instruction_seed stringlengths 2 10.7k | solution stringlengths 0 10.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
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6. Does there exist an infinite set of positive integers $A$, such that the sum of elements of any non-empty finite subset is composite, and any two elements in $A$ are coprime? | 6. Existence. Define the sequence $a_{1}, a_{2}, \cdots, a_{n} \cdots \cdots$ by induction as follows: $a_{1}=4$, and if $a_{1}, a_{2}, \cdots, a_{n-1}$ are all determined, then let $a_{n}=\left[\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)!\right]^{3}-1$.
This constructs the sequence $a_{1}, a_{2}, \cdots, a_{n} \cdots \cdo... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
According to meteorological data, for every increase of $0.5$ km in height of a certain mountain peak, the temperature decreases by $3^{\circ}\mathrm{C}$. One day, before a mountaineering team set off, the temperature at the foot of the mountain was $-2^{\circ}\mathrm{C}$. $(1)$ After the mountaineering team ascends $... | Let's break down the solution step by step, following the given rules and format.
### For the first part of the problem:
Given that for every increase of $0.5$ km in height, the temperature decreases by $3^{\circ}\mathrm{C}$, and the initial temperature at the foot of the mountain is $-2^{\circ}\mathrm{C}$.
- The te... | 4.5\, \text{km} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
No matter what the value of $m$ is, the line $l: (2m+1)x + (m+1)y - 7m - 4 = 0$ always passes through a fixed point $P$. The coordinates of point $P$ are \_\_\_\_\_\_. | Solution: The line $l: (2m+1)x + (m+1)y - 7m - 4 = 0$ can be rewritten as $m(2x+y-7) + (x+y-4) = 0$,
thus, line $l$ must pass through the intersection point of the lines $2x+y-7=0$ and $x+y-4=0$.
By solving the system of equations $$\begin{cases} 2x+y-7=0 \\ x+y-4=0 \end{cases}$$, we obtain $$\begin{cases} x=3 \\ y... | (3,1) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
6.1. Solve the equation $\frac{n!}{2}=k!+l!$ in natural numbers, where $n!=1 \cdot 2 \ldots n$. If there are no solutions, indicate 0; if there is one solution, indicate $n$; if there are multiple solutions, indicate the sum of the values of $n$ for all solutions. Recall that a solution is a triplet ( $n, k, l$); if so... | Answer: 10 (all triples of solutions $(n, k, l):(3,1,2),(3,2,1),(4,3,3)$).
Solution. Note that if $k4$, then $n!>4 \cdot(n-1)!\geqslant 2 \cdot(k!+l!)$, so there are no such solutions. If $n=2$, we get $1=k!+l!$ - no solutions; if $n=3$, the equation $3=k!+l!$ has two solutions: $k=1, l=2$ and $k=2, l=1$; if $n=4$, th... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A can finish a piece of work in 5 days. B can do it in 16 days. They work together for some days and then A goes away. B finishes the work in 6 days. For how many days did A and B work together? | Let's calculate the work done by A and B per day.
A can finish the work in 5 days, so A's work rate is \( \frac{1}{5} \) of the work per day.
B can finish the work in 16 days, so B's work rate is \( \frac{1}{16} \) of the work per day.
When A and B work together, their combined work rate per day is \( \frac{1}{5} + \... | 2 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Jim ran 16 miles in 2 hours, Frank ran 20 miles in 2.5 hours, and Susan ran 12 miles in 1.5 hours. Calculate each runner's speed in miles per hour. Then, find the difference in speed between Frank and Jim, and the difference in speed between Susan and Jim. | To calculate each runner's speed in miles per hour (mph), we divide the distance they ran by the time it took them to run that distance.
For Jim:
Speed = Distance / Time
Speed = 16 miles / 2 hours
Speed = 8 mph
For Frank:
Speed = Distance / Time
Speed = 20 miles / 2.5 hours
Speed = 8 mph
For Susan:
Speed = Distance ... | 0 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
152. On the chord $AB$ of the circle $O$, as on a diameter, a second circle $O_{1}$ is described. Through the point $A$, a secant is drawn, intersecting the circles $O_{1}$ and $O$ at points $M_{i}$ and $N_{i}$, respectively. Prove that the geometric locus of points symmetric to $N_{i}$ with respect to points $M_{i}$, ... | 152. The reflected point $P$ has the property that $\angle A P B = \angle A N B$, or $\angle A P B + \angle A N B = 180^{\circ}$. The desired geometric locus of points is a circle symmetric to the given one with respect to the line $A B$. | proof | Geometry | proof | Yes | Yes | olympiads | false |
Theorem 4 (E.Klein) In a plane, given any 5 points, no three of which are collinear, there must be 4 points that are the vertices of a convex quadrilateral. | Proof: If the convex hull of 5 points is a convex pentagon or a convex quadrilateral, the conclusion is obviously true. Therefore, without loss of generality, assume the convex hull of 5 points is $\triangle A B C$, and the other 2 points $D$ and $E$ are inside $\triangle A B C$. Since no three points are collinear, li... | proof | Geometry | proof | Yes | Yes | olympiads | false |
(1) Simplify: $\dfrac{\sin(\pi -\alpha)\cos(\pi +\alpha)\sin(\dfrac{\pi}{2}+\alpha)}{\sin(-\alpha)\sin(\dfrac{3\pi}{2}+\alpha)}$.
(2) Given $\alpha \in (\dfrac{\pi}{2}, \pi)$, and $\sin(\pi -\alpha) + \cos \alpha = \dfrac{7}{13}$, find $\tan \alpha$. | ### Problem 1: Simplify the Expression
Given expression: $\dfrac{\sin(\pi -\alpha)\cos(\pi +\alpha)\sin(\dfrac{\pi}{2}+\alpha)}{\sin(-\alpha)\sin(\dfrac{3\pi}{2}+\alpha)}$
Step-by-step simplification:
- $\sin(\pi - \alpha) = \sin \alpha$ because $\sin$ is co-terminal with $\pi - \alpha$.
- $\cos(\pi + \alpha) = -\co... | -\dfrac{12}{5} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given a point $P$ and a line $e$. We have a ruler and various objects suitable for drawing circles (coin, plate, can). Using these tools, construct the line passing through $P$ and parallel to $e$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the tran... | The diversity of usable objects - circle templates - from the perspective of construction means that their diameters are different. Just as with a straight ruler, we can achieve that the drawn circle passes through one or two prescribed - given or constructed - points.
We will use the following auxiliary theorem: if t... | notfound | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
In a right triangle, if the length of the side opposite a $30^{\circ}$ angle is $2$ cm, then the length of the hypotenuse is ____. | In a right triangle, the side opposite the $30^{\circ}$ angle is half the length of the hypotenuse. Given that the length of the side opposite the $30^{\circ}$ angle is $2$ cm, we can find the length of the hypotenuse by doubling this length.
Let's denote the length of the hypotenuse as $H$. We know that:
$$H = 2 \ti... | 4 \text{ cm} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
Find all real numbers $c$ such that there exists a function $f: \mathbb{R}_{ \ge 0} \rightarrow \mathbb{R}$ which satisfies the following.
For all nonnegative reals $x, y$, $f(x+y^2) \ge cf(x)+y$.
Here $\mathbb{R}_{\ge 0}$ is the set of all nonnegative reals. | 1. **Case \( c \le 0 \):**
- Consider the function \( f(x) = \sqrt{x} \).
- For \( x, y \ge 0 \), we need to check if \( f(x + y^2) \ge cf(x) + y \).
- Substituting \( f(x) = \sqrt{x} \), we get:
\[
\sqrt{x + y^2} \ge c\sqrt{x} + y
\]
- Since \( c \le 0 \), \( c\sqrt{x} \le 0 \). Therefore, t... | c < 1 | Inequalities | math-word-problem | Yes | Yes | aops_forum | false |
In Grade 4, there are two sections -- Diligence and Industry. The two sections have an equal number of students after 2 students were transferred from section Industry to Diligence. How many students were there in section Diligence before the transfer if, in both sections combined, there are a total of 50 students? | Let's denote the number of students in section Diligence before the transfer as D and the number of students in section Industry before the transfer as I.
According to the information given, after transferring 2 students from Industry to Diligence, both sections have an equal number of students. This means that:
D + ... | 23 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Given that $f(x)$ is an even function defined on $\mathbb{R}$, and $f(x)$ is monotonically decreasing on $(-\infty, 0]$, determine the solution set of the inequality $f(\log x) > f(-2)$.
A: $(\frac{1}{100}, 100)$
B: $(100, +\infty)$
C: $(\frac{1}{100}, +\infty)$
D: $(0, \frac{1}{100}) \cup (100, +\infty)$ | Since $f(x)$ is an even function defined on $\mathbb{R}$, and it is monotonically decreasing on $(-\infty, 0]$,
Then, $f(x)$ is an increasing function on $(0, +\infty)$.
Thus, the inequality $f(\log x) > f(-2)$ is equivalent to $f(|\log x|) > f(2)$.
This implies that $|\log x| > 2$,
Hence, either $\log x 2$,
So, ... | \text{D} \; (0, \frac{1}{100}) \cup (100, +\infty) | Inequalities | MCQ | Yes | Yes | cn_k12 | false |
1. We have 3 kg of a copper-tin alloy, in which $40\%$ is copper, and 7 kg of another copper-tin alloy, in which $30\%$ is copper. What masses of these alloys should be taken to obtain 8 kg of an alloy containing $p\%$ copper after melting? Find all $p$ for which the problem has a solution. | Answer: $0.8 p-24$ kg; $32-0.8 p$ kg; $31.25 \leqslant p \leqslant 33.75$.
Solution. If the first alloy is taken $x$ kg, then the second $(8-x)$ kg. The conditions of the problem impose restrictions on the possible values of $x$:
$$
\left\{\begin{array}{l}
0 \leqslant x \leqslant 3 ; \\
0 \leqslant 8-x \leqslant 7
\e... | 31.25\leqslantp\leqslant33.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Tatuya, Ivanna, and Dorothy took a quiz together. Tatuya scored twice as much as Ivanna, and Ivanna scored 3/5 times as many marks as Dorothy. The average marks scored by the three is 84. How many marks did Dorothy score? | Let's denote Dorothy's score as D, Ivanna's score as I, and Tatuya's score as T.
According to the information given:
T = 2I (Tatuya scored twice as much as Ivanna)
I = (3/5)D (Ivanna scored 3/5 times as many marks as Dorothy)
The average marks scored by the three is 84, so the total marks scored by the three is 3 ... | 90 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Given that $\binom{3n+1}{23} = \binom{n+6}{23}$ where $n \in \mathbb{N}^*$, and $(3-x)^n = a\_0 + a\_1x + a\_2x^2 + \dots + a\_nx^n$, find the value of $a\_0 - a\_1 + a\_2 - \dots + (-1)^na\_n$. | 1. First, let's find the expression for $\binom{3n+1}{23} = \binom{n+6}{23}$. Since both binomial coefficients are equal, we can deduce that $3n+1 = n+6 + 23$, which simplifies to $n = 12$.
2. Now that we have the value of $n$, we can substitute it back into the equation for $(3-x)^n$:
$$
(3-x)^{12} = a\_0 + a\_1... | 4^{12} | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false |
Jack received 3 emails in the morning, 4 emails in the afternoon, and some emails in the evening. Jack received 11 emails in the morning and evening. How many emails did Jack receive in the evening? | Jack received 11 emails in the morning and evening combined. Since we know he received 3 emails in the morning, we can subtract that number from the total to find out how many emails he received in the evening.
11 emails (morning and evening) - 3 emails (morning) = 8 emails (evening)
Jack received $\boxed{8}$ emails... | 8 | Logic and Puzzles | math-word-problem | Incomplete | Yes | orca_math | true |
6. [9] On the number line, infinitely many points with natural coordinates are marked. When a wheel rolls along the line, each marked point that the wheel passes over leaves a point trace on it. Prove that one can choose such a real $R$ that if the wheel of radius $R$ is rolled along the line, starting from zero, then ... | Solution 1. Divide the wheel into 360 arcs and number them, starting from point $O$ in the order of "rolling". We will prove by induction on $k$ that we can choose a circle of length $360 s$, rolling which along a straight line, we will get traces within the first $k$ arcs.
Base case $(k=1)$. It is sufficient to choos... | proof | Number Theory | proof | Yes | Incomplete | olympiads | false |
Given the function $f(x)=|2x-1|-|x+2|$:
1. Solve the inequality: $f(x) > 0$;
2. If $f(x)+3|x+2| \geqslant |a-1|$ holds for all real numbers $x$, find the range of values for $a$. | 1. The function $f(x)$ can be rewritten as:
$$
f(x) =
\begin{cases}
-x + 3, & x \leqslant -2 \\
-3x - 1, & -2 0$ implies $-x + 3 > 0$, which solves to $x 0$ implies $-3x - 1 > 0$, which solves to $-2 0$ implies $x - 3 > 0$, which solves to $x > 3$.
Therefore, the solution set for $f(x) > 0$ is $\boxed{x 3}$.
2. ... | -4 \leqslant a \leqslant 6 | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false |
Emily went to the store and bought art supplies for $20 and 2 skirts that cost the same amount of money. If Emily paid $15 for each of the skirts, what was the total amount she spent at the store? | Emily bought 2 skirts at $15 each, so the total cost for the skirts is 2 * $15 = $30.
She also bought art supplies for $20.
Therefore, the total amount she spent at the store is the cost of the skirts plus the cost of the art supplies, which is $30 + $20 = $\boxed{\$50}$ . | \$50 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
6. (10 points) April has 30 days. If there are 5 Saturdays and Sundays, then April 1st is a
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The note above is not part of the translation but an instruction that was given.... | 6. (10 points) April has 30 days. If there are 5 Saturdays and Sundays, then April 1st is a Saturday $\qquad$
【Solution】Solution: April has 30 days;
$$
30 \div 7=4 \text { (weeks) } \cdots 2 \text { (days); }
$$
The remaining 2 days are Saturday and Sunday;
So April 1st is a Saturday.
The answer is: Saturday. | Saturday | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let \(N\) be the sum and product of all odd primes less than \(2^5\). First, calculate the product of these primes. Then, add all these primes together. What is the remainder when the sum is divided by \(2^5\)? | The odd primes less than \(2^5=32\) are \(3, 5, 7, 11, 13, 17, 19, 23, 29, 31\).
1. Calculate the product:
\[
P = 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31
\]
Using modulo \(2^5=32\), note that anything multiplied by an even factor effectively becomes 0 modulo 32. The... | 30 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Given two points \\(A(0,0)\\) and \\(B(2,2)\\), the distances from these points to line \\(l\\) are \\(1\\) and \\(2\\) respectively. The number of such lines \\(l\\) is ( ).
A: \\(1\\) line
B: \\(2\\) lines
C: \\(3\\) lines
D: \\(4\\) lines | The solution is as follows: The line at a distance of \\(1\\) from point \\(A(0,0)\\) can be considered as the tangent line to a circle centered at \\(A\\) with a radius of \\(1\\).
Similarly, the line at a distance of \\(2\\) from point \\(B(2,2)\\) can be considered as the tangent line to a circle centered at \\(B\... | \text{B} | Geometry | MCQ | Yes | Yes | cn_k12 | false |
11.007. Find the ratio of the volume of a cube to the volume of a regular tetrahedron, the edge of which is equal to the diagonal of the face of the cube. | ## Solution.
For a tetrahedron where all edges are equal, denote the edges as $a$. The radius of the circumscribed circle around the base of the tetrahedron is $R=\frac{a \sqrt{3}}{3}$. The foot of the height $H$ is the center of the circumscribed circle around $\triangle A B C$ (see Fig. 11.5). Therefore, $H=\sqrt{a^... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The diagonals of a rhombus measure 30 units and 16 units. What is the area of this rhombus in square units? | To find the area of the rhombus, we utilize the formula:
\[
\text{Area} = \frac{d_1 \times d_2}{2}
\]
where \(d_1\) and \(d_2\) are the lengths of the diagonals. Here, \(d_1 = 30\) units and \(d_2 = 16\) units. Substituting these values into the formula gives:
\[
\text{Area} = \frac{30 \times 16}{2} = \frac{480}{2} = ... | 240 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
Given vectors $\overrightarrow{a}=(x,1),\overrightarrow{b}=(-2,3)$.
$(1)$ If $(2\overrightarrow{a}+\overrightarrow{b})⊥\overrightarrow{b}$, find the value of the real number $x$.
$(2)$ If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is acute, find the range of values for the real number $x$. | ### Solution:
#### Part (1):
Given vectors $\overrightarrow{a}=(x,1)$ and $\overrightarrow{b}=(-2,3)$, we need to find $x$ such that $(2\overrightarrow{a}+\overrightarrow{b})$ is perpendicular to $\overrightarrow{b}$.
First, calculate $2\overrightarrow{a}+\overrightarrow{b}$:
\[
2\overrightarrow{a}+\overrightarrow{b... | (-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, \frac{3}{2}) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given the sequence ${a_n}$ satisfying $a_1=1$, $a_{n+1}=1-\frac{1}{4a_n}$ (n∈N*).
1. Let $b_n=\frac{2}{2a_n-1}$, prove that the sequence ${b_n}$ is an arithmetic sequence, and find the general term formula for ${a_n}$;
2. Let $c_n=\frac{4a_n}{n+1}$, find the sum of the first n terms $T_n$ of the sequence ${c_nc_{n+1}}$... | 1. Proof:
We will show that the sequence ${b_n}$ is an arithmetic sequence by demonstrating that the difference between consecutive terms is constant.
$b_{n+1} - b_n = \frac{2}{2a_{n+1}-1} - \frac{2}{2a_n-1} = \frac{4a_n}{2a_{n}-1} - \frac{2}{2a_n-1} = 2$
Since the difference between consecutive terms is constant (2)... | \frac{4n}{n+1} | Algebra | proof | Yes | Yes | cn_k12 | false |
A man, a woman and a boy can together complete a piece of work in 3 days. A woman alone can do it in 6 days and a boy alone in 18 days. How long will a man take to complete the work? | Let's denote the work rate of the man, woman, and boy as M, W, and B respectively, where these rates are in units of work per day.
According to the problem, the combined work rate of the man, woman, and boy is such that they can complete the work in 3 days. Therefore, their combined work rate is 1/3 of the work per da... | 1/9 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
The average of 10 numbers is calculated as 46. It is discovered later on that while calculating the average, the number 75 was incorrectly read as 25, and this incorrect number was used in the calculation. What is the correct average of the 10 numbers? | If the average of the 10 numbers was calculated as 46, the sum of those numbers would be:
Average = Sum of numbers / Number of numbers
46 = Sum of numbers / 10
Sum of numbers = 46 * 10
Sum of numbers = 460
However, since the number 75 was incorrectly read as 25, the sum of the numbers used in the incorrect calculatio... | 51 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
[b]p1.[/b] If $P$ is a (convex) polygon, a triangulation of $P$ is a set of line segments joining pairs of corners of $P$ in such a way that $P$ is divided into non-overlapping triangles, each of which has its corners at corners of $P$. For example, the following are different triangulations of a square.
(a) Prove that... | null | null | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
224. Prove the inequality $M\left[X-\left(x_{i}+x_{k}\right) / 2\right]^{2} \geqslant D(X)$, where $x_{i}$ and $x_{k}$ are any two possible values of the random variable $X$. | Solution. 1) Suppose that $\left(x_{i}+x_{k}\right) / 2=M(X)$. Then
$$
M\left[X-\frac{x_{i}-x_{k}}{2}\right]^{2}=D(X)
$$
2) Suppose that $\left(x_{i}+x_{k}\right) / 2 \neq M(X)$. We will prove that in this case
$$
M\left[X-\frac{x_{i}+x_{k}}{2}\right]^{2}>D(X)
$$
Transform the left side of the inequality using the ... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
Calen originally had 5 more pencils than does Caleb, and Caleb has 3 less than twice as many pencils as does Candy. Calen lost 10 pencils. Candy has 9 pencils. How many pencils does Calen have now? | Let's start by figuring out how many pencils Caleb has. We know that Caleb has 3 less than twice as many pencils as Candy. Since Candy has 9 pencils, twice as many would be 9 * 2 = 18 pencils. Therefore, Caleb has 18 - 3 = 15 pencils.
Now, we know that Calen originally had 5 more pencils than Caleb. Since Caleb has 15... | 10 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A perpendicular line is drawn from point P(2, 4) to the line $ax+y+5=0$, and the foot of the perpendicular is Q(4, 3). The modulus of $z=a+4i$ is
A: $\sqrt {29}$
B: $2 \sqrt {5}$
C: $\frac { \sqrt {101}}{2}$
D: $\frac { \sqrt {485}}{2}$ | Since a perpendicular line is drawn from point P(2, 4) to the line $ax+y+5=0$, and the foot of the perpendicular is Q(4, 3),
the product of the slopes of these two lines equals -1.
The slope of line PQ is $\frac {3-4}{4-2}=- \frac {1}{2}$,
thus, $- \frac {1}{2} \times (-a) = -1$
Therefore, $a=-2$,
So, $z=-2+4... | \text{B} | Geometry | MCQ | Yes | Yes | cn_k12 | false |
Given the curve $f(x)= \frac {2}{3}x^{3}-x^{2}+ax-1$ has two tangent lines with a slope of $3$, and the x-coordinates of the tangent points are all greater than zero, then the range of the real number $a$ is \_\_\_\_\_\_. | The derivative of $f(x)= \frac {2}{3}x^{3}-x^{2}+ax-1$ is $f′(x)=2x^{2}-2x+a$.
According to the problem, we have $2x^{2}-2x+a=3$, which means $2x^{2}-2x+a-3=0$ has two distinct positive roots.
Therefore, $\triangle =4-8(a-3) > 0$ and $a-3 > 0$,
Solving these gives $3 < a < \frac {7}{2}$.
Thus, the answer is: $\... | (3, \frac {7}{2}) | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
Define a sequence $\langle f(n)\rangle^{\infty}_{n=1}$ of positive integers by $f(1) = 1$ and \[f(n) = \begin{cases} f(n-1) - n & \text{ if } f(n-1) > n;\\ f(n-1) + n & \text{ if } f(n-1) \leq n, \end{cases}\]
for $n \geq 2.$ Let $S = \{n \in \mathbb{N} \;\mid\; f(n) = 1993\}.$
[b](i)[/b] Prove that $S$ is an infini... | ### Part (i): Prove that \( S \) is an infinite set.
1. **Define the sequence and initial conditions:**
\[
f(1) = 1
\]
For \( n \geq 2 \):
\[
f(n) = \begin{cases}
f(n-1) - n & \text{if } f(n-1) > n \\
f(n-1) + n & \text{if } f(n-1) \leq n
\end{cases}
\]
2. **Observe the behavior of the... | 12417 | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
If the function $f(x) = x^2 - 2(a-1)x + 2$ is a decreasing function on $(-\infty, 2]$ and an increasing function on $[2, \infty)$, then the value of $a$ is ( ).
A. $-3$
B. $-1$
C. $3$
D. $1$ | For the quadratic function $f(x) = x^2 - 2(a-1)x + 2$:
1. The axis of symmetry is given by $x = a-1$.
2. The function is said to be decreasing on $(-\infty, 2]$ and increasing on $[2, \infty)$.
The axis of symmetry of a parabola is the x-coordinate of the vertex. The vertex form of a quadratic function is $f(x) = a... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
We will prove that a cube can be divided into exactly five tetrahedra, disregarding the congruence of the resulting pieces. | Let the cube be $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, where $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are two parallel faces. Without loss of generality, assume that its volume is one unit. Furthermore, suppose that the cube is divided into the tetrahedra $\Delta_{1}, \Delta_{2}, \Del... | proof | Geometry | proof | Yes | Yes | olympiads | false |
If the asymptotes of a hyperbola are $y = \pm \sqrt{3}x$, then its eccentricity could be ( )
A: $\sqrt{3}$
B: 2
C: $\sqrt{3}$ or $\frac{2\sqrt{3}}{3}$
D: $\frac{2\sqrt{3}}{3}$ or 2 | Given the asymptotes of the hyperbola are $y = \pm \sqrt{3}x$,
- When the foci are on the x-axis, we have $\frac{b}{a} = \sqrt{3}$, which leads to $\frac{c^{2}-a^{2}}{a^{2}} = 3$. Thus, its eccentricity is 2.
- When the foci are on the y-axis, we have $\frac{a}{b} = \sqrt{3}$, which leads to $\frac{a^{2}}{c^{2}-a^{2}... | \text{D} | Geometry | MCQ | Yes | Yes | cn_k12 | false |
A positive integer is [i]totally square[/i] is the sum of its digits (written in base $10$) is a square number. For example, $13$ is totally square because $1 + 3 = 2^2$, but $16$ is not totally square.
Show that there are infinitely many positive integers that are not the sum of two totally square integers. | 1. **Define the problem and notation:**
- A positive integer is called *totally square* if the sum of its digits (written in base $10$) is a square number.
- We need to show that there are infinitely many positive integers that are not the sum of two totally square integers.
2. **Consider the number $10^k - 1$:*... | null | Number Theory | proof | Yes | Yes | aops_forum | false |
Given that $y=f(x)$ is an odd function defined on $\mathbb{R}$, and when $x > 0$, $f(x)=2^{x}-1$, then $f(-2)$ equals to __( ).
A: $3$
B: $-3$
C: $-\dfrac{3}{4}$
D: $-\dfrac{11}{4}$ | According to the problem, when $x > 0$, $f(x)=2^{x}-1$, then $f(2)=2^{2}-1=3$,
Since the function $f(x)$ is an odd function defined on $\mathbb{R}$,
then $f(-2)=-f(2)=-3$;
Therefore, the answer is: $\boxed{\text{B}}$.
According to the problem, by calculating the value of $f(2)$ from the function's expression, and s... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Given a circle \\({x}^{2}+{y}^{2}+2ax-2ay+2{a}^{2}-4a=0(0 < a \leqslant 4)\\) with center \\(C\\), and a line \\(l\\): \\(y=x+4\\).
\\((1)\\) Write down the coordinates of the center of the circle and its radius;
\\((2)\\) Find the maximum length of the chord cut by the circle \\(C\\) from line \\(l\\). | Solution:
\\((1)\\) Given the standard equation of the circle is \\((x+a)^{2}+(y-a)^{2}=4a(0 < a \leqslant 4)\\),
then the coordinates of the center \\(C\\) are \\((-a,a)\\), and the radius is \\(2\sqrt{a}\\).
\\((2)\\) The equation of line \\(l\\) can be rewritten as: \\(x-y+4=0\\). The distance from the center ... | 2\sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
1. The number of different integer pairs $(x, y)$ that satisfy $0<x<y$ and $\sqrt{1992}=\sqrt{x}+\sqrt{y}$ is ( ).
(A) 0
(B) 1
(C) 3
(D) 4 | $-, 1 . A$ | A | Number Theory | MCQ | Yes | Problem not solved | cn_contest | false |
Let
\[x^8 - 50x^4 + 1 = p(x) q(x),\]
where $p(x)$ and $q(x)$ are monic, non-constant polynomials with integer coefficients. Find $p(1) + q(1)$. | We start by using a similar approach to factorize the polynomial:
\begin{align*}
x^8 - 50x^4 + 1 &= (x^8 + 2x^4 + 1) - 52x^4 \\
&= (x^4 + 1)^2 - (\sqrt{52}x^2)^2 \\
&= (x^4 + \sqrt{52}x^2 + 1)(x^4 - \sqrt{52}x^2 + 1).
\end{align*}
However, the coefficients $\sqrt{52}$ are not integers, which violates the condition that... | 4 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
A group of 10 people decided to take off their shoes outside the library to avoid making sound of footsteps. Out of these 10 people, 4 were wearing regular shoes, 3 were wearing sandals, and 3 were wearing slippers. Regular shoes have 2 shoes, sandals have 1 shoe and slippers have 1 shoe. How many shoes were kept outsi... | The 4 people wearing regular shoes would each have 2 shoes, so that's 4 x 2 = 8 shoes.
The 3 people wearing sandals would each have 2 sandals, so that's 3 x 2 = 6 shoes (since each sandal is considered a shoe).
The 3 people wearing slippers would each have 2 slippers, so that's 3 x 2 = 6 shoes (since each slipper is ... | 20 | Logic and Puzzles | math-word-problem | Yes | Incomplete | orca_math | true |
Given an ellipse $C\_1$: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b > 0)$ and a hyperbola $C\_2$ with common left and right foci $F\_1$ and $F\_2$. The product of the eccentricities of the two curves is $e\_1 \cdot e\_2 = 1$. $D$ is the intersection point of the two curves in the first quadrant, and $F\_1D$ ... | Let the hyperbola be represented as $\frac{x^{2}}{A^{2}} - \frac{Y^{2}}{B^{2}} = 1 (A > 0, B > 0)$. The semi-focal distance of both the ellipse and hyperbola is $c$. Let $PF\_1 = m$ and $PF\_2 = n$. Therefore, $m + n = 2a$ and $m - n = 2A$.
Since $e\_1e\_2 = 1$, we have $\frac{c}{a} \cdot \frac{c}{A} = 1$. This implie... | \frac{2a^{2} - b^{2}}{2a} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
A train 140 meters long takes 28.997680185585153 seconds to cross a bridge of a certain length if the speed of the train is 36 kmph. How long is the bridge? | To find the length of the bridge, we first need to calculate the distance the train travels while crossing the bridge. Since the speed of the train is given in kilometers per hour (kmph), we need to convert it to meters per second (m/s) to match the units of the train's length and the time taken to cross the bridge.
T... | 149.98 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A proposition is related to the natural number $n$. If the proposition is true when $n=k$ ($k \in \mathbb{N}^*$), then it can be deduced that the proposition is also true when $n=k+1$. It is known that the proposition is false when $n=5$. Therefore, we can conclude that
A: The proposition is true when $n=4$
B: The p... | The correct answer is $\boxed{\text{B}}$.
Explanation is omitted. | \text{B} | Logic and Puzzles | MCQ | Yes | Problem not solved | cn_k12 | false |
Lindsey saved $50 in September, $37 in October, and $11 in November. Lindsey's mom gave her $25 since she saved more than $75. Lindsey spent some money on a video game and had $36 left. How much did the video game cost? | First, let's calculate the total amount Lindsey saved over the three months:
September: $50
October: $37
November: $11
Total saved = $50 + $37 + $11 = $98
Lindsey's mom gave her an additional $25 because she saved more than $75. So now, Lindsey has:
Total with mom's gift = $98 + $25 = $123
After buying the video g... | \$87 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A semicircle $(O)$ is drawn with the center $O$, where $O$ lies on a line $\ell$. $C$ and $D$ lie on the circle $(O)$, and the tangent lines of $(O)$ at points $C$ and $D$ intersects the line $\ell$ at points $B$ and $A$, respectively, such that $O$ lies between points $B$ and $A$. Let $E$ be the intersection point be... | null | null | Geometry | proof | Yes | Yes | aops_forum | false |
A cylindrical frustum has its height extended to $n$ times the original length, and both radii of the two bases are reduced to $\frac{1}{n}$ of their original sizes. What is the ratio of its lateral surface area to the original one? | **Analysis**
This problem primarily tests our understanding of the formula for the lateral surface area of a cylindrical frustum.
**Step-by-Step Solution**
1. Let's denote the original heights as $h$, the radii of the two bases as $R$ and $r$, and the slant height as $l$. The original lateral surface area can be exp... | 1 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
Maria, Juan, Carlos, and Manu take turns flipping a coin in the given order. The first one to flip tails wins. What is the probability that Manu will win? Express your answer as a common fraction. | For Manu to win on his first turn, the sequence of flips would need to be HHHH, which has a probability of $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. For Manu to win on his second turn, the sequence of flips would need to be HHHHHHHH, which has a probability of $\left(\frac{1}{2}\right)^8 = \frac{1}{256}$.
The prob... | \frac{1}{15} | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
11.1. The quadratic trinomial $y=x^{2}+a x+b$ has real roots $t$ and $l$, where $-1<t<0$. Prove that if $\kappa$ is added to the coefficients $a$ and $b$ of the root $t$, then the resulting trinomial will also have two distinct real roots. | Solution: By Vieta's theorem $b=t \cdot s, a=-t-s$. Then the obtained quadratic trinomial has the form $g(x)=x^{2}+(a+t) x+(b+t)=x^{2}-s x+s t+t$. Its discriminant is $D=s^{2}-4 s t-4 t=(s-2 t)^{2}-4 t(1+t) \geqslant-4 t(1+t)$. Since $-10$, the discriminant of the new quadratic trinomial is positive, which proves the s... | proof | Algebra | proof | Yes | Yes | olympiads | false |
Which of the following statements are true?
A. 4 is a factor of 24.
B. 19 is a divisor of 209 but not of 57.
C. 30 is neither a divisor of 90 nor 68.
D. 15 is a divisor of 30 but not of 60.
E. 9 is a factor of 180.
Write your answer with the letters in alphabetical order, separated by commas. For example, i... | A. Since $24 = 4 \cdot 6$, there is an integer $n$ such that $24 = 4 \cdot n$. Therefore, by the definition of a factor, 4 is a factor of 24, and statement A is true.
B. We can list the divisors of 209. They are 1, 11, 19, and 209. Therefore, 19 is a divisor of 209. Listing the divisors of 57, they are 1, 3, 19, and 5... | \text{A,E} | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Problem 4.1. On nine cards, the numbers from 1 to 9 are written (each one only once). These cards were laid out in a row such that there are no three consecutive cards with numbers in ascending order, and there are no three consecutive cards with numbers in descending order. Then, three cards were flipped over, as show... | Answer: The number 5 is written on card $A$, the number -2 on card $B$, and the number -9 on card $C$.
Solution. The missing numbers are $-2, 5$, and 9.
If the number 5 is on card $B$, then we get the consecutive numbers 3, 4, 5. If the number 5 is on card $C$, then we get the consecutive numbers 8, 7, 5. Therefore, ... | 5,2,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. A castle is surrounded by a fortress wall with nine towers, where knights guard the towers. Every hour, each knight moves to an adjacent tower, moving either only in the direction of the clock's hands or only in the opposite direction. During one night, each knight has been on every tower. It is known that t... | Solution. We will call a "group" the knights who move together. Since there was a moment when each of the five towers had only one knight, there are five groups of one knight each (single groups). At this moment, each of the remaining four towers had at most two groups, so there are at most 13 groups, with at most four... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
Let $g(x)$ be a function defined for all positive real numbers satisfying the conditions $g(x) > 0$ for all $x > 0$ and
\[g(x - y) = \sqrt{g(xy) + 4}\]for all $x > y > 0.$ For some positive $z$, determine $g(z)$ given that there exist $x$ and $y$ such that $x - y = xy = z = 3056.$ | First, assume there exist $x$ and $y$ such that $x - y = xy = 3056.$ From these equations,
\[x^2 - 2xy + y^2 = 3056^2,\]so $x^2 + 2xy + y^2 = 3056^2 + 8 \cdot 3056.$ Then, $x + y = \sqrt{3056^2 + 8 \cdot 3056},$ so by Vieta's formulas, $x$ and $y$ are the roots of
\[t^2 - (\sqrt{3056^2 + 8 \cdot 3056})t + 3056 = 0.\]
(... | 2 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
In $\triangle ABC$, prove:
$(1) \frac{a^{2}+b^{2}}{c^{2}}= \frac{\sin ^{2}A+\sin ^{2}B}{\sin ^{2}C}$
$(2) a^{2}+b^{2}+c^{2}=2(bc\cos A+ca\cos B+ab\cos C)$ | Proof:
$(1)$ By the sine law, we have $a=2R\sin A$, $b=2R\sin B$, and $c=2R\sin C$, where $R$ is the circumradius of $\triangle ABC$.
Then,
$$\begin{align*}
\frac{a^{2}+b^{2}}{c^{2}} &= \frac{(2R\sin A)^{2}+(2R\sin B)^{2}}{(2R\sin C)^{2}} \\
&= \frac{4R^{2}\sin ^{2}A+4R^{2}\sin ^{2}B}{4R^{2}\sin ^{2}C} \\
&= \frac{\si... | a^{2}+b^{2}+c^{2}=2(bc\cos A+ca\cos B+ab\cos C) | Geometry | proof | Yes | Yes | cn_k12 | false |
Pipe A can fill a tank in 2 hours. Due to a leak at the bottom, it takes 3 hours for Pipe A to fill the tank. In what time can the leak alone empty the full tank? | Let's denote the rate at which Pipe A fills the tank as \( A \) and the rate at which the leak empties the tank as \( L \).
When Pipe A is filling the tank without the leak, it takes 2 hours to fill the tank. Therefore, the rate of Pipe A is:
\[ A = \frac{1 \text{ tank}}{2 \text{ hours}} \]
When the leak is present, ... | 6 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Let $p-$ be a prime number and $p>5$. Prove that if the congruence $x^{4}+x^{3}+x^{2}+x+1 \equiv 0(\bmod p)$ is solvable, then $p \equiv 1(\bmod 5)$.
Derive from this the infinitude of the set of prime numbers of the form $5 n+1$.
# | See the solution of problem $\underline{60756}$. | proof | Number Theory | proof | Yes | Incomplete | olympiads | false |
Given that point $P(7m, -3m) (m \neq 0)$ is on the terminal side of angle $\alpha$, calculate:
1) The value of $\frac{\cos(\frac{\pi}{2} + \alpha)\sin(-\pi - \alpha)}{\cos(\frac{11\pi}{2} - \alpha)\sin(\frac{9\pi}{2} + \alpha)}$.
2) The value of $2 + \sin\alpha\cos\alpha - \cos^2\alpha$. | 1) From the given information, we have $\tan\alpha = -\frac{3}{7}$. Now let's simplify the given expression step-by-step:
$$\frac{\cos(\frac{\pi}{2} + \alpha)\sin(-\pi - \alpha)}{\cos(\frac{11\pi}{2} - \alpha)\sin(\frac{9\pi}{2} + \alpha)}$$
Using cofunction identities, we have $\cos(\frac{\pi}{2} + \alpha) = -\sin\a... | 2 + \sin\alpha\cos\alpha - \cos^2\alpha = \frac{23}{29} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Elise had $8. Then she saved $13 from her allowance and spent $2 on a comic book and some money on a puzzle. Now, Elise has $1 left. How much money did Elise spend on the puzzle? | Elise started with $8 and saved an additional $13 from her allowance, so she had a total of:
$8 + $13 = $21
She then spent $2 on a comic book, so she had:
$21 - $2 = $19
After buying the puzzle, Elise has $1 left, so the amount spent on the puzzle is:
$19 - $1 = $18
Therefore, Elise spent $\boxed{\$18}$ on the p... | \$18 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
The contrapositive of the proposition "If $a>b$, then $a-5>b-5$" is ( )
A. If $ab-5$, then $a>b$
C. If $a\leq b$, then $a-5\leq b-5$
D. If $a-5\leq b-5$, then $a\leq b$ | To find the contrapositive of a given proposition, you negate both the hypothesis and the conclusion, and then swap their positions.
Given the original proposition "If $a>b$, then $a-5>b-5$",
the negation of the hypothesis is "$a\leq b$",
and the negation of the conclusion is "$a-5\leq b-5$".
Swapping these, we... | \text{D} | Inequalities | MCQ | Yes | Yes | cn_k12 | false |
2. Let $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=3$. Prove:
$$\sum \frac{1}{a \sqrt{2\left(a^{2}+b c\right)}} \geqslant \sum \frac{1}{a+b c}$$
where, " $\sum$ " denotes the cyclic sum. | 2. Without loss of generality, let $a \geqslant b \geqslant c$. Then $\frac{(c-a)(c-b)}{3(c+a b)} \geqslant 0$ and $\frac{(a-b)(a-c)}{3(a+b c)}+ \frac{(b-a)(b-c)}{3(b+a c)}=\frac{c(a-b)^{2}}{3}\left[\frac{1+a+b-c}{(a+b c)(b+a c)}\right] \geqslant 0$. Therefore, $\sum \frac{(a-b)(a-c)}{3(a+b c)} \geqslant 0$. And $\sum ... | proof | Inequalities | proof | Yes | Yes | inequalities | false |
Given the function $f(x)=\left\{{\begin{array}{l}{{2^x}({x＜1})}\\{{{log}_2}x({x≥1})}\end{array}}\right.$, if $f\left(a\right)=2$, then $a=\_\_\_\_\_\_$. | ### Step-by-Step Solution:
To find the value of $a$ such that $f(a) = 2$, we consider the two cases given by the function $f(x)$.
#### Case 1: $a < 1$
For $a < 1$, the function is defined as $f(x) = 2^x$. Setting $f(a) = 2$, we have:
\[2^a = 2\]
This simplifies to:
\[2^a = 2^1\]
By comparing the exponents, we find... | 4 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
If two dice are thrown consecutively and the numbers obtained are $m$ and $n$, then the probability that point $P(m, n)$ lies on the line $x+y=4$ is ( )
A: $\frac {1}{3}$
B: $\frac {1}{4}$
C: $\frac {1}{6}$
D: $\frac {1}{12}$ | Solution: When throwing two dice consecutively, there are a total of $6 \times 6 = 36$ outcomes, each with an equal chance of occurring.
The points $P(m, n)$ that lie on the line $x+y=4$ include the outcomes $(1, 3)$, $(2, 2)$, and $(3, 1)$, which are three in total.
Therefore, the probability that point $P(m, n)$ ... | D | Combinatorics | MCQ | Yes | Yes | cn_k12 | false |
Given the equation \( x + y = 50 \), where \( x \) and \( y \) are positive integers, how many distinct ordered-pair solutions \( (x, y) \) exist? | The equation can be analyzed by setting \( x \) to each positive integer value starting from 1 up to 49. For each value of \( x \), \( y \) can be determined uniquely as \( y = 50 - x \).
- When \( x = 1 \), \( y = 49 \).
- When \( x = 2 \), \( y = 48 \).
- ...
- When \( x = 49 \), \( y = 1 \).
Thus, each value of \(... | 49 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
3. As shown in Figure 1, in $\triangle A B C$, point $D$ is on side $B C$. It is known that $A B=A D=2, A C$ $=4$, and $B D: D C=2: 3$. Then $\triangle A B C$ is (
(A) Acute triangle
(B) Right triangle
(C) Obtuse triangle
(D) Cannot be determined | 3. (B).
Let $BD=2x$, then $DC=3x$. Draw $AH \perp BD$ at $H$, then $BH=x, CH=4x$. From $2^{2}-x^{2}$. $=AB^{2}-BH^{2}=AH^{2}=$ $AC^{2}-CH^{2}=4^{2}-(4x)^{2}$, solving gives $x^{2}=\frac{4}{5}$.
At this point, $BC^{2}=(5x)^{2}=25 \times \frac{4}{5}=20=2^{2}+4^{2}=$ $AB^{2}+AC^{2}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false |
Emily's quiz scores so far are: 91, 94, 86, 88, and 101. What score does she need to get on the sixth quiz to make the arithmetic mean of the six scores equal 94? | 1. Subtract 90 from every score to simplify the arithmetic. The new scores are $91-90=1$, $94-90=4$, $86-90=-4$, $88-90=-2$, and $101-90=11$.
2. For the average of the six scores to be 94, the sum after reducing each by 90 must be $6 \times (94-90) = 6 \times 4 = 24$.
3. The sum of her first five adjusted scores is $1 ... | 104 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
There are x kids waiting for the swings and twice as many kids waiting for the slide. If each kid waits 2 minutes for the swings and 15 seconds for the slide, The shorter wait is 270 seconds shorter. What is the value of unknown variable x? What is the value of unknown variable x? | To solve for the number of kids \( x \) waiting for the swings, we need to set up equations based on the given information and then solve them step by step.
1. **Define the variables:**
- Let \( x \) be the number of kids waiting for the swings.
- The number of kids waiting for the slide is \( 2x \).
2. **Calcu... | 6 | Algebra | math-word-problem | Yes | Yes | metamath | true |
The third and fourth terms of a decreasing geometric sequence are 18 and 12, respectively. What is the first term of the sequence? | 1. The common ratio of the geometric sequence is $\frac{12}{18} = \frac{2}{3}$. Since the sequence is decreasing and each term is less than the previous, the common ratio must be less than one, which holds true in this scenario.
2. The $n^{\text{th}}$ term of the sequence can be expressed as $a \left(\frac{2}{3}\right... | 40.5 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
17. What is the maximum number of parts into which the plane can be divided by:
a) two triangles;
b) two rectangles;
c) two convex $n$-gons? | 17. Let's immediately provide the solution to problem (v), namely, we will prove that the maximum number of parts into which two convex $n$-gons can divide the plane is $2n+2$; from this it will follow that two triangles can divide the plane into no more than 8 parts (problem (a)); see Fig. 54, a), and two rectangles c... | 2n+2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
714. A discrete two-dimensional random variable ( $X, Y$ ), the components of which are independent, is given by the distribution law:
| $Y$ | $X$ | | |
| :---: | :---: | :---: | :---: |
| | $x_{1}$ | $x_{2}$ | $x_{3}$ |
| | | | |
| $y_{1}$ | 0.18 | 0.30 | 0.12 |
| $y_{2}$ | 0.12 | 0.20 | 0.08 |
Simulate the r... | Solution. Let's find the distribution law of the component $X$:
$$
\begin{aligned}
& p_{1}=P\left(X=x_{1}\right)=0.18+0.12=0.30 \\
& p_{2}=P\left(X=x_{2}\right)=0.30+0.20=0.50, \\
& p_{3}=P\left(X=x_{3}\right)=0.12+0.08=0.20 .
\end{aligned}
$$
Thus, the required distribution law is
$$
\begin{array}{cccc}
X & x_{1} &... | Other | math-word-problem | Yes | Yes | olympiads | false | |
A square has a diagonal of 6 inches, and a circle has a diameter of 6 inches. Determine by how many square inches the area of the circle is greater than the area of the square. Express your answer as a decimal to the nearest tenth. | 1. **Square Calculation**:
Let the side length of the square be \( s \). Then, from the Pythagorean Theorem, we have \( s^2 + s^2 = 6^2 \) leading to \( 2s^2 = 36 \) and \( s^2 = 18 \). Hence, the area of the square is \( 18 \) square inches.
2. **Circle Calculation**:
The radius of the circle is half its diamet... | 10.3 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
Given $f(x) = x^2 + 2(a-2)x + 5$ is an increasing function on the interval $[4, +\infty)$, then the range of the real number $a$ is ( )
A: $(-\infty, -2]$
B: $[-2, +\infty)$
C: $[-6, +\infty)$
D: $(-\infty, -6]$ | Since the graph of the function $f(x) = x^2 + 2(a-2)x + 5$ is a parabola opening upwards, with the axis of symmetry at $x = 2-a$,
if the function $f(x) = x^2 + 2(a-2)x + 5$ is increasing on the interval $[4, +\infty)$,
then $2-a \leq 4$,
solving this yields $a \geq -2$.
Therefore, the answer is: $\boxed{\text{B... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Distribute 5 volunteers into 4 groups, with one group consisting of 2 individuals and the other three groups each having 1 individual. They are to be assigned to 4 different intersections to assist the traffic police. Calculate the number of different distribution methods. | To solve this problem, we have to consider two steps: first, grouping the volunteers, and then assigning these groups to the 4 intersections.
First, let's choose 2 out of the 5 volunteers to form the group of two. There are $\binom{5}{2}$ ways to do this.
Next, we need to assign the formed groups to 4 different inter... | 240 | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false |
8. According to the pattern in the table below, 2020 is located at ( ).
A. Row 505, Column 505
B. Row 505, Column 506
C. Row 505, Column 508
D. Row 504, Column 504
E. Row 504, Column 506 | $\mathrm{C}$ | C | Number Theory | MCQ | Yes | Yes | olympiads | false |
Example 24 (2007 Bulgarian National Mathematical Olympiad) Two players, A and B, play the following game: A first divides a pile of $n$ stones into three piles, with each pile having at least one stone, and one pile having more stones than each of the other two piles. Then, B divides the pile with the most stones in th... | When $n=1,2,3$, the stones cannot be divided.
When $n=4,5,6,7$, player A has a winning strategy (because A can make the number of stones in the largest pile 2 or 3).
When $n=8,9$, after one move by A, the number of stones in the largest pile is between 4 and 7 (inclusive), so B has a winning strategy.
Similarly, when... | 3^{k}or3^{k}-1fork>1 | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
Let be given a triangle $ ABC$. Points $ P$ on side $ AC$ and $ Y$ on the production of $ CB$ beyond $ B$ are chosen so that $ Y$ subtends equal angles with $ AP$ and $ PC$. Similarly, $ Q$ on side $ BC$ and $ X$ on the production of $ AC$ beyond $ C$ are such that $ X$ subtends equal angles with $ BQ$ and $ QC$. Lines... | null | null | Geometry | proof | Yes | Yes | aops_forum | false |
Two trains are running in opposite directions at the same speed. The length of each train is unknown. If they cross each other in 16 seconds, and the speed of each train (in km/hr) is 27, what is the length of each train in meters? | First, we need to convert the speed of the trains from km/hr to m/s because the time they take to cross each other is given in seconds and we want to find the length in meters.
The conversion factor from km/hr to m/s is:
1 km/hr = 1000 m / 3600 s
So, the speed of each train in m/s is:
27 km/hr * (1000 m / 3600 s) = 2... | 120 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
The point $P$ on a real number line is moved according to the rule as follows.
Flip a coin, if heads is the spot on the face, then $P$ is moved by $+1$, and if tails is the spot on the face, then $P$ is moved to the point of reflection for the origin. At first $P$ is set on the origin on the real number, let $a_n$ be t... | Let's solve each part of the problem step-by-step.
### Part (1): Probability that \( a_3 = 0 \)
1. **Initial Position**: \( P \) starts at the origin, so \( a_0 = 0 \).
2. **Possible Moves**: Each coin flip can result in either \( +1 \) or a reflection through the origin.
3. **Sequence Analysis**: We need to find the... | \frac{n - 1}{2^n} | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
15.21. Cut the cube into three equal pyramids. | 15.21. Let's take one vertex of the cube as the common vertex of these tetrahedra, and three non-adjacent faces of the cube as their bases. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $\alpha=\left(\alpha_{1}, \ldots, \alpha_{n}\right)$ and $\beta=\left(\beta_{1}, \ldots, \beta_{n}\right)$ be two sets of exponents with equal sums.
Prove that, if $\alpha \neq \beta$, then for all non-negative $x_{1}, \ldots, x_{n}$, the inequality $T_{\alpha}\left(x_{1}, \ldots, x_{n}\right) \geq$ $T_{\beta}\lef... | It is evident that it suffices to prove the inequality in the case when the set $\beta$ is obtained from the set $\alpha$ by removing one "brick" from the Young diagram. Let's prove the case when the transition is made from $\beta=\left(\alpha_{1}-\right.$ $\left.1, \alpha_{2}+1, \alpha_{3}, \ldots, \alpha_{n}\right)$ ... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
Two people, A and B, are making a certain type of mechanical part. It is known that the time it takes for A to make 240 parts is equal to the time it takes for B to make 280 parts. Together, they make 130 parts per day. If A makes x parts per day, which of the following equations is correct?
A: $\frac{240}{x}$=$\frac{... | To solve this problem, we start by understanding that the ratio of parts made by A and B in the same amount of time is given. Specifically, A makes 240 parts in the same time it takes B to make 280 parts. This implies that their rates of work are in proportion, which can be expressed as a ratio of parts made per day.
... | \text{A: } \frac{240}{x}=\frac{280}{130-x} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
27. [15] Let $\left\{a_{i}\right\}_{i \geq 0}$ be a sequence of real numbers defined by
$$
a_{n+1}=a_{n}^{2}-\frac{1}{2^{2020 \cdot 2^{n}-1}}
$$
for $n \geq 0$. Determine the largest value for $a_{0}$ such that $\left\{a_{i}\right\}_{i \geq 0}$ is bounded. | Answer: $\quad 1+\frac{1}{2^{2020}}$
Solution: Let $a_{0}=\frac{1}{\sqrt{2}^{2020}}\left(t+\frac{1}{t}\right)$, with $t \geq 1$. (If $a_{0}<\frac{1}{\sqrt{2}^{2018}}$ then no real $t$ exists, but we ignore these values because $a_{0}$ is smaller.) Then, we can prove by induction that
$$
a_{n}=\frac{1}{\sqrt{2}^{2020 \c... | 1+\frac{1}{2^{2020}} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5 Given complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=2,\left|z_{2}\right|=3$. If the angle between the vectors they correspond to is $60^{\circ}$, find $\frac{\left|z_{1}+z_{2}\right|}{\left|z_{1}-z_{2}\right|}$.
(2002 National High School League Question) | As shown in Figure $7-3$, by the geometric meaning of the addition and subtraction of complex numbers and the cosine theorem, we easily get:
$$
\left|z_{1}+z_{2}\right|=\sqrt{19},\left|z_{1}-z_{2}\right|=\sqrt{7} \text {, so } \frac{\left|z_{1}+z_{2}\right|}{\left|z_{1}-z_{2}\right|}=\frac{\sqrt{19}}{\sqrt{7}}=\frac{\s... | \frac{\sqrt{133}}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
68*. a) Prove that if the centers of three spheres do not lie on the same straight line, then the set of points having the same power with respect to all three spheres is a straight line—the radical axis of the three spheres. This line is perpendicular to the plane passing through the centers of the spheres.
b) Prove ... | 68. a) If the centers of three spheres do not lie on the same straight line, then their radical planes are not parallel. Therefore, the radical planes of the first and second spheres intersect with the radical plane of the second and third spheres. The line of their intersection contains all points that have the same p... | proof | Geometry | proof | Yes | Yes | olympiads | false |
Calculate $\log_{10} 50 + \log_{10} 45 - \log_{10} 9$. | Using the logarithmic property $\log_b x + \log_b y = \log_b(xy)$ and $\log_b x - \log_b y = \log_b\left(\frac{x}{y}\right)$, we can proceed as follows:
1. Combine the logs of multiplication:
\[
\log_{10} 50 + \log_{10} 45 = \log_{10} (50 \cdot 45)
\]
2. Convert the product:
\[
50 \cdot 45 = 2250
\]
... | 1 + 2 \log_{10} 5 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
(1) Solve the inequality: $5x-12\leqslant 2\left(4x-3\right)$;
(2) Solve the system of equations: $\left\{\begin{array}{l}x-y=5\\ 2x+y=4\end{array}\right.$. | ### Step-by-Step Solution
#### (1) Solve the inequality: $5x-12\leqslant 2\left(4x-3\right)$
- **Step 1:** Expand the brackets on the right side of the inequality.
\[
5x - 12 \leqslant 8x - 6
\]
- **Step 2:** Rearrange the terms to bring like terms to one side and constants to the other.
\[
5x - 8x \leqsla... | null | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A sign painter paints individual numerals for a row of 70 houses. The houses are numbered with consecutive integers from 1 to 70. How many times does the digit 9 appear in the house numbers? | 1. Review house numbering range: 1 to 70.
2. Count occurrences of the digit 9 within this range:
- The tens: From 1 to 70, the tens digits range from 00 to 70, crossing through the 9s when housing numbers are in the 90s (which are beyond our range). However, for the range 1-70, the tens place only includes 9 for the... | 7 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
What is the product of all possible values for $b$, if the length of the segment between the points $(3b, b+2)$ and $(6, 3)$ is $3\sqrt{5}$ units? | To solve this, we will use the distance formula, which calculates the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ as $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
For our points $(3b, b+2)$ and $(6, 3)$:
\[
\sqrt{(6 - 3b)^2 + (3 - (b+2))^2} = 3\sqrt{5}
\]
Squaring both sides, we have:
\[
(6 - 3b)^2 + (... | -0.8 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
## 79.
On charges of robbery, A, B, and C were brought to court. The following has been established:
1) At least one of the three defendants is guilty.
2) If A is guilty and B is not guilty, then C is not guilty.
This information is insufficient to prove the guilt of each of the three defendants individually, but it... | 79. Two defendants, one of whom must be guilty, are V and C. Indeed, suppose A is not guilty. Then, by statement (1), V or C must be guilty. On the other hand, suppose A is guilty. If V is guilty, then at least one of V and C is definitely guilty. But suppose V is not guilty. Then A is guilty, and V is not guilty. Ther... | VorC | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
a batsman scored 120 runs whichincluded 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets ? | First, let's calculate the total runs scored by the batsman through boundaries and sixes.
A boundary is worth 4 runs, so 3 boundaries would give:
3 boundaries * 4 runs/boundary = 12 runs
An six is worth 6 runs, so 8 sixes would give:
8 sixes * 6 runs/six = 48 runs
Now, let's add the runs from boundaries and sixes to... | 50\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
The win records for a soccer league with five teams are displayed in the graph, but the team names are not shown. Use the clues below to identify how many games the Eagles won:
1. The Lions won more games than the Eagles.
2. The Sharks won more games than the Wolves, but fewer games than the Falcons.
3. The Wolves won... | - **Analyzing clue 3**: The Wolves won more than 25 games; thus, the possible wins for the Wolves are 30, 35, 40, or 45 games.
- **Analyzing clue 2**: The Sharks won more games than the Wolves but fewer than the Falcons. This implies the Wolves cannot have won the highest number of games (45), and the Sharks cannot hav... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | synthetic_math | true |
Heather is going to sew some aprons for a kiddie crew program. She already was able to sew 13 aprons, and today, she sewed three times as many aprons. Heather needs to sew 49 aprons tomorrow if she wants to sew half of the remaining number of aprons needed. How many aprons does Heather need to sew in total for the kidd... | Heather sewed 13 aprons initially and then sewed three times as many today. So today, she sewed 3 * 13 = 39 aprons.
The total number of aprons she has sewn so far is 13 (initial) + 39 (today) = 52 aprons.
If Heather needs to sew 49 aprons tomorrow to sew half of the remaining number of aprons needed, this means there... | 150 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Given functions $f(x) = x + \frac{a^2}{x}$ and $g(x) = x + \ln(x)$, where $a > 0$.
(1) If $x = 1$ is an extremum point of the function $h(x) = f(x) + g(x)$, find the value of the real number $a$.
(2) If for any $x_1, x_2 \in (1, e)$ ($e$ is the base of the natural logarithm), the inequality $f(x_1) \geq g(x_2)$ holds, ... | (1) Since $h(x) = f(x) + g(x) = 2x + \frac{a^2}{x} + \ln(x)$, the domain of $h(x)$ is $(0, +\infty)$. Therefore, the derivative of $h(x)$ is $h'(x) = 2 - \frac{a^2}{x^2} + \frac{1}{x}$. Since $x=1$ is an extremum point of $h(x)$, we have $h'(1)=0$, which implies $3 - a^2 = 0$. Since $a>0$, we have $a = \sqrt{3}$. Upon ... | \left[\frac{e+1}{2}, +\infty\right) | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
9. Let $A B C D$ be a trapezoid with $A B \| C D$ and $A D=B D$. Let $M$ be the midpoint of $A B$, and let $P \neq C$ be the second intersection of the circumcircle of $\triangle B C D$ and the diagonal $A C$. Suppose that $B C=27, C D=25$, and $A P=10$. If $M P=\frac{a}{b}$ for relatively prime positive integers $a$ a... | Answer: 2705
Solution 1: As $\angle P B D=\angle P C D=\angle P A B, D B$ is tangent to $(A B P)$. As $D A=D B, D A$ is also tangent to $(A B P)$. Let $C B$ intersect $(A B P)$ again at $X \neq B$; it follows that $X D$ is the $X$-symmedian of $\triangle A X B$. As $\angle A X C=\angle D A B=\angle A D C, X$ also lies ... | 2705 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kuzneiov A.
A square box of candies is divided into 49 equal square cells. In each cell, there is a chocolate candy - either black or white. In one sitting, Sasha can eat two candies if they are of the same color and lie in adjacent cells, either by side or by corner. What is the maximum number of candies that Sasha c... | Evaluation. For the layout shown in the left image, none of the 16 black candies can be eaten, and out of the 33 white candies, no more than 32 can be eaten due to parity.
Algorithm. We will... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $E: \dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1 \left(a > 0, b > 0\right)$, $P$ is a point on the right branch of the hyperbola $E$, $M$ is the midpoint of the line segment $F_{1}P$, and $O$ is the origin. If the perimeter of $\triangle OF_... | To solve this problem, we start by analyzing the given information and applying the properties of a hyperbola.
1. **Given Information and Equation (1):**
From the definition of a hyperbola, we know that for any point $P$ on the hyperbola, the difference in distances to the foci $F_{1}$ and $F_{2}$ is constant and eq... | C: y = \pm \sqrt{2}x | Geometry | MCQ | Yes | Yes | cn_k12 | false |
Given that $\sin \alpha= \frac{3}{5} $ and $\alpha \in (\frac{\pi}{2}, \pi)$, find the value of $\sin 2\alpha$ $(\ )$
A: $\ \frac{24}{25} $
B: $- \frac{24}{25} $
C: $\ \frac{12}{25} $
D: $- \frac{12}{25} $ | **Analysis**
First, determine the value of $\cos \alpha$, then use the double-angle formula to find the value of $\sin 2\alpha$.
**Step-by-step Solution**
1. Find the value of $\cos \alpha$. Given that $\sin \alpha = \frac{3}{5}$ and $\alpha$ is in the second quadrant (where cosine is negative), we can determine $\c... | - \frac{24}{25} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
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