(Imre Kürti... | A=400K,B=640K,C=600K,D=480K,E=750K | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Zoe, along with her four friends, formed a recycling team for their class. They collected paper, plastic, and aluminum cans for recycling. For every 12 pounds of paper, 6 pounds of plastic, and 4 pounds of aluminum they recycled, they earned one point. Zoe recycled 35 pounds of paper, 15 pounds of plastic, and 5 pounds... | To calculate the total points earned by the recycling team, we need to determine the points earned by each member for each type of recyclable material and then sum them up.
For Zoe:
- Paper: 35 pounds / 12 pounds per point = 2 points (with 11 pounds of paper left over)
- Plastic: 15 pounds / 6 pounds per point = 2 poi... | 28 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
the average age of a group of persons going for picnic is 16 years . 20 new persons with an average age of 15 years join the group on the spot due to which their average age becomes 15.5 years . the number of persons initially going for picnic is ? | Let's denote the initial number of persons going for the picnic as \( N \) and the total age of the group as \( T \).
From the given information, we have:
1. The average age of the initial group is 16 years, so the total age of the group is \( T = 16N \).
2. When 20 new persons with an average age of 15 years join t... | 20 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
11.4. In a right-angled trapezoid $A B C D$, with right angles at vertices $A$ and $B$, there are two circles. One of them is tangent to the lateral sides and to the larger base $A D$, while the other is tangent to the lateral sides, to the smaller base $B C$, and to the first circle.
1) The line passing through the c... | ## Solution.
1) Since the circles are tangent to each other, we construct their common tangent $L Z$. Let: $m(\angle D)=\alpha, R$ - the radius of the larger circle with center at $O_{1}, r$ - the radius of the smaller circle with center at $O_{2}, M$ - the foot of the perpendicular from $O_{1}$ to $A D, N$ - the foot... | \frac{116}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A student needs to investigate and analyze the changes in math scores of eighth-grade students in the school. The following are the disordered statistical steps: ① Draw a line chart to represent the changes in scores; ② Collect the math scores of each student who moved from seventh grade to eighth grade; ③ Analyze the ... | To solve the problem of determining the correct order of statistical steps for analyzing the changes in math scores of eighth-grade students, we follow these logical steps:
1. **Collect the Math Scores:** The first step in any data analysis is to gather the necessary data. In this context, the data needed is the math ... | D | Other | MCQ | Yes | Yes | cn_k12 | false |
(1) Given the set $A=\{x|3\leq x<7\}$, and $B=\{x|2<x<10\}$, find $(\complement_R A) \cap B$
(2) Let the non-empty set $B=\{x|x=\log_{2}m\}$, if $B \subseteq \{1,2\}$, find the values of the real number $m$. | (1) Since the set $A=\{x|3\leq x<7\}$,
then $\complement_R A=\{x|x<3 \text{ or } x\geq7\}$,
Since $B=\{x|2<x<10\}$,
then $(\complement_R A) \cap B=\{x|2<x<3 \text{ or } 7\leq x<10\}$.
(2) Since the non-empty set $B=\{x|x=\log_{2}m\}$, if $B \subseteq \{1,2\}$,
then $\log_{2}m=1$ or $\log_{2}m=2$,
thus $m=2^... | 2 \text{ or } 4 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Find the largest prime divisor of the number represented by $100111001_6$ in decimal form. | First, we convert $100111001_6$ from base-6 to a decimal representation. This can be written as:
$$
100111001_6 = 6^8 + 6^5 + 6^4 + 6^3 + 6 + 1
$$
Letting $x = 6$, the number is:
$$
x^8 + x^5 + x^4 + x^3 + x + 1
$$
Again, using the Rational Root Theorem, $x+1$ is a factor. The polynomial simplifies as:
$$
x^8 + x^5 +... | 43 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Wario is a field goal kicker on the high school football team. He attempts 60 field goals throughout the season. He misses a certain fraction of the field goals. Of the field goals missed, 20 percent were wide right. Wario missed 3 field goals wide right. What fraction of the field goals did Wario miss? | If Wario missed 3 field goals wide right and those represent 20 percent of the field goals he missed, we can set up the following equation to find the total number of missed field goals:
0.20 * Total Missed Field Goals = 3
To find the total number of missed field goals, we divide both sides of the equation by 0.20:
... | 1/4 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Factorize: \\({a}^{3}-16a=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\) | **Step 1: Identify the problem**
This problem primarily tests the understanding of factoring by grouping and the difference of squares formula. Familiarity with the structure of the difference of squares formula is key to solving this problem.
**Step 2: Factorize by grouping**
Factorize the expression \\({a}^{3}-16a\\... | a(a+4)(a-4) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
If "x∈\[2,5\] or x∈{x|x4}" is a false statement, then the range of x is \_\_\_\_\_\_\_\_\_\_\_. | The statement "x∈\[2,5\] or x∈{x|x4}" is false. This means x cannot be in the interval \[2,5\] nor can it be less than 1 or greater than 4. Therefore, the range of x is from 1 to 2, not including 2.
Thus, the range of x is $\boxed{[1,2)}$. | [1,2) | Algebra | math-word-problem | Incomplete | Yes | cn_k12 | false |
\section*{Problem 1 - 081241}
Every nonnegative periodic decimal fraction represents a rational number that can also be expressed in the form \(\frac{p}{q}\) (where \(p\) and \(q\) are natural numbers and coprime, \(p \geq 0, q>0\)).
Let \(a_{1}, a_{2}, a_{3}\) and \(a_{4}\) be digits used to represent numbers in the... | }
For digits \(a, b, c, d \in\{0,1, \ldots, 9\}\), we denote with \([a b c d]\) the (up to) four-digit number
\[
[a b c d]:=1000 a+100 b+10 c+d
\]
Then \(10000 z_{1}-z_{1}=\left[a_{1} a_{2} a_{3} a_{4}\right]\), so \(z_{1}=\frac{\left[a_{1} a_{2} a_{3} a_{4}\right]}{9999}\). Similarly, \(z_{2}=\frac{\left[a_{4} a_{1... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
Given $f(x)= \begin{cases} f(x-5),x\geqslant 0 \\ \log _{3}(-x),x < 0\\ \end{cases}$, find the value of $f(2017)$. | From $f(x)= \begin{cases} f(x-5),x\geqslant 0 \\ \log _{3}(-x),x < 0\\ \end{cases}$,
We can see that when $x\geqslant 0$, $f(x+5)=f(x+5-5)=f(x)$,
This implies that $f(x+5k)=f(x)$, where $k$ is a positive integer,
Thus, $f(2017)=f(403\times5+2)=f(2)=f(-3)=\log _{3}3=1$.
So the answer is: $\boxed{1}$.
This is becaus... | 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
After moving the line $y=3x$ down by $2$ units, the resulting line is ( )
A: $y=3x+2$
B: $y=3x-2$
C: $y=3\left(x+2\right)$
D: $y=3\left(x-2\right)$ | To solve this problem, let's analyze the given information and the transformation applied to the line.
The original line is given by the equation $y = 3x$. This equation is in the slope-intercept form $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. For the given line, the slope $m = 3$ and the y-inte... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Iterations $\quad]$ [ Binary Number System ]
A flock of geese was flying. On each lake, half of the geese and another half-goose landed. The rest continued flying. All the geese landed on $n$ lakes.
How many geese were there in the flock? | ## Answer
$2^{n}-1$. | 2^{n}-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. If $M=\left\{(x, y)|| \tan \pi y \mid+\sin ^{2} \pi x=0\right\}, N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant 2\right\}$, then the number of elements in $M \cap N$ is
A. 4
B. 5
C. 8
D. 9 | $M \cap N=\{(x, y) \mid x, y=-1,0,1\}, M \cap N$ has 9 elements. Therefore, the answer is $D$. | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
An aquarium holds an equal number of clownfish and blowfish. Some of the blowfish stay in their own tank, and the remaining blowfish swim into a display tank. An equal number of clownfish join the blowfish in the display tank, but then a third of these clownfish swim back into their own tank. The aquarium holds a combi... | Let's denote the total number of clownfish as C and the total number of blowfish as B. Since the aquarium holds an equal number of clownfish and blowfish, we have:
C = B
The problem states that the aquarium holds a combined total of 100 fish, so:
C + B = 100
Since C = B, we can say:
2C = 100
Dividing both sides b... | 26 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A pool can be filled by one particular pipe in some hours, whereas a different pipe requires 6 h to fill the pool. It takes 3.75 hours to fill the pool with both pipes turned on. How long would it take for the first pipe to fill the pool alone? | Let's denote the time it takes for the first pipe to fill the pool alone as \( T \) hours. The rate at which the first pipe fills the pool is \( \frac{1}{T} \) of the pool per hour.
The second pipe fills the pool in 6 hours, so its rate is \( \frac{1}{6} \) of the pool per hour.
When both pipes are turned on, their c... | 8.18 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A train, initially traveling at 100 kmph, accelerates at a rate of 0.5 m/s^2. Meanwhile, a motorbike, initially traveling at 64 kmph, accelerates at a rate of 0.3 m/s^2. Given that the train overtakes the motorbike in 85 seconds, what is the length of the train in meters? | First, let's convert the initial speeds from km/h to m/s to be consistent with the acceleration units.
For the train:
100 km/h = (100 * 1000 m) / (3600 s) = 27.78 m/s
For the motorbike:
64 km/h = (64 * 1000 m) / (3600 s) = 17.78 m/s
Now, let's calculate the distance each vehicle travels in 85 seconds using the formu... | 1572.5 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Ariel is collecting fish. She has 45 fish. Some fraction of them are male, and she has 15 female fish. What fraction of her fish are male? | To find the fraction of male fish, we need to subtract the number of female fish from the total number of fish to get the number of male fish. Then we can express that number as a fraction of the total number of fish.
Number of male fish = Total number of fish - Number of female fish
Number of male fish = 45 - 15
Numb... | 2/3 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
## Task 3
In ski jumping, the longest jump was 80 meters. The shortest jump was 67 meters. What is the difference in meters between these two jumps? | $67+x=80, x=13$. Between the two jumps, there was a difference of 13 meters. | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
If you double the length of each edge of a cube, how many times bigger does the cube's surface area become? | If you double the length of each edge of a cube, the new edge length will be 2 times the original edge length.
Let's say the original edge length of the cube is "a". The surface area of the original cube is given by the formula:
Surface area of original cube = 6 * a^2
When you double the edge length, the new edge le... | 4 | Geometry | math-word-problem | Yes | Yes | orca_math | true |
The equation of the asymptotes of the hyperbola $\frac{x^2}{4}-\frac{y^2}{3}=1$ is ____. | To find the equations of the asymptotes of the hyperbola $\frac{x^2}{4} - \frac{y^2}{3} = 1$, we start with the given equation and rewrite it for the asymptotes by setting the right side to $0$, resulting in $\frac{x^2}{4} - \frac{y^2}{3} = 0$. We can then solve for $y$ in terms of $x$:
\[
\begin{align*}
\frac{x^2}{4}... | \sqrt{3}x \pm 2y = 0 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Each successive term in the sequence $4096, 1024, 256, z, w, 4, 1, \frac{1}{4},...$ is obtained by multiplying the previous term by a constant. What is the value of $z + w$? | Let the constant multiplicand in this sequence be $r$.
1. **From 4 to 1:** \(4r = 1\)
\[ r = \frac{1}{4} \]
2. **From 256 to \(z\):** \(256r = z\)
\[ z = 256 \cdot \frac{1}{4} = 64 \]
3. **From z(64) to \(w\):** \(64r = w\)
\[ w = 64 \cdot \frac{1}{4} = 16 \]
Summing \(z\) and \(w\),
\[ z + w = 64 + 16 = \... | 80 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Combine like terms: $m-(m^{2}n+3m-4n)+(2nm^{2}-3n)$. | To combine like terms in the expression $m-(m^{2}n+3m-4n)+(2nm^{2}-3n)$, we follow these steps:
1. Distribute the negative sign into the parentheses:
$$= m - m^{2}n - 3m + 4n + 2m^{2}n - 3n$$
2. Combine like terms by adding or subtracting coefficients of the same variables:
$$= (2m^{2}n - m^{2}n) + (-3m + m) + (4n -... | m^{2}n - 2m + n | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A chemist mixes one liter of pure water with 1 liter of a 30% salt solution. What is the concentration of the resulting mixture? | To find the concentration of the resulting mixture, we need to determine the total amount of salt and the total volume of the mixture.
First, let's calculate the amount of salt in the 1 liter of 30% salt solution:
Amount of salt = Volume of solution × Concentration of salt
Amount of salt = 1 liter × 30%
Amount of salt... | 15\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Let the center of the circumscribed circle of the equilateral triangle $ABC$ be $O$, and the other endpoint of the diameter starting from point $A$ be $D$. Draw the (smaller than a semicircle) arcs with centers at $A$ and $D$, respectively, passing through $B$ and $C$. Let one of the points of these two arcs be $P$, an... | Let the projections of $P$ and $Q$ onto the diameter $AD$ be $P'$ and $Q'$, respectively, and denote the radius of the circumcircle of the triangle by $r$. Thus, the radius of the arc $BC$ centered at $D$ is also $r$, and the length of the side of the triangle and the radius of the arc $BC$ centered at $A$ is $AB = r \... | proof | Geometry | proof | Yes | Yes | olympiads | false |
In triangle $ABC$, the sides opposite to angles $A$, $B$, and $C$ are denoted by $a$, $b$, and $c$ respectively. Given that angle $A = \frac{\pi}{4}$, $\sin A + \sin (B - C) = 2\sqrt{2}\sin 2C$, and the area of triangle $ABC$ is $1$. Find the length of side $BC$. | In triangle $ABC$, since
$$\sin A + \sin (B - C) = 2\sqrt{2}\sin 2C,$$
and using the angle addition formula for sine, we have
$$\sin A + \sin B \cos C - \cos B \sin C = 2\sqrt{2}\sin C \cos C.$$
From angle $A = \frac{\pi}{4}$, we know that $\sin A = \cos A = \frac{\sqrt{2}}{2}$, thus the equation simplifies to
$$\frac... | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
6. As shown in Figure 1, quadrilateral $ABCD$ is a unit square, $BD=BE$, $CE \parallel BD$. Then the length of $CE$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{3}$
(C) $\frac{\sqrt{3}}{4}$
(D) $\frac{\sqrt{6}-\sqrt{2}}{2}$ | 6. D.
As shown in Figure 3, draw $C O \perp$
$B D$ at point $O$, and $E F \perp B D$
at point $F$. Then $O$ is the center of the square $A B C D$.
It is easy to see that quadrilateral $C E F O$ is a rectangle.
$$
\text { Also, } B E=B D=\sqrt{2}, B O=C O=E F=\frac{\sqrt{2}}{2} \text {, }
$$
Thus, the three interior ... | \frac{\sqrt{6}-\sqrt{2}}{2} | Geometry | MCQ | Yes | Yes | olympiads | false |
Among the following calculations, which one is correct?
A: $2+\sqrt {5}=2\sqrt {5}$
B: $4\sqrt {5}-3\sqrt {5}=1$
C: $\sqrt {x^{2}+y^{2}}=x+y$
D: $\sqrt {45}-\sqrt {20}=\sqrt {5}$ | To evaluate each option step-by-step:
**Option A:** $2+\sqrt{5}$ cannot be simplified directly to $2\sqrt{5}$ because one term is a rational number and the other is an irrational number. They are not "like terms" and thus cannot be combined in this manner. Therefore, option A is incorrect.
**Option B:** When simplify... | \text{D} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
The problem involves a curve $C_1$ in the rectangular coordinate system $xOy$ with the polar coordinate equation $\rho \cos \theta = 4$. Point $M$ moves on the curve $C_1$, and point $P$ is on the line segment $OM$ satisfying $|OM| \cdot |OP| = 16$. The tasks are:
(1) Find the rectangular coordinate equation of the tra... | (1) First, we find the rectangular coordinate equation of curve $C_1$. Since $\rho \cos \theta = 4$, we have $x = 4$ (as $x = \rho \cos \theta$ in polar coordinates). Let $P$ have coordinates $(x, y)$ and $M$ have coordinates $(4, y_0)$. Then, the slope of line $OM$ is $\frac{y_0}{4}$, which equals the slope of line $O... | \frac{1}{2}(2)(2 + \sqrt{3}) \cdot \frac{\sqrt{3}}{2} = 1 + \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
Given the pattern: $1-\frac{1}{2}=\frac{1}{2}$; $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}=\frac{1}{3}+\frac{1}{4}$; $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}$; $\cdots \quad \quad \cdots$. Determine the $n$th equation based on this pattern: $\_\_\_\_\_\_\_\_\_\_\_\_... | **Analysis**
This problem tests the ability to observe, analyze, conjecture, and induce the general term formula of a sequence, and it also tests reasoning and computational abilities. It is a basic problem.
From the given information, it can be determined that the $n$th equation contains $2n$ terms. The odd terms are... | 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n} = \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$, $5$, and $8$, while those of $T'$ have lengths $a$, $a$, and $b$. Which of the following numbers is closest to $b$?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{... | The area of $T$ is $\dfrac{1}{2} \cdot 8 \cdot 3 = 12$ and the perimeter is 18.
The area of $T'$ is $\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ and the perimeter is $2a + b$.
Thus $2a + b = 18$, so $2a = 18 - b$.
Thus $12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$, so $48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the number of square units in the area of the irregular hexagon shown below?
[asy]
unitsize(0.5cm);
defaultpen(linewidth(0.7)+fontsize(10));
dotfactor = 4;
int i,j;
for(i=0;i<=5;++i)
{
for(j=-4;j<=4;++j)
{
dot((i,j));
}
}
draw((0,0)--(1,3)--(4,4)--(5,0)--(4,-4)--(1,-3)--cycle,linewidth(2)... | Break the hexagon into two trapezoids and calculate their areas:
- For the upper trapezoid, bases are 3 units and 2 units, the height is 4 units. Area = $\frac{1}{2} \cdot (3 + 2) \cdot 4 = 10$ square units.
- For the lower trapezoid, bases are 3 units and 2 units, the height is 4 units. Area = $\frac{1}{2} \cdot (3 + ... | 20 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
Example 28 Several people gather, some of whom know each other. It is known that if two people in the gathering have the same number of acquaintances, then they do not have any common acquaintances. Prove: If there is someone in the gathering who has at least 2008 acquaintances, then there must also be someone who has ... | We consider the person with the most acquaintances (if there are several such people, we choose any one of them), denoted as $A$, and suppose $A$ knows $n$ people, who are sequentially denoted as $B_{1}, B_{2}, \cdots, B_{n}$. Since any two people $B_{i}$ and $B_{j}$ among $B_{1}, B_{2}, \cdots, B_{n}$ know $A$, i.e., ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
[ Average values
Vasya, in a rage, cuts a rectangular sheet of paper with scissors. Every second, he cuts the first piece he comes across with a random straight cut into two parts.
a) Find the expected number of sides of the polygon that will randomly fall to Vasya after an hour of such work
b) Solve the same proble... | b) Let as a result of Vasya's activity, there are $k+1$ polygons on the table, including $p_{3}$ triangles, $p_{4}$ quadrilaterals, $\ldots, p_{m} m$-gons ($m$ - the largest possible number of sides of the resulting polygon). Then the expected number of sides of the selected polygon is $1 / k+1\left(3 p_{3}+4 p_{4}+\ld... | 4;\frac{n+14400}{3601} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Given vectors $\overrightarrow{a}=(1,0)$, $\overrightarrow{b}=(-1,1)$, then ( )
A: $\overrightarrow{a}// \overrightarrow{b}$
B: $\overrightarrow{a}\perp \overrightarrow{b}$
C: $(\overrightarrow{a}- \overrightarrow{b})// \overrightarrow{b}$
D: $(\overrightarrow{a}+ \overrightarrow{b})\perp \overrightarrow{a}$ | According to the problem, we analyze the options in order:
For option A, with vectors $\overrightarrow{a}=(1,0)$, $\overrightarrow{b}=(-1,1)$, since $1\times1 \neq 0\times(-1)$, it implies that $\overrightarrow{a}// \overrightarrow{b}$ does not hold, so A is incorrect;
For option B, with vectors $\overrightarrow{a}... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false |
10,11 |
| :---: | :---: | :---: |
| | [ Graph theory (other) | |
| | Divisibility of numbers. General properties | |
| | Chessboard coloring | |
Author: Shapovostov $A . B$.
Can natural numbers be placed at the vertices of a cube so that in each pair of numbers connected by an edge, one of them is divisible by ... | Let's mark four vertices of a cube, none of which are connected by an edge, and place four different prime numbers in them (for example, 2, 3, 5, 7). In each of the remaining vertices, place the product of the numbers in the marked vertices connected to it by edges (there are exactly three of them). It is easy to see t... | proof | Number Theory | proof | Incomplete | Yes | olympiads | false |
A bag contains eight blue balls and five yellow balls. A ball is taken from the bag, its color noted, and then it is placed back in the bag. A second ball is then taken and its color noted. What is the probability that the two balls are of the same color? | We can have either two blue balls or two yellow balls.
- The probability of drawing two blue balls is \(\left(\frac{8}{13}\right)^2 = \frac{64}{169}\).
- The probability of drawing two yellow balls is \(\left(\frac{5}{13}\right)^2 = \frac{25}{169}\).
- Therefore, the probability that both balls drawn are of the same c... | \frac{89}{169} | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
Suppose that $4^a = 5$, $5^b = 6$, $6^c = 7$, and $7^d = 8$. What is $a \cdot b\cdot c \cdot d$?
$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{3}{2} \qquad \textbf{(C) } 2 \qquad \textbf{(D) } \frac{5}{2} \qquad \textbf{(E) } 3$ | \begin{align*} 8&=7^d \\ 8&=\left(6^c\right)^d\\ 8&=\left(\left(5^b\right)^c\right)^d\\ 8&=\left(\left(\left(4^a\right)^b\right)^c\right)^d\\ 8&=4^{a\cdot b\cdot c\cdot d}\\ 2^3&=2^{2\cdot a\cdot b\cdot c\cdot d}\\ 3&=2\cdot a\cdot b\cdot c\cdot d\\ a\cdot b\cdot c\cdot d&=\boxed{\textbf{(B) }\dfrac{3}{2}}\\ \end{align... | \textbf{(B)}\frac{3}{2} | NaN | NaN | NaN | NaN | amc_aime | false |
Consider the 2004th, 2005th, and 2006th rows of Pascal's triangle. Let $(a_i)$, $(b_i)$, and $(c_i)$ represent the sequence, from left to right, of elements in these rows, respectively, with the leftmost element occurring at $i = 0.$ Compute
\[
\sum_{i = 0}^{2005} \frac{b_i}{c_i} - \sum_{i = 0}^{2004} \frac{a_i}{b_i}.
... | From Pascal's triangle, the entries in rows $n-1$, $n$, and $n+1$ are given by
\[
a_i = \binom{n-1}{i}, \ b_i = \binom{n}{i}, \ c_i = \binom{n+1}{i}.
\]
For $n = 2005$, we have:
\[
\frac{a_i}{b_i} = \frac{\binom{2004}{i}}{\binom{2005}{i}} = 1 - \frac{i}{2005}
\]
and
\[
\frac{b_i}{c_i} = \frac{\binom{2005}{i}}{\binom{20... | 0.5 | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
Wario is a field goal kicker on the high school football team. He attempts 60 field goals throughout the season. He misses 1/4 of the field goals. Of the field goals missed, a certain percentage were wide right. Wario missed 3 field goals wide right. What percentage of the missed field goals went wide right? | First, we need to determine how many field goals Wario missed in total. Since he misses 1/4 of the field goals and he attempts 60, we calculate:
Missed field goals = 1/4 * 60 = 15
Now, we know that Wario missed 3 field goals wide right. To find the percentage of the missed field goals that went wide right, we divide ... | 20\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A boy needs to cover a distance of 1800 meters in 30 minutes. Find his average speed in meters per second. | To find the average speed, we need to divide the total distance by the total time taken.
The total distance is 1800 meters.
The total time taken is 30 minutes. To convert minutes to seconds, we multiply by 60 (since there are 60 seconds in a minute).
30 minutes * 60 seconds/minute = 1800 seconds
Now, we can calcula... | 1 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Let's inscribe an equilateral cone into a sphere. At what distance from the center of the sphere should we cut these two bodies with a plane parallel to the base of the cone so that the difference in the areas of the sections is equal to the area of the base of the cone? | Let the sought distance be $x$, the radius of the sphere be $r$, the radius of the spherical cap be $y$, and the radius of the conical section be $z$. According to the problem:
$$
y^{2} \pi - z^{2} \pi = \frac{3}{4} r^{2} \pi
$$
or
$$
y^{2} - z^{2} = \frac{3}{4} r^{2}
$$
But
$$
y^{2} = r^{2} - x^{2}
$$
and
$$
z^... | \frac{r}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $y$ be an angle such that $\tan y = \frac{p}{q}$ and $\tan 3y = \frac{q}{p + q}$. Then the least positive value of $y$ equals $\tan^{-1} m$. Compute $m$. | Given $\tan y = \frac{p}{q}$ and $\tan 3y = \frac{q}{p+q}$, we can express $\tan 3y$ using $\tan y$:
\[\tan 3y = \frac{q}{p + q} = \frac{1}{\frac{p}{q} + 1} = \frac{1}{\tan y + 1},\]
so $(\tan y + 1) \tan 3y = 1$. Using the triple angle formula for tangent,
\[\tan 3y = \frac{3\tan y - \tan^3 y}{1 - 3\tan^2 y},\]
we ge... | 1 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Let's determine the positive integers $n$ for which we can write one of the letters $I, M, O$ in each cell of an $n \times n$ table such that:
- in every row and every column, one third of the cells contain the letter $I$, one third contain the letter $M$, and one third contain the letter $O$; and
- in every diagonal,... | Lajkó Kálmán's solution. Let's find such $n \times n$ squares! If a square has a valid filling, then $3 \mid n$, which is obvious.
It's not hard to see that if we have a filling for an $n \times n$ square, then we also have one for any $(k n) \times (k n)$ square, since we can pack $k^2$ well-filled $n \times n$ squar... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $x_{1}, x_{2}, \ldots, x_{n}$ be a sequence where each term is 0, 1, or -2. If
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=-5 \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=19
\end{array}\right.
$$
determine $x_{1}^{5}+x_{2}^{5}+\cdots+x_{n}^{5}$. | Let $a$ be the number of terms equal to 1 and $b$ be the number of terms equal to -2. We can write:
$$
\left\{\begin{array} { l }
{ a \cdot 1 + b \cdot ( - 2 ) = - 5 } \\
{ a \cdot 1 ^ { 2 } + b \cdot ( - 2 ) ^ { 2 } = 1 9 }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
a-2 b=-5 \\
a+4 b=19
\end{array}\righ... | -125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 2-50 Find the number of n-bit binary numbers in which the pattern 010 appears for the first time in the last 3 bits. | To find the number of $n$-bit binary numbers $b_{1} b_{2} \cdots b_{n}$ that, when scanned from left to right, have the pattern 010 appear for the first time in the last 3 bits. Naturally, excluding the last 3 bits, no other 3 consecutive bits should form the pattern 010.
Let $a_{n}$ denote the number of $n$-bit numbe... | a_{3}=1,a_{4}=2,a_{5}=3,a_{6}=5,a_{7}=9,a_{8}=16,a_{9}=28,a_{10}=49,\cdots | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Rthea, a distant planet, is home to creatures whose DNA consists of two (distinguishable) strands of bases with a fixed orientation. Each base is one of the letters H, M, N, T, and each strand consists of a sequence of five bases, thus forming five pairs. Due to the chemical properties of the bases, each pair must c... | Answer: $12 \cdot 7^{4}$ or 28812
There are $4 \cdot 3=12$ ways to choose the first base pairs, and regardless of which base pair it is, there are 3 possibilities for the next base on one strand and 3 possibilities for the next base on the other strand. Among these possibilities, exactly 2 of them have identical bases ... | 28812 | NaN | NaN | NaN | NaN | olympiads | false |
Given lines $l$ and $m$, and planes $\alpha$ and $\beta$, which of the following conditions can conclude that $\alpha \parallel \beta$? ( )
A: $l \subset \alpha$, $m \subset \alpha$ and $l \parallel \beta$, $m \parallel \beta$
B: $l \subset \alpha$, $m \subset \beta$ and $l \parallel m$
C: $l \perp \alpha$, $m \perp \... | Since $l \subset \alpha$, $m \subset \alpha$ and $l \parallel \beta$, $m \parallel \beta$,
when $l$ intersects with $m$, $\alpha \parallel \beta$; when $l$ does not intersect with $m$, $\alpha$ and $\beta$ either intersect or are parallel, hence option A is incorrect;
Since $l \subset \alpha$, $m \subset \beta$ and... | C | Geometry | MCQ | Yes | Yes | cn_k12 | false |
in a class , there are 18 boys who are over 160 cm tall . if these constitute 3 - fourths of the boys and the total number of boys is two - thirds of the total number of students in the class , what is the number of girls in the class ? | Let's denote the total number of boys in the class as B and the total number of students in the class as S. According to the information given:
1. The 18 boys who are over 160 cm tall constitute three-fourths of the boys in the class. Therefore, we can write:
(3/4) * B = 18
2. The total number of boys is two-third... | 12 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
the maximum number of student amoung them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | To find the maximum number of students among whom 1001 pens and 910 pencils can be distributed equally, we need to find the greatest common divisor (GCD) of the two numbers, 1001 and 910. The GCD will give us the largest number of students that can receive the same number of pens and pencils.
Let's calculate the GCD o... | 91 | Number Theory | math-word-problem | Yes | Yes | orca_math | true |
Determine the roots of unity in the field of $ p$-adic numbers.
[i]L. Fuchs[/i] | null | null | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Given \\(a \in \mathbb{R}\\), the equation \\(a^{2}x^{2}+(a+2)y^{2}+4x+8y+5a=0\\) represents a circle. The coordinates of the center of the circle are \_\_\_\_\_\_, and the radius is \_\_\_\_\_\_. | Since the equation \\(a^{2}x^{2}+(a+2)y^{2}+4x+8y+5a=0\\) represents a circle,
it follows that \\(a^{2}=a+2 \neq 0\\). Solving this, we get \\(a=-1\\) or \\(a=2\\).
When \\(a=-1\\), the equation becomes \\(x^{2}+y^{2}+4x+8y-5=0\\).
Completing the square, we get \\((x+2)^{2}+(y+4)^{2}=25\\). The center of the circl... | 5 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Solve the inequality:
\[\frac{2}{x - 1} - \frac{3}{x - 2} + \frac{5}{x - 3} - \frac{2}{x - 4} < \frac{1}{20}.\] | Start by subtracting $\frac{1}{20}$ from both sides:
\[\frac{2}{x - 1} - \frac{3}{x - 2} + \frac{5}{x - 3} - \frac{2}{x - 4} - \frac{1}{20} < 0.\]
Combine all terms over a common denominator:
\[\frac{40(x-2)(x-3)(x-4) - 60(x-1)(x-3)(x-4) + 100(x-1)(x-2)(x-4) - 40(x-1)(x-2)(x-3) - (x-1)(x-2)(x-3)(x-4)}{20(x-1)(x-2)(x-3)... | (-\infty, -2) \cup (-1, 1) \cup (2, 3) \cup (4, 6) \cup (7, \infty) | Inequalities | math-word-problem | Yes | Yes | synthetic_math | true |
In triangle $ABC$ where $\angle BAC = 60^\circ$, side $AB = 20$ and side $BC = 21$. Find $\tan A$. | First, we apply the Law of Cosines to find side $AC$:
\[ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos A \]
\[ AC^2 = 20^2 + 21^2 - 2 \cdot 20 \cdot 21 \cdot \cos 60^\circ \]
\[ AC^2 = 400 + 441 - 840 \cdot \frac{1}{2} \]
\[ AC^2 = 841 - 420 \]
\[ AC^2 = 421 \]
\[ AC = \sqrt{421} \]
Now, apply the Law of Sines t... | \tan A = \frac{21\sqrt{3}}{2\sqrt{421 - \frac{1323}{4}}} | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
Given the function $f(x)= \begin{cases} (\frac{1}{2})^{x},x f(k)$. | First, compute $f(-2)$. According to the definition of the function, when $x f(k)$ is equivalent to $f(k) < 9$.
We now consider two cases:
1. If $k < 0$, then $f(k) = (\frac{1}{2})^{k}$. Solving the inequality $(\frac{1}{2})^{k} < 9$ gives $\log_{\frac{1}{2}}9 < k < 0$.
2. If $k \geqslant 0$, then $f(k) = (k-1)^{2}... | \log_{\frac{1}{2}}9 < k < 4 | Algebra | math-word-problem | Incomplete | Yes | cn_k12 | false |
The monotonic increasing interval of the function $y=\log_{2}(1-x^{2})$ is. | Since the function $y=\log_{2}(1-x^{2})$ is defined, we have $1-x^{2}>0 \Rightarrow (x+1)(x-1)1$, the monotonic decreasing interval of the function $y=\log_{2}(1-x^{2})$ is the same as the monotonic decreasing interval of $g(x)=1-x^{2}$.
For $y=g(x)=1-x^{2}$, which opens downwards and has the axis of symmetry at $x=0$... | (-1, 0) | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
## Aufgabe 2
$$
\begin{array}{ll}
2+2+2 & 5+5-5 \\
8+0-8 & 9-2-6
\end{array}
$$
|
$2+2+2=6 ; 5+5-5=5 ; 8+0-8=0 ; 9-2-6=1$
| 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The left and right foci of a hyperbola are $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. If $\triangle F_{1} A B$ is an equilateral triangle, then the eccentricity of the hyperbola is $\qquad$ | 2. $\sqrt{3}$.
Let the equation of the hyperbola be
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0) \text {, }
$$
with the semi-focal distance being $c$.
Then $c^{2}=a^{2}+b^{2}$.
By $\left|F_{1} A\right|-\left|F_{2} A\right|=\left|F_{1} B\right|-\left|F_{2} B\right|=2 a$,
$$
\left|F_{1} A\right|=\left|F_{2} A... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Solve the equation $4^x - 2^{x+2} - 12 = 0$. | Let $2^x = t$ (where $t > 0$), then the original equation can be transformed into: $t^2 - 4t - 12 = 0$.
Solving this equation, we get: $t = 6$ or $t = -2$ (discard $t = -2$ since $t > 0$).
Therefore, $x = \log_2{6} = 1 + \log_2{3}$.
Thus, the solution set of the original equation is $\boxed{\{x | x = 1 + \log_2{3}\}... | \{x | x = 1 + \log_2{3}\} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Find the maximum value of $\tan^6 x + \sec^4 x.$ | Start with the identity $\sec^2 x = 1 + \tan^2 x$. Raise each side of this identity to the power of 2 to get:
\[\sec^4 x = (1 + \tan^2 x)^2 = 1 + 2\tan^2 x + \tan^4 x.\]
Now create an expression to calculate with $\tan^6 x$ and $\sec^4 x$:
\[\tan^6 x + \sec^4 x = \tan^6 x + 1 + 2\tan^2 x + \tan^4 x.\]
Rearrange and fac... | \infty | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
A square with horizontal and vertical sides is drawn on the plane. It held several segments parallel to the sides, and there are no two segments which lie on one line or intersect at an interior point for both segments. It turned out that the segments cuts square into rectangles, and any vertical line intersecting the ... | 1. **Restate the Problem:**
Consider a square \( U = [0,1]^2 \) in the Cartesian coordinate system. The square is partitioned into rectangles by several segments parallel to the sides of the square. The conditions are:
- Any vertical line \( x = a \) intersects exactly \( k \) rectangles.
- Any horizontal line... | k\ell | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
How many positive four-digit integers of the form $\_\_20$ are divisible by 40? | Let the four-digit integer be $ab20$, where $a$ and $b$ are digits. We can simplify our check by considering $ab00$ instead of $ab20$, as subtraction of 20 does not affect divisibility by 40.
A number is divisible by 40 if and only if it is divisible by both 8 and 5. Since $ab00 = ab \times 100 = ab \times (2^2 \cdot... | 12 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Four. (This question is worth 18 points) Let $\triangle A B C$ be an equilateral triangle with side length 6. Draw a line $l$ through vertex $A$, and let the distances from vertices $B$ and $C$ to $l$ be denoted as $d_{1}, d_{2}$. Find the maximum value of $d_{1} + d_{2}$.
保留源文本的换行和格式，直接输出翻译结果。 | Extend $B A$ to $B^{\prime}$, such that $A B^{\prime}=A B$, and connect $B^{\prime} C$. Then a line $l$ passing through vertex $A$ either intersects $B C$ or intersects $B^{\prime} C$.
(1) If $l$ intersects $B C$ at $D$. Then
$$
\begin{array}{l}
\frac{1}{2}\left(d_{1}+d_{2}\right) \cdot A D \\
=S_{\triangle A B D}+S_{\... | 6 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Suppose for some $x, y, z$ we have $x+y+z = 2$, $xy+yz+zx = -3$, and $xyz = 2$. What is $x^3+y^3+z^3$? | Given the conditions $x+y+z = 2$, $xy+yz+zx = -3$, and $xyz = 2$, we can write the polynomial whose roots are $x, y, z$ as:
\[ t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz = t^3 - 2t^2 - 3t + 2. \]
Now, we use the identity for the sum of cubes:
\[ x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) + 3xyz. \]
Using the given values:
... | 32 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Given the sequence ${a_n}$ satisfies $a_1=10$, $a_{n+1}-a_{n}=n (n\in\mathbb{N^*})$, find the minimum value of $\frac{a_{n}}{n}$, that is, $n=$ \_\_\_\_\_\_. | Since $a_{n+1}-a_{n}=n (n\in\mathbb{N^*})$,
We have $a_{n}-a_{n-1}=n-1$,
$a_{n-1}-a_{n-2}=n-2$,
...
$a_{2}-a_{1}=1$.
By adding these equations together, we get $a_{n}-a_{1}=1+2+...+(n-1)=\frac{n(n-1)}{2}$.
Given that $a_{1}=10$,
We obtain $a_{n}=\frac{n(n-1)}{2}+10=\frac{1}{2}n^{2}-\frac{1}{2}n+10$,
Hence, $\fr... | 5 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Laticia knitted 4 pairs of socks for her nephew. She did such a good job that everyone asked her to start selling them. In the first week, she knitted some pairs of socks. In the second week, she knitted 4 more pairs than the week before. On the third week, she only knitted half of the total of the first two weeks. In ... | Let's denote the number of pairs of socks Laticia knitted in the first week as x.
In the second week, she knitted x + 4 pairs.
In the third week, she knitted half of the total of the first two weeks, which is (x + x + 4)/2 = (2x + 4)/2 = x + 2 pairs.
In the fourth week, she knitted 3 fewer pairs than the week before... | 13 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
15. Three tired cowboys entered a saloon and hung their hats on a buffalo horn at the entrance. When the cowboys left deep into the night, they were unable to distinguish one hat from another, and therefore picked three hats at random. Find the probability that none of them took their own hat. | # Solution.
Note that the cowboys have 6 different ways to take the hats. Indeed, the first cowboy can take any of the three hats, the second one can take one of the remaining two, and the third will take the last hat.
At the same time, in only two out of the six scenarios, none of the cowboys will take their own hat... | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A rectangular sheet of paper whose dimensions are $12 \times 18$ is folded along a diagonal, creating the $M$-shaped region drawn in the picture (see below). Find the area of the shaded region.
[img]https://cdn.artofproblemsolving.com/attachments/4/7/d82cde3e91ab83fa14cd6cefe9bba28462dde1.png[/img] | 1. **Understanding the Problem:**
We have a rectangular sheet of paper with dimensions \(12 \times 18\). When folded along a diagonal, it creates an \(M\)-shaped region. We need to find the area of the shaded region.
2. **Identify the Diagonal:**
The diagonal of the rectangle can be calculated using the Pythagor... | 138 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
If 30% of the 880 students at a certain college are enrolled in biology classes, how many students at the college are not enrolled in a biology class? | If 30% of the 880 students are enrolled in biology classes, then 70% are not enrolled in biology classes (since 100% - 30% = 70%).
To find out how many students are not enrolled in biology classes, we calculate 70% of 880:
0.70 * 880 = 616
So, $\boxed{616}$ students at the college are not enrolled in a biology clas... | 616 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
In triangle $ABC$, $P$ and $Q$ are points on sides $\overline{AC}$ and $\overline{AB}$ respectively, such that $AP = \frac{1}{3}AC$ and $AQ = \frac{1}{3}AB$. The area of $\triangle ABC$ is given as 36 square units. Determine the area of $\triangle CPQ$. | First, determine the areas of new segments:
1. Since $AP = \frac{1}{3}AC$ and $AQ = \frac{1}{3}AB$, triangles $\triangle APQ$, $\triangle BAP$, and $\triangle APB$ are one-ninth the area of $\triangle ABC$ because the height from $P$ and $Q$ to line $AB$ and $AC$ respectively, will also be one-third of the triangle's h... | 8 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
The operation $\&$ as defined by $a \& b = (a+b)(a-b)$. Calculate the value of $8 \& 5$. | Substituting $a$ and $b$ with 8 and 5 respectively into $(a+b)(a-b)$,
\[ 8 \& 5 = (8 + 5)(8 - 5) = (13)(3) = \boxed{39}. \] | 39 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
After studying the chapter on triangles in the eighth grade, a class organized a math activity lesson. The teacher asked the students to discuss their understanding of triangle-related knowledge. Xiaofeng said, "There exists a triangle for which the ratios of its three altitudes are $1:1:2$, $1:2:3$, $2:3:4$, $3:4:5$."... | To solve this problem, we need to understand that the altitudes of a triangle are inversely proportional to the lengths of their corresponding sides. This means that if the altitudes have a certain ratio, the sides of the triangle will have the inverse of that ratio. Let's analyze each option step by step.
**Option A:... | B | Geometry | MCQ | Yes | Yes | cn_k12 | false |
$\underline{\text { Problem } A}$.
In triangle $\$ \mathrm{ABC} \$$, \$ $\$$ is the center of the inscribed circle, and the excenter opposite to side \$ is \$ В \$. The exircle touches side \$ В \$ at point $\$ K \$$, which lies on the midline parallel to \$BC\$. Prove that \$ \angle B \leq 60^{\circ} \$$.
# | 
Therefore, the quadrilateral $BKNH_A$ is cyclic and $\angle BKH_A... | proof | Geometry | proof | Incomplete | Incomplete | olympiads | false |
Given $a \in \mathbb{R}$, if the equation $a^2x^2 + (a+2)y^2 + 4x + 8y + 5a = 0$ represents a circle, then the coordinates of the center of this circle are ( )
A: $(-2, -4)$
B: $\left(-\frac{1}{2}, -1\right)$
C: $(-2, -4)$ or $\left(-\frac{1}{2}, -1\right)$
D: Indeterminate | Given the equation $a^2x^2 + (a+2)y^2 + 4x + 8y + 5a = 0$ represents a circle,
- For the equation to represent a circle, the coefficients of $x^2$ and $y^2$ must be equal and non-zero. Hence $a^2 = a + 2 \neq 0$. Solving this, we get $a = -1$ or $a = 2$.
**Case 1:** $a = -1$
Plugging $a = -1$ into the equation, we ge... | A: (-2, -4) | Geometry | MCQ | Yes | Yes | cn_k12 | false |
In 1935, the famous mathematicians P. Erdos and G. Szekeres proved a well-known proposition:
For any positive integer $n \geqslant 3$, there exists a number $f(n)$, such that if and only if $m \geqslant f(n)$, any set of $m$ points in the plane, with no three points collinear, contains $n$ points that form a convex $n... | Lemma 1: Let the vertex set of an arbitrary convex hull $K$ be denoted as $S$. If there are $m$ points inside the convex hull forming a convex $m$-gon, and this convex $m$-gon has a side $A_{2} A_{3}$, with its adjacent sides $A_{1} A_{2}$ and $A_{3} A_{4}$ (where $A_{1}, A_{2}, A_{3}, A_{4}$ are all vertices), and if ... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false |
If the equation $2ax^2-x-1=0$ has exactly one solution in the interval $(0,1)$, then the range of $a$ is. | When $a>0$, the corresponding function $f(x)=2ax^2-x-1$ has exactly one solution in the interval $(0,1)$. This requires that $f(0) \cdot f(1) 1$.
When $a \leq 0$, the function $f(x)=2ax^2-x-1$ has no solution in the interval $(0,1)$.
Therefore, the answer is $a > 1$, or $\boxed{a > 1}$. | a > 1 | Algebra | math-word-problem | Yes | Incomplete | cn_k12 | false |
If the quadratic equation $\left(k-1\right)x^{2}+3x+k^{2}-1=0$ has one solution $x=0$, then $k=$____. | Given that the quadratic equation $\left(k-1\right)x^{2}+3x+k^{2}-1=0$ has one solution $x=0$, we substitute $x=0$ into the equation to find the value of $k$.
Starting with the equation:
\[
\left(k-1\right)x^{2}+3x+k^{2}-1=0
\]
Substituting $x=0$:
\[
\left(k-1\right)\cdot 0^{2}+3\cdot 0+k^{2}-1=0
\]
Simplifying the... | -1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A man can row upstream at 15 kmph and downstream at 35 kmph. What is the speed of the man in still water? | The speed of the man in still water is the average of his upstream and downstream speeds. This is because the upstream speed is reduced by the current of the water, and the downstream speed is increased by the current. Therefore, the speed in still water would be the speed without any current influence.
To find the sp... | 25 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Two numbers have a certain relationship. Their sum is 124, and the smaller number is 31. What is the relationship between the two numbers? | Let's denote the smaller number as S and the larger number as L. We are given that S = 31 and the sum of the two numbers S + L = 124.
To find the relationship between the two numbers, we need to find the value of L. We can do this by rearranging the sum equation:
L = 124 - S
L = 124 - 31
L = 93
Now we have the value... | 62 | Algebra | math-word-problem | Yes | Problem not solved | orca_math | true |
Two identical rectangular crates are packed with cylindrical pipes, each pipe having a diameter of 12 cm. In Crate A, the pipes are packed in a square grid pattern while in Crate B, the pipes are packed in a hexagonal grid pattern. Each crate is packed with 200 pipes. Determine the positive difference in the total heig... | **Crate A:**
- In Crate A, the pipes are arranged in a square grid pattern. Since the diameter of each pipe is 12 cm, the height added by each new row of pipes is equal to the diameter of one pipe.
- Suppose there are 20 rows of pipes (which is reasonable for close packing in the large crate with space for 200 pipes). ... | 228 - 66\sqrt{3} \text{ cm} | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
The interstellar vehicle Gibraltar is a monstrosity on par with mountains, created for the purpose of transporting machinery, construction materials, and families for the establishment of colonies on distant planets. With its full capacity, it accommodates 300 family units each consisting of four members. Nonetheless, ... | First, let's calculate the full capacity of the Gibraltar in terms of people. Since it can accommodate 300 family units, each consisting of four members, the full capacity is:
300 families * 4 members/family = 1200 members
Now, Gibraltar will start its journey with 100 people less than one-third of its full capacity.... | 300 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
For a matinee, a movie theater charges 4.50 dollars for children and a certain amount for adults. At today's matinee, there are 20 more children than adults and the total receipts are 405 dollars. There are 48 children at today's matinee. How much does the movie theater charge for adults? | Let's denote the amount the movie theater charges for adults as A dollars.
According to the information given, there are 48 children at the matinee. Since there are 20 more children than adults, we can express the number of adults as (48 - 20) or 28 adults.
The total receipts from the children's tickets are 48 childr... | \$6.75 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
$F(x)=(x^{3}-2x)f(x)(x\neq 0)$ is an odd function, and $f(x)$ is not always equal to zero, then $f(x)$ is ($\quad$)
A: Odd function
B: Even function
C: Odd function or even function
D: Neither odd nor even function | Solution: $F(x)=(x^{3}-2x)f(x)(x\neq 0)$ is an odd function, and $f(x)$ is not always equal to zero,
We can get $F(-x)=(-x^{3}+2x)f(-x)=-F(x)$
$=-(x^{3}-2x)f(x)$,
We can get $f(-x)=f(x)$,
So, $f(x)$ is an even function.
Hence, the answer is $\boxed{B}$.
From $F(x)$ being an odd function, we can get $F(-x)=-F(x)$,... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Each vertex of a convex polygon has integer coordinates, and no side of this polygon is horizontal or vertical. Prove that the sum of the lengths of the segments of lines of the form $x = m$, $m$ an integer, that lie within the polygon is equal to the sum of the lengths of the segments of lines of the form $y = n$, $n$... | null | null | Geometry | proof | Yes | Yes | aops_forum | false |
Compute the series given by
\[
\frac{2 + 8}{5^{99}} + \frac{2 + 2 \cdot 8}{5^{98}} + \frac{2 + 3 \cdot 8}{5^{97}} + \dots + \frac{2 + 98 \cdot 8}{5^2}.
\] | Let
\[
S = \frac{2 + 8}{5^{99}} + \frac{2 + 2 \cdot 8}{5^{98}} + \frac{2 + 3 \cdot 8}{5^{97}} + \dots + \frac{2 + 98 \cdot 8}{5^2}.
\]
Then,
\[
5S = \frac{2 + 8}{5^{98}} + \frac{2 + 2 \cdot 8}{5^{97}} + \frac{2 + 3 \cdot 8}{5^{96}} + \dots + \frac{2 + 98 \cdot 8}{5}.
\]
Subtracting these equations, we get
\[
4S = \frac... | \frac{1}{8} + \frac{786}{25} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
To be proven, if $a$ and $b$ are positive numbers, then
$$
\frac{(a-b)^{2}}{2(a+b)} \leq \sqrt{\frac{a^{2}+b^{2}}{2}}-\sqrt{a b} \leq \frac{(a-b)^{2}}{a+b}
$$ | The difference standing in the middle, its first term is called the quadratic mean of $a$ and $b$, and the second term is called the geometric mean of $a$ and $b$. We denote them by $Q$ and $M$ respectively:
$$
Q=\sqrt{\frac{a^{2}+b^{2}}{2}}, \quad M=\sqrt{a b}
$$
These are also positive, and
$$
\begin{aligned}
& (a... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
Given that the coordinates of the two foci of an ellipse are $(-2,0)$ and $(2,0)$, and it passes through the point $(2\sqrt{3},\sqrt{3})$, what is the standard equation of the ellipse? | The coordinates of the two foci of the ellipse are $(-2,0)$ and $(2,0)$, hence $c=2$.
Let the equation of the ellipse be: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-4}=1$. The ellipse passes through the point $(2\sqrt{3},\sqrt{3})$,
Thus, we have: $\frac{12}{a^{2}}+\frac{3}{a^{2}-4}=1$. Solving for $a$ gives $a=4$.
Theref... | \frac{x^{2}}{16}+\frac{y^{2}}{12}=1 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
Let $a,b,c$ be constants and $a,b,c$ are positive real numbers. Prove that the equations
$2x+y+z=\sqrt{c^2+z^2}+\sqrt{c^2+y^2}$
$x+2y+z=\sqrt{b^2+x^2}+\sqrt{b^2+z^2}$
$x+y+2z=\sqrt{a^2+x^2}+\sqrt{a^2+y^2}$
have exactly one real solution $(x,y,z)$ with $x,y,z \geq 0$. | 1. **Rewrite the equations:**
We start by rewriting the given equations in a more convenient form. Consider the first equation:
\[
2x + y + z = \sqrt{c^2 + z^2} + \sqrt{c^2 + y^2}
\]
Isolate \(x\):
\[
2x = \sqrt{c^2 + z^2} + \sqrt{c^2 + y^2} - y - z
\]
\[
x = \frac{\sqrt{c^2 + z^2} - z + \... | null | Inequalities | proof | Yes | Yes | aops_forum | false |
9. As shown in Figure 1, each corner of a cube is cut off to form a new solid figure. Then, among the lines connecting the vertices of the new solid figure, the number of lines located inside the original cube is .
| 9.120.
According to the problem, the new solid figure has a total of 24 vertices. Connecting every two points results in $\mathrm{C}_{24}^{2}=276$ lines, of which 36 lines are all on the surface of the original cube.
Each face of the new solid figure has 8 vertices. Excluding the edges, there can be $\frac{5 \times 8... | 120 | Geometry | math-word-problem | Incomplete | Yes | cn_contest | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.