question stringlengths 12 1.61k | answer stringlengths 0 2.23k | openr1_source stringclasses 8
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In $\triangle ABC$, $B= \frac {\pi}{4}$, $C= \frac {\pi}{3}$, $c=1$, find the length of the shortest side. | First, we find angle $A$ using the triangle angle sum theorem: $A=\pi-B-C= \frac {5\pi}{12}$.
Since $B < C$, by the property that the larger angle is opposite the larger side, $b$ is the shortest side.
Next, we apply the Law of Sines: $\frac {b}{\sin B}= \frac {c}{\sin C}$.
Substituting the given values, we get: $\f... | cn_k12 | 5283b907-8ea9-5348-a268-4fb1fc02daf7 | open-r1/OpenR1-Math-220k | [
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] |
8. When evaluated, which of the following is not an integer?
A $1^{-1}$
B $4^{-\frac{1}{2}}$
C $6^{0}$
D $8^{\frac{2}{3}}$
E $16^{\frac{3}{4}}$ | Solution
B
We have
$$
\begin{array}{l}
1^{-1}=\frac{1}{1}=1, \\
4^{-\frac{1}{2}}=\frac{1}{4^{\frac{1}{2}}}=\frac{1}{\sqrt{4}}=\frac{1}{2}, \\
6^{0}=1, \\
8^{\frac{2}{3}}=\left(8^{\frac{1}{3}}\right)^{2}=(\sqrt[3]{8})^{2}=2^{2}=4,
\end{array}
$$
and
$$
16^{\frac{3}{4}}=\left(16^{\frac{1}{4}}\right)^{3}=(\sqrt[4]{16})^{3... | olympiads | 7e153a19-1432-5759-b0a4-4fa8963dba41 | open-r1/OpenR1-Math-220k | [
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] |
Given $f\_1(x)=(x^{2}+2x+1)e^{x}$, $f\_2(x)=[f\_1(x)]′$, $f\_3(x)=[f\_2(x)]′$, …, $f_{n+1}(x)=[f\_n(x)]′$, $n∈N^{*}.$ Let $f\_n(x)=(a\_nx^{2}+b\_nx+c\_n)e^{x}$, find $b_{2015}=($ $)$
A: $4034$
B: $4032$
C: $4030$
D: $4028$ | Since $f\_1(x)=(x^{2}+2x+1)e^{x}$,
We have $f\_2(x)=[f\_1(x)]′=(x^{2}+4x+3)e^{x}$,
$f\_3(x)=[f\_2(x)]′=(x^{2}+6x+7)e^{x}$,
$f\_4(x)=[f\_3(x)]′=(x^{2}+8x+13)e^{x}$,
The sequence ${c\_n}$ is $2$, $4$, $6$, $8$, …,
Hence, $b\_n=2n$,
So, $b_{2015}=2015×2=4030$,
Thus, the answer is: $\boxed{C}$.
First, find the derivative, ... | cn_k12 | a1469479-7e94-5c33-a354-09dc9c246bea | open-r1/OpenR1-Math-220k | [
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] |
The sum of the squares of the first $n$ natural numbers is divisible by $n$ when? | I. solution: It is known that
$$
1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
$$
Since $n, n+1$ and $2 n+1$ are relatively prime numbers, our expression is clearly divisible by $n$ if and only if $\frac{(n+1)(2 n+1)}{6}$ is an integer, that is, $(n+1)(2 n+1)$ is divisible by 2 and 3. Since $(2 n+1)$ is odd,... | olympiads | c86ddcea-0fb5-53b7-9493-d8c8632e8519 | open-r1/OpenR1-Math-220k | [
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] |
5. As shown in the figure, a large triangle $A B C$ is divided into seven parts by three line segments, four of which are triangles and the other three are quadrilaterals. The sum of the perimeters of the three quadrilaterals is 25 cm, the sum of the perimeters of the four triangles is 20 cm, and the perimeter of trian... | 【Answer】 13
【Solution】 If we add the sum of the perimeters of the three quadrilaterals and the sum of the perimeters of the four triangles, then the middle segments are all added twice. For example, in the figure below, $GH$ is both a side of quadrilateral $GF BH$ and a side of $\triangle GHI$. And $AB, BC, CA$ each a... | olympiads | 531f8534-5b40-5e15-9caa-5934d6642935 | open-r1/OpenR1-Math-220k | [
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] |
The constant term in the expansion of $$(x^{2}+ \frac {1}{x^{2}}-2)^{3}$$ is ( )
A: 20
B: -20
C: 15
D: -15 | Given the binomial $$(x^{2}+ \frac {1}{x^{2}}-2)^{3}$$ = $$(x- \frac {1}{x})^{6}$$, the general term formula for its expansion is $T_{r+1}$ = $C_{6}^{r}$ • $(-1)^{r}$ • $x^{6-2r}$.
Let $6-2r=0$, we find $r=3$. Therefore, the constant term in the expansion is $-C_{6}^{3}$ = $-20$.
Hence, the correct answer is: $\boxed... | cn_k12 | 2a90c8b8-1af1-5aed-babe-aa1ebd9b88c5 | open-r1/OpenR1-Math-220k | [
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] |
Given the function $f(x)=\left\{{\begin{array}{l}{{x^{\frac{1}{2}}},x≥0}\\{f(x+2),x＜0}\end{array}}\right.$, then the value of $f({-\frac{5}{2}})$ is ( )
A: $-\frac{1}{2}$
B: $\frac{1}{2}$
C: $\frac{{\sqrt{2}}}{2}$
D: $\frac{{\sqrt{6}}}{2}$ | Given the function $f(x)=\left\{{\begin{array}{l}{{x^{\frac{1}{2}}},x≥0}\\{f(x+2),x＜0}\end{array}}\right.$, we want to find the value of $f({-\frac{5}{2}})$.
1. Since $x = -\frac{5}{2} < 0$, we apply the recursive definition of $f$ for $x < 0$: $f({-\frac{5}{2}}) = f\left({-\frac{5}{2}+2}\right) = f\left({-\frac{1}{2... | cn_k12 | a9a0f10c-696b-584a-94ca-2bda8bd58e24 | open-r1/OpenR1-Math-220k | [
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] |
Among the following statements about the linear function $y = -2x + 1$:
① $y$ decreases as $x$ increases. ② The graph is parallel to the line $y = -2x$. ③ The intersection point of the graph with the y-axis is $(0, 1)$. ④ The graph passes through the first, second, and fourth quadrants.
The correct ones are ( )
A: 4
B... | For ①, since $-2 < 0$, $y$ decreases as $x$ increases. Therefore, this statement is correct.
For ②, the linear function $y = -2x + 1$ and the line $y = -2x$ have the same slope, which is $-2$. Thus, the linear function $y = -2x + 1$ is parallel to the line $y = -2x$. Therefore, this statement is correct.
For ③, by se... | cn_k12 | cd33ba4c-3672-5afa-862b-6b0204eca4cb | open-r1/OpenR1-Math-220k | [
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] |
6. Given $x \in[0,3]$, then the maximum value of $\frac{\sqrt{2 x^{3}+7 x^{2}+6 x}}{x^{2}+4 x+3}$ is $\qquad$ | 6. $\frac{1}{2}$.
When $x=0$, the value of the original expression is 0; when $x \neq 0$, since
$$
\frac{\sqrt{2 x^{3}+7 x^{2}+6 x}}{x^{2}+4 x+3}=\frac{\sqrt{2 x+7+\frac{6}{x}}}{x+4+\frac{3}{x}}=\frac{t}{\frac{t^{2}+1}{2}}=\frac{2}{t+\frac{1}{t}},
$$
where $t=\sqrt{2 x+7+\frac{6}{x}}$. Since $x \in(0,3]$, the range o... | olympiads | 044f832d-1564-5415-aed7-63838f13efa5 | open-r1/OpenR1-Math-220k | [
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] |
If $k$ is a positive integer, then the meaning of $(k^{5})^{3}$ is:
A: Adding 3 $k^{5}$ together
B: Adding 5 $k^{3}$ together
C: Multiplying 3 $k^{5}$ together
D: Multiplying 8 $k$ together | To understand the meaning of $(k^{5})^{3}$, let's break it down step by step according to the rules of exponentiation:
1. The base is $k^{5}$, which means we have $k$ raised to the power of 5.
2. The exponent outside the parentheses is 3, which means we need to multiply $k^{5}$ by itself 3 times.
So, $(k^{5})^{3}$ ca... | cn_k12 | 000fad4c-a5b1-5cfb-ad3c-0308bb20a72e | open-r1/OpenR1-Math-220k | [
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] |
The price of paper A is 3 coins for 2 sheets, while the price of paper B is 2 coins for 3 sheets. What is the ratio of the unit prices of paper A and B?
A: 9:4
B: 4:9
C: 3:3
D: | Solution: (3÷2):(2÷3),
\= $\frac {3}{2}$:$\frac {2}{3}$,
\= 9:4,
Answer: The ratio of the unit prices of paper A and B is 9:4.
Hence, the correct option is A.
In this problem, we first need to find the unit prices of both types of paper to determine their price ratio. According to the basic properties of a ratio, we ca... | cn_k12 | 2786072d-c65c-506c-a38d-ce68a0dfd4de | open-r1/OpenR1-Math-220k | [
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] |
A number's $\frac {5}{9}$ is 60, its $\frac {1}{4}$ is \_\_\_\_\_\_. | Solution: 60 ÷ $\frac {5}{9}$ × $\frac {1}{4}$,
\= 60 × $\frac {9}{5}$ × $\frac {1}{4}$,
\= 27;
Answer: This number's $\frac {1}{4}$ is 27.
So the answer is: $\boxed{27}$.
Given that a number's $\frac {5}{9}$ is 60, the number itself is 60 ÷ $\frac {5}{9}$. To find its $\frac {1}{4}$, we use multiplication.
This ... | cn_k12 | 2848530b-d6de-50fc-b61c-c78aa972b574 | open-r1/OpenR1-Math-220k | [
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] |
## Task 3 - 030723
A wooden cube with an edge length of $30 \mathrm{~cm}$ is to be cut into cubes with an edge length of $10 \mathrm{~cm}$.
a) How many cuts must be made? (Sawing in a package is not allowed.)
b) How many cubes will you get? | a) You need 26 cuts $(2+6+18)$.
b) You get 27 cubes $(3 \cdot 3 \cdot 3)$. | olympiads | 376fe86b-df53-5eb5-8f5a-77a2a0f146fb | open-r1/OpenR1-Math-220k | [
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] |
Bakayev E.V.
About the group of five people, it is known that:
Alyosha is 1 year older than Alekseev,
Borya is 2 years older than Borisov,
Vasya is 3 years older than Vasiliev,
Grisha is 4 years older than Grigoryev, and there is also Dima and Dmitriev in this group.
Who is older and by how many years: Dima or Dm... | The sum of the ages of Alyosha, Borya, Vasya, Grisha, and Dima is equal to the sum of the ages of Alexeev, Borisov, Vasilyev, Grigoryev, and Dmitriev. Therefore, Dmitriev is older than Dima by \(1+2+3+4=10\) years.
## Answer
\[
\begin{aligned} &\({[\quad\) The Pigeonhole Principle (other). \(] } \\ &\) Problem \(\und... | olympiads | 54352c80-de24-5f09-a7c2-1f63debe9503 | open-r1/OpenR1-Math-220k | [
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] |
Calculate $2\sin 390^{\circ}-\tan \left(-45^{\circ}\right)+5\cos 360^{\circ}=\_\_\_\_\_\_$. | To solve the given expression $2\sin 390^{\circ}-\tan \left(-45^{\circ}\right)+5\cos 360^{\circ}$, we can break it down as follows:
1. **Simplifying $\sin 390^{\circ}$**:
Since $390^{\circ} = 360^{\circ} + 30^{\circ}$, and $\sin$ is periodic with a period of $360^{\circ}$, we have
$\sin 390^{\circ} = \sin (360^{\c... | cn_k12 | a5f9369d-e0a5-53ce-9957-7985550328f1 | open-r1/OpenR1-Math-220k | [
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] |
Given the function $f(x)=x^{3}+ax^{2}+bx+c$, it reaches a maximum value of $7$ when $x=-1$, and it reaches a minimum value when $x=3$. Find the values of $a$, $b$, $c$, and this minimum value. | Since $f(x)=x^{3}+ax^{2}+bx+c$, then $f'(x)=3x^{2}+2ax+b$.
Since the function reaches a maximum value when $x=-1$ and a minimum value when $x=3$, $-1$ and $3$ are the roots of the equation $f'(x)=0$, i.e., $-1$ and $3$ are the two roots of the equation $3x^{2}+2ax+b=0$.
Therefore, we have the system of equations:
$... | cn_k12 | 1dc3d3bf-031d-58d4-a55a-d07c3611874e | open-r1/OpenR1-Math-220k | [
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] |
3. Given $M=2^{7} t=3^{5} s$, where $t$ is odd, and $s$ cannot be divided by 3. Then the sum of all divisors of $M$ in the form of $2^{P} 3^{q}$ is $\qquad$ . | 3. The sum of all divisors of $M$ is
$$
\begin{array}{l}
\left(1+2+\cdots+2^{7}\right)\left(1+3+\cdots+3^{5}\right) \\
=92820 .
\end{array}
$$ | cn_contest | 5118c70f-f544-5df6-be36-8e8c430b56db | open-r1/OpenR1-Math-220k | [
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] |
Mark started the day with 14 buttons. His friend Shane gave him 3 times that amount of buttons. Then his other friend Sam asked if he could have half of Mark’s buttons. How many buttons did Mark end up with? | Shane gave Mark 14*3=<<14*3=42>>42 buttons
After that Mark had 42+14=<<42+14=56>>56 buttons
Then Mark gave 56/2=<<56/2=28>>28 buttons to Shane
Which left Mark with 28 buttons
#### 28 | null | null | openai/gsm8k | [
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] |
1. Let $a, b$, and $c$ be positive real numbers. Determine the largest total number of real roots that the following three polynomials may have among them: $a x^{2}+b x+c, b x^{2}+c x+a$, and $c x^{2}+a x+b$. | Answer: 4 If all the polynomials had real roots, their discriminants would all be nonnegative: $a^{2} \geq$ $4 b c, b^{2} \geq 4 c a$, and $c^{2} \geq 4 a b$. Multiplying these inequalities gives $(a b c)^{2} \geq 64(a b c)^{2}$, a contradiction. Hence one of the quadratics has no real roots. The maximum of 4 real root... | olympiads | 92b4104f-9bfc-5f86-bcca-85a5f2526b2c | open-r1/OpenR1-Math-220k | [
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] |
(Optional question) Given $a$, $b$, $c \in (0, +\infty)$, and $$\frac {1}{a}+ \frac {2}{b}+ \frac {3}{c}=2$$, find the minimum value of $a+2b+3c$ and the values of $a$, $b$, $c$ when the minimum is achieved. | Solution: Since
$$( \frac {1}{a}+ \frac {2}{b}+ \frac {3}{c})(a+2b+3c)=[( \sqrt { \frac {1}{a}})^{2}+( \sqrt { \frac {2}{b}})^{2}+( \sqrt { \frac {3}{c}}) ^{2}][( \sqrt {a})^{2}+( \sqrt {2b})^{2}+( \sqrt {3c})^{2}]$$
$$≥( \sqrt { \frac {1}{a}} \sqrt {a}+ \sqrt { \frac {2}{b}} \sqrt {2b}+ \sqrt { \frac {3}{c}} \sqrt {3c... | cn_k12 | 81da14d9-9762-5416-b0e5-ba54668e6a7f | open-r1/OpenR1-Math-220k | [
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] |
Convert the binary number 11001 to a decimal number. | To convert the binary number 11001 to its decimal equivalent, we can use the place value method where each digit in the binary number is multiplied by the corresponding power of 2 based on its position from the right (starting from position 0). Let's compute each position's value and then sum them up:
Starting from th... | cn_k12 | 47c7d4df-7d91-506f-bffa-eea0ff8ecbda | open-r1/OpenR1-Math-220k | [
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Let the universal set $U=\{1,2,3,4\}$, and let $A=\{x|x^2-5x+m=0, x\in U\}$. If $\complement_U A=\{1,4\}$, find the value of $m$. | Given the universal set $U=\{1,2,3,4\}$ and $\complement_U A=\{1,4\}$,
we can deduce that $A=\{2,3\}$.
Substituting $x=2$ into the equation $x^2-5x+m=0$, we solve to find $m=6$.
Therefore, the answer is $\boxed{6}$. | cn_k12 | c34fb70c-0a67-5db4-86ab-d6889d51ac3b | open-r1/OpenR1-Math-220k | [
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] |
$$
\sqrt{7+4 \sqrt{3}}+\sqrt{7-4 \sqrt{3}}=?
$$ | I. solution. The question is about a positive number and its square:
$$
\begin{gathered}
(\sqrt{7+4 \sqrt{3}}+\sqrt{7-4 \sqrt{3}})^{2}=7+4 \sqrt{3}+2 \sqrt{(7+4 \sqrt{3})(7-4 \sqrt{3})}+ \\
+7-4 \sqrt{3}=14+2 \sqrt{49-48}=16
\end{gathered}
$$
thus, the value of the expression is $\sqrt{16}=4$.
II. solution. We can n... | olympiads | e7ea069e-a147-549b-b6ca-9ef5818227c6 | open-r1/OpenR1-Math-220k | [
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] |
In a certain land, all Arogs are Brafs, all Crups are Brafs, all Dramps are Arogs, and all Crups are Dramps. Which of the following statements is implied by these facts?
$\textbf{(A) } \text{All Dramps are Brafs and are Crups.}\\ \textbf{(B) } \text{All Brafs are Crups and are Dramps.}\\ \textbf{(C) } \text{All Arogs a... | It may be easier to visualize this by drawing some sort of diagram. From the first statement, you can draw an Arog circle inside of the Braf circle, since all Arogs are Brafs, but not all Brafs are Arogs. Ignore the second statement for now, and draw a Dramp circle in the Arog circle and a Crup circle in the Dramp circ... | amc_aime | fa9b8819-e1cc-5942-9d6a-dc50ec74a736 | open-r1/OpenR1-Math-220k | [
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] |
In triangle $ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given that $3a\cos C=2c\cos A$ and $\tan A= \frac {1}{3}$, find angle $B$. | 1. From the given equation $3a\cos C=2c\cos A$, we can use the Sine Law to rewrite it as $3\sin A\cos C=2\sin C\cos A$.
2. Rearranging the equation, we get $3\tan A=2\tan C$.
3. Substituting the given value of $\tan A= \frac {1}{3}$, we get $2\tan C=3\times \frac {1}{3}=1$, which simplifies to $\tan C= \frac {1}{2}$.
4... | cn_k12 | 30a9e373-dbbd-51e8-8855-2da5c4919399 | open-r1/OpenR1-Math-220k | [
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] |
## Task 1
A bear can live to be 50 years old, a fox can live to be one fifth of that; a wolf can live 5 years longer than a fox.
How old can a wolf, how old can a fox become? | A fox can live for 10 years and a wolf can live for 15 years. | olympiads | aa18fe9b-cd87-5389-9c1b-6e889e6d304c | open-r1/OpenR1-Math-220k | [
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] |
5. 138 Find all polynomials \( P(x) \) that satisfy
\[
x P(x-1) \equiv (x-2) P(x), \quad x \in \mathbb{R}
\] | [Solution] Substituting $x=0,2$ into the given identity, we can see that the polynomial $P(x)$ has roots 0 and 1, meaning it is divisible by the polynomial $x^{2}-x$. Next, substituting $P(x) \equiv\left(x^{2}-x\right) Q(x)$ into the identity, we get
$$
\begin{array}{l}
x\left[(x-1)^{2}-(x-1)\right] Q(x-1) \equiv(x-2)\... | olympiads | 4fda16f7-85c8-5982-bedd-d6253e661a5d | open-r1/OpenR1-Math-220k | [
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Given ${\log_4}a=\frac{1}{2}$, find the value of $a$. | Given ${\log_4}a=\frac{1}{2}$, we want to find the value of $a$.
Starting with the given equation:
$${\log_4}a=\frac{1}{2},$$
We use the definition of a logarithm to rewrite this equation in exponential form. The base of the logarithm becomes the base of the exponent, the right side of the equation becomes the expon... | cn_k12 | 1c54f41c-008e-5d65-844b-22d50bb834f7 | open-r1/OpenR1-Math-220k | [
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In $\triangle ABC$, the sides opposite to angles A, B, and C are $a$, $b$, and $c$ respectively, with $b=2 \sqrt {3}$, $c=3$, and $\cos B=- \frac {1}{3}$.
Find:
(I) The value of $\sin C$;
(II) The area of $\triangle ABC$. | (This problem is worth 13 points)
(I) In $\triangle ABC$, we have $\cos B=- \frac {1}{3}$.
Therefore, $\sin B= \sqrt {1- \cos^{2}B}= \sqrt {1-( \frac {1}{3})^{2}}= \frac {2 \sqrt {2}}{3}$.
Since $b=2 \sqrt {3}$ and $c=3$, by the sine law, we have:
$$\frac {b}{\sin B}= \frac {c}{\sin C}$$
which implies
$$\frac {2 \sqrt ... | cn_k12 | 2df2b2ef-8324-5d1c-b29f-bfd29b4ad34c | open-r1/OpenR1-Math-220k | [
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] |
Given vectors $\overrightarrow{a}=(-3,m)$ and $\overrightarrow{b}=(2,-2)$ are perpendicular to each other, then $|\overrightarrow{a}|$ is equal to
A: $3$
B: $3\sqrt{2}$
C: $9$
D: $18$ | Given vectors $\overrightarrow{a}=(-3,m)$ and $\overrightarrow{b}=(2,-2)$ are perpendicular, we must use the dot product to express this condition algebraically. The formula for the dot product of two vectors $\overrightarrow{a}=(a_1,a_2)$ and $\overrightarrow{b}=(b_1,b_2)$ is $\overrightarrow{a} \cdot \overrightarrow{... | cn_k12 | 06d67fbe-aa96-5557-b3da-e40d120a4519 | open-r1/OpenR1-Math-220k | [
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] |
The positional relationship between the line $4x-3y=0$ and the circle $x^{2}+y^{2}-18x-45=0$ is ( ).
A: Intersect.
B: Separate.
C: Tangent.
D: Uncertain. | **Analysis**
This question examines the general equation of a circle and the positional relationship between a line and a circle. It is not difficult and is considered a basic problem.
**Solution**
Given the circle $x^{2}+y^{2}-18x-45=0$, we can find the center of the circle is $(9,0)$, and the radius is $\sqrt{126}... | cn_k12 | 0c7985fe-0c2c-5c51-9ccf-b7b6ec9bccbf | open-r1/OpenR1-Math-220k | [
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] |
3B. In triangle $ABC$, a median $BD$ is drawn. Points $E$ and $F$ divide the median into three equal segments $\overline{BE}=\overline{EF}=\overline{FD}$. If $\overline{AB}=1$ and $\overline{AF}=\overline{AD}$, determine the length of segment $CE$. | Solution. Since $\overline{A F}=\overline{A D}$, triangle $D A F$ is isosceles, so $\measuredangle A D F = \measuredangle A F D$, and from this we have $\measuredangle B F A = \measuredangle C D E$. Segment $B D$ is a median, so $\overline{C D} = \overline{A F}$, and from the condition of the problem $\overline{B F} = ... | olympiads | 4a390c16-65e1-5c9f-8b8b-aff410eb3724 | open-r1/OpenR1-Math-220k | [
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] |
8.26 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}, n \geqslant 1 .
$$
Find the general term formula for $a_{n}$. | [Solution] When $n \geqslant 2$, since
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}, \\
a_{1}+a_{2}+\cdots+a_{n-1}=(n-1)^{2} a_{n-1},
\end{array}
$$
Therefore,
From this, we can get
Thus, we obtain
$$
\begin{array}{l}
a_{n}=n^{2} a_{n}-(n-1)^{2} a_{n-1} . \\
a_{n}=\frac{n-1}{n+1} a_{n-1}, n=2,3, \cdots \\
... | olympiads | b85c7eb3-b0f2-51ce-b62d-7e15aa501ed7 | open-r1/OpenR1-Math-220k | [
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] |
22. Two numbers in the $4 \times 4$ grid can be swapped to create a Magic Square (in which all rows, all columns and both main diagonals add to the same total).
What is the sum of these two numbers?
A 12
B 15
C 22
D 26
E 28
\begin{tabular}{|c|c|c|c|}
\hline 9 & 6 & 3 & 16 \\
\hline 4 & 13 & 10 & 5 \\
\hline 14 & 1 & 8 ... | Solution: $\mathbf{E}$
We have included, in italics, the row and column totals. The circled numbers are the totals of the numbers in the two main diagonals.
The sum of all the numbers in the square is $1+2+\ldots+15+16$. This is the 16th triangular number, $\frac{1}{2}(16 \times 17)=136$. So the total of each row, col... | olympiads | 73ceaa53-b43a-51c4-834d-1a3f56df85e0 | open-r1/OpenR1-Math-220k | [
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] |
Given \\(\alpha: x \geqslant a\\) and \\(\beta: |x-1| < 1\\). If \\(\alpha\\) is a necessary but not sufficient condition for \\(\beta\\), then the range of the real number \\(a\\) is ( ).
A: \\(a \geqslant 0\\)
B: \\(a \leqslant 0\\)
C: \\(a \geqslant 2\\)
D: \\(a \leqslant 2\\) | **Analysis**
This question tests the knowledge of necessary conditions, sufficient conditions, and necessary and sufficient conditions. The key to solving this question is to construct an inequality about \\(a\\) based on the principle that the smaller condition is sufficient and the larger condition is necessary.
**... | cn_k12 | 8f505f33-2259-5fe8-925d-a83363c8b9c4 | open-r1/OpenR1-Math-220k | [
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If $|a+1|+(b-2)^{2}=0$.
$(1)$ Find the values of $a$ and $b$;
$(2)$ Find the value of $(a+b)^{2020}+a^{2019}$. | ### Step-by-Step Solution
#### Part 1: Finding the values of $a$ and $b$
Given that $|a+1|+(b-2)^{2}=0$, we can analyze this equation in steps:
1. For the sum of a modulus and a square to be zero, both terms must individually be zero because both are always non-negative. Therefore, we have two separate equations:
... | cn_k12 | bee89dc6-bb07-574b-ac7d-362922fc49f3 | open-r1/OpenR1-Math-220k | [
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In triangle $\triangle ABC$, given $a=4\sqrt{3}$, $c=12$, $C=\frac{π}{3}$, then $A=\left(\ \ \right)$
A: $\frac{π}{3}$
B: $\frac{π}{6}$
C: $\frac{π}{6}$ or $\frac{{5π}}{6}$
D: $\frac{π}{6}$ or $\frac{π}{3}$ | Given in triangle $\triangle ABC$ that $a=4\sqrt{3}$, $c=12$, and $C=\frac{\pi}{3}$, we apply the Law of Sines to find angle $A$.
Using the Law of Sines, we get:
\[ \frac{a}{\sin A} = \frac{c}{\sin C} \]
Substituting the given values into this equation, we have:
\[ \frac{4\sqrt{3}}{\sin A} = \frac{12}{\sin\left(\frac{... | cn_k12 | 9ef4d5df-b147-59a6-8d59-43d69465a2c0 | open-r1/OpenR1-Math-220k | [
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10.1. Consider functions of the form $y=x^{2}+a x+b$, where $a+b=2021$. Prove that the graphs of all such functions have a common point. | Solution. $y$ (1) $=1+a+b=2022$. Therefore, each of the given graphs passes through the point with coordinates $(1 ; 2022)$. | olympiads | c5bace95-7e32-5966-b565-da35bb5dc8a0 | open-r1/OpenR1-Math-220k | [
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Alfred likes to save $1,000.00 over 12 months for his Christmas shopping. He has $100.00 left over from last year's holiday to put towards this year's goal. How much money does Alfred now need to save each month in order to reach his goal in 12 months? | He wants to save $1,000.00 and he has $100.00 left over so he still needs to save 1000-100= $<<1000-100=900.00>>900.00
He wants to save $900.00 over 12 months so that means he needs to save 900/12 = $<<900/12=75.00>>75.00 per month
#### 75 | null | null | openai/gsm8k | [
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Given that point $F(2,0)$ is a focus of the ellipse $3kx^{2}+y^{2}=1$, find the value of the real number $k$. | The ellipse $3kx^{2}+y^{2}=1$ can be rewritten as $\frac{x^{2}}{\frac{1}{3k}}+y^{2}=1$.
From the problem, we know that the focus is at $(2,0)$, so the distance $c$ from the center of the ellipse to the focus is $2$.
The relationship between $a$, $b$ and $c$ in an ellipse is given by $a^{2}=b^{2}+c^{2}$. In this case,... | cn_k12 | c5799df3-d517-5cbe-a21a-e7520a0f06e5 | open-r1/OpenR1-Math-220k | [
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Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence $\{a_n\}$. If $S_7 = 35$, then $a_4 =$
A: $8$
B: $7$
C: $6$
D: $5$ | Given that $S_7 = 35$, we know that the sum of the first 7 terms of the arithmetic sequence is 35. The formula for the sum of the first $n$ terms of an arithmetic sequence is $S_n = \frac{n}{2}(a_1 + a_n)$. Since we are looking for $a_4$, which is the middle term of the first 7 terms, it also represents the average val... | cn_k12 | 1820c6b8-713b-5194-9022-8d3d083b2f90 | open-r1/OpenR1-Math-220k | [
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Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$. | Answer: -2015.
Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{31}{65} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{31}{65} \overrightarrow{AO}$.
Let the v... | olympiads | c8bed7bd-4e41-5fef-8d7f-f9b360bffb31 | open-r1/OpenR1-Math-220k | [
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Which of the following functions is an even function and decreases on the interval $(0, +\infty)$?
A: $y=x^2+1$
B: $y=|x|$
C: $y=-x^2+1$
D: $y= \frac {1}{x}$ | For option A, $y=x^2+1$ is an even function but increases on the interval $(0, +\infty)$, so A is excluded.
For option B, $y=|x|$ is an even function but also increases on the interval $(0, +\infty)$, so B is excluded.
For option C, the graph of $y=-x^2+1$ is symmetric about the y-axis, making it an even function, an... | cn_k12 | aba92387-e8fe-5243-83a8-9b5f8b5ed4ab | open-r1/OpenR1-Math-220k | [
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Given the function $f(x) = x \ln x - a(x - 1)$.
(I) Determine the monotonicity of the function $f(x)$;
(II) If $f(x) \geqslant 0$ always holds, find the value of $a$. | (I) The domain of the function $f(x)$ is $(0, +\infty)$,
$f'(x) = \ln x + 1 - a$,
Solving $f'(x) = 0$ gives $x = e^{a-1}$,
When $x \in (0, e^{a-1})$, $f'(x) 0$,
Therefore, $f(x)$ is monotonically decreasing in $(0, e^{a-1})$, and monotonically increasing in $(e^{a-1}, +\infty)$;
(II) From (I), $f(x)$ has a minimu... | cn_k12 | 2b85e7a5-3df1-5f7c-addc-1e2389f3cb39 | open-r1/OpenR1-Math-220k | [
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1.83 Mark 10 points on a circle. How many different convex polygons can be constructed using some of these points as vertices? (Polygons are considered the same only if all their vertices coincide)
| [Solution] For positive integers $k, 3 \leqslant k \leqslant 10$, every selection of $k$ points can form a convex polygon, and different sets of points form different polygons. There are $C_{10}^{k}$ different ways to choose $k$ points. Since
$$
\begin{aligned}
& C_{10}^{3}+C_{10}^{4}+C_{10}^{5}+\cdots+C_{10}^{10} \\
=... | olympiads | 9d505aa6-c1f4-5f20-9156-d465f68f7a21 | open-r1/OpenR1-Math-220k | [
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In regression analysis, the term that represents the difference between a data point and its corresponding position on the regression line is ( )
A: Total sum of squares
B: Residual sum of squares
C: Regression sum of squares
D: Coefficient of determination $R^2$ | **Analysis of the Problem:** From the analysis of residuals, it is known that the residual sum of squares represents the difference between a data point and its corresponding position on the regression line. Therefore, the answer is B.
**Key Point:** Residual analysis
$\boxed{\text{B}}$ | cn_k12 | 88b51f9a-9b71-5084-80e5-6268f2a9124f | open-r1/OpenR1-Math-220k | [
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10. A cell phone with a completely drained battery needs to be charged for 2 hours to fully recharge if it is not used in the meantime. If, however, it is used during charging, half of the introduced energy is immediately consumed, and only the remaining part accumulates in the battery. Knowing that it took 2 and a hal... | (10) The correct answer is $(B)$.
Let $E$ be the total energy that can be stored in the mobile phone. The energy stored in one minute without using it will be equal to $\frac{E}{120}$: the energy stored in one minute of usage will be equal to $\frac{1}{2} \cdot \frac{E}{120}=\frac{E}{240}$. If the time required for ch... | olympiads | b5a492f2-a307-5aaf-87ca-010ca101bb28 | open-r1/OpenR1-Math-220k | [
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Given the set $A = \{y | y = x^2 + 1, x \in \mathbb{R}\}$, and the domain of the function $y = \log(4x - x^2)$ is $B$, then $A \cap B = \ $. | $A = \{y | y = x^2 + 1, x \in \mathbb{R}\} = \{y | y \geq 1\}$
For $4x - x^2 > 0$, solving this inequality yields $x \in (0, 4)$. Therefore, the domain is $B = \{x | 0 < x < 4\}$.
Thus, $A \cap B = [1, 4)$.
Therefore, the answer is $\boxed{[1, 4)}$. | cn_k12 | c26d8b19-177f-567c-a768-5bc42f7621bc | open-r1/OpenR1-Math-220k | [
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12. List all positive integers that are coprime with 105 in ascending order, and find the 100th term of this sequence. | 12. First, it can be proven that if $a$ is a number coprime with 105, then $a+105k\left(k \in \mathbf{N}_{+}\right)$ is also coprime with 105. In the set $S_{1}=\{1,2, \cdots, 105\}$, we can see that there are 48 numbers $a_{1}=1, a_{2}=2, a_{3}=4, a_{4}=8 \cdots, a_{48}=104$ that are coprime with 105. Thus, $S_{n}=\{1... | olympiads | 53d0422d-a011-5163-8820-e6d2ca6e1ea6 | open-r1/OpenR1-Math-220k | [
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] |
The sequence $\{a\_n\}$ satisfies $a_{n+1}= \begin{cases} 2a_{n}, & 0\leqslant a_{n}\leqslant \frac {1}{2} \\ 2a_{n}-1, & \frac {1}{2} < a_{n} < 1 \end{cases}$, with $a_{1}= \frac {3}{5}$. $S_{n}$ represents the sum of the first $n$ terms of the sequence. Find $S_{2016}$. | 1. First, we calculate the initial terms of the sequence to identify any pattern:
- $a_{1}= \frac {3}{5}$
- $a_{2}=2a_{1}-1= \frac {1}{5}$
- $a_{3}=2a_{2}= \frac {2}{5}$
- $a_{4}=2a_{3}= \frac {4}{5}$
- $a_{5}=2a_{4}-1= \frac {3}{5}$
2. We observe that the sequence is cyclic with a period of 4: $\{a_n\}... | cn_k12 | 1aae82a0-07f9-58bc-ba00-da0bc186f4f5 | open-r1/OpenR1-Math-220k | [
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If the function $f\left(x\right)=x^{2}-2ax+2$ is an increasing function on the interval $\left[3,+\infty \right)$, then the range of $a$ is ______. | To determine the range of $a$ for which the function $f(x) = x^2 - 2ax + 2$ is increasing on the interval $[3, +\infty)$, we first rewrite the function in a form that makes it easier to analyze its behavior. The function can be rewritten as:
\[
f(x) = (x - a)^2 + 2 - a^2
\]
This form reveals that the function is a pa... | cn_k12 | 5d85b154-243c-53e6-b8ac-4380a3d2db1c | open-r1/OpenR1-Math-220k | [
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4. (10 points) The number of students in Xiaoming's class is less than 40, but more than 30. Therefore, the ratio of boys to girls in the class cannot be ( )
A. 2: 3
B. 3: 4
C. 4: 5
D. 3: 7 | 【Analysis】First, consider the ratio as parts, find out the total number of people in terms of parts, since the number of people must be an integer, the total number of people must be a multiple of the total parts. Identify the option where there is no multiple of the total parts in the numbers greater than 30 and less ... | olympiads | 7d7973ab-6938-5d14-ad4c-9ef18755fb79 | open-r1/OpenR1-Math-220k | [
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4- 120 If $\operatorname{tg} x+\operatorname{tg} y=25$, and $\operatorname{ctg} x+\operatorname{ctg} y=30$. Find $\operatorname{tg}(x+y)$. | [Solution] From $\operatorname{ctg} x+\operatorname{ctg} y=30$ we get
then
$$
\frac{1}{\operatorname{tg} x}+\frac{1}{\operatorname{tg} y}=30 \text {, }
$$
then $\frac{\operatorname{tg} x+\operatorname{tg} y}{\operatorname{tg} x \cdot \operatorname{tg} y}=30$,
which means $\frac{25}{\operatorname{tg} x \cdot \operator... | olympiads | ea0d2995-3f70-5e4f-9ae2-84c1b97c489c | open-r1/OpenR1-Math-220k | [
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298. An unknown polynomial gives a remainder of 2 when divided by $x-1$, and a remainder of 1 when divided by $x-2$. What remainder does this polynomial give when divided by $(x-1)(x-2)$? | 298. Let $p(x)$ be our unknown polynomial, $q(x)$ the quotient from dividing this polynomial by $(x-1)(x-2)$, and $r(x)=a x+b$ the sought remainder:
$$
p(x)=(x-1)(x-2) q(x)+a x+b
$$
According to the problem, we have:
$$
\begin{aligned}
& p(x)=(x-1) q_{1}(x)+2, \text { hence } p(1)=2 \\
& p(x)=(x-2) q_{2}(x)+1, \text... | olympiads | 1df044d1-4761-5249-8dee-18cf1c169176 | open-r1/OpenR1-Math-220k | [
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A local restaurant was offering a "Build Your Own Hot Brownie" dessert. The brownie cost $2.50 and each scoop of ice cream costs $1.00. Any syrup costs $0.50 and nuts cost $1.50. If Juanita orders the brownie with 2 scoops of ice cream, double syrup, and nuts, how much will her dessert cost? | She orders 2 scoops of ice cream that's $1.00 per scoop so the ice cream costs 2*1 = $<<2*1=2.00>>2.00.
She wants double syrup. The syrup is $0.50 so that will cost 2*.50 = $<<2*.50=1.00>>1.00.
The brownie is $2.50, the ice cream will be $2.00, the syrup will be $1.00 and the nuts will be $1.50 bringing her total to 2.... | null | null | openai/gsm8k | [
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## Problem Statement
Find the $n$-th order derivative.
$y=\frac{4 x+7}{2 x+3}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{4 x+7}{2 x+3}\right)^{\prime}=\frac{4 \cdot(2 x+3)-(4 x+7) \cdot 2}{(2 x+3)^{2}}=\frac{8 x+12-8 x-14}{(2 x+3)^{2}}=-\frac{2}{(2 x+3)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-\frac{2}{(2 x+3)^{2}}\right)^{\prime}=-\frac{2 \cdot(-2)}{(2 x... | olympiads | 04b86e37-6fac-528a-8d20-d5ec3b016bf3 | open-r1/OpenR1-Math-220k | [
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5.4. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
y=x^{2}+x\left(\log _{5} 2-\log _{2} 10\right)-\log _{2} 25-3 \log _{5} 2
$$
do not exceed 7. | Solution. Let $a=\log _{2} 5$. Then the condition of the problem will turn into the inequality
$$
x^{2}-\left(a-\frac{1}{a}+1\right) x-\left(2 a+\frac{3}{a}+7\right) \leqslant 0 .
$$
Considering that $a \in(2,3)$, we get $x \in\left[-\frac{1}{a}-2, a+3\right]$. Since $-3<-\frac{1}{a}-2<-2,5<a+3<6$, the integer soluti... | olympiads | 9aaf45ac-5483-5f88-8dac-625c3117add9 | open-r1/OpenR1-Math-220k | [
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] |
The three sides of a triangle are $3$, $2a+1$, and $8$. Find the range of $a$. | To find the range of $a$ for the sides of a triangle with lengths $3$, $2a+1$, and $8$, we apply the triangle inequality theorem. This theorem states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.
Given the sides $3$, $2a+1$, and $8$, we have two ... | cn_k12 | 2692ab4e-19f2-5f2f-a469-9849a3cd7203 | open-r1/OpenR1-Math-220k | [
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2. If real numbers $x, y$ satisfy $x^{2}+2 \cos y=1$, then the range of $x-\cos y$ is | 2.
Answer: $[-1, \sqrt{3}+1]$.
Solution: Since $x^{2}-1-2 \cos y \in[-1,3]$, it follows that $x \in[-\sqrt{3}, \sqrt{3}]$. From $\cos y=\frac{1-x^{2}}{2}$, we have $x-\cos y-x-\frac{1-x^{2}}{2}-\frac{1}{2}(x+1)^{2}-1$. Therefore, when $x=-1$, $x-\cos y$ has a minimum value of -1 (at this time, $y$ can be $\frac{\pi}{2}... | olympiads | b8d86f9e-34e9-502a-bd8a-143dc4641054 | open-r1/OpenR1-Math-220k | [
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4. A, B, and C buy study materials together. It is known that A and B paid a total of 67 yuan, B and C paid a total of 64 yuan, and A and C paid a total of 63 yuan. Therefore, A paid $\qquad$ yuan. | 【Analysis】A and B paid a total of 67 yuan, B and C paid a total of 64 yuan, and A and C paid a total of 63 yuan. Adding these three amounts together gives twice the total amount paid by the three people. Dividing this by 2 gives the total amount paid by the three people. Subtracting the amount paid by B and C from this... | olympiads | 6c8ae736-3140-5ead-8a18-5778f96bef2c | open-r1/OpenR1-Math-220k | [
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] |
Evochkimov M.A.
Vasya was given a quadratic equation $x^{2}+p_{1} x+q_{1}=0$ for homework, where $p_{1}$ and $q_{1}$ are integers. He found its roots $p_{2}$ and $q_{2}$ and wrote a new equation $x^{2}+p_{2} x+q_{2}=0$. Repeating the operation three more times, Vasya noticed that he had solved four quadratic equations... | The fifth equation with integer coefficients should not have distinct real roots. Therefore, if its coefficients are denoted by $p_{5}$ and $q_{5}$, then $p_{5}^{2} \leq 4 q_{5}$ and $q_{5}^{2} \leq 4 p_{5}$. Both numbers are positive, and by squaring the first inequality and substituting the condition from the second,... | olympiads | b47ea051-a4a4-5900-9018-103c168eabf8 | open-r1/OpenR1-Math-220k | [
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From a pocket containing 2 red balls and 2 black balls, two balls are drawn. The mutually exclusive but not complementary events are $(\quad)$.
A: At least one black ball and both are black balls
B: At least one black ball and at least one red ball
C: Exactly one black ball and exactly two black balls
D: At least one b... | For option $A$: The event "at least one black ball" and the event "both are black balls" can occur simultaneously, such as when both balls are black. Therefore, these two events are not mutually exclusive, so option $A$ is incorrect.
For option $B$: The event "at least one black ball" and the event "at least one red b... | cn_k12 | 5af4696f-4b12-53f6-b98d-7b38f7a6a2bd | open-r1/OpenR1-Math-220k | [
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14. Given that $\tan \alpha+\cot \alpha=4$, find $\sqrt{\sec ^{2} \alpha+\csc ^{2} \alpha-\frac{1}{2} \sec \alpha \csc \alpha}$. | 14. $\sqrt{14}$ | olympiads | affa9753-84ea-5f05-b82e-6dfa64160980 | open-r1/OpenR1-Math-220k | [
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] |
In the Cartesian coordinate system, the graph corresponding to the equation $x^{2}+y^{2}=1$ undergoes a scaling transformation $\begin{cases} x'=5x \\ y'=3y \end{cases}$. The equation corresponding to the graph after this transformation is ______. | Solution: Given the scaling transformation $\begin{cases} x'=5x \\ y'=3y \end{cases}$, we can derive: $\begin{cases} x= \frac {1}{5}x' \\ y= \frac {1}{3}y' \end{cases}$
Substituting into the equation $x^{2}+y^{2}=1$, we get: $\frac {x'^{2}}{25}+ \frac {y'^{2}}{9}=1$, which simplifies to $\frac {x^{2}}{25}+ \frac {y^{... | cn_k12 | 87b777f4-2578-5a96-9eea-f769915d29b4 | open-r1/OpenR1-Math-220k | [
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] |
Which of the following conditions implies that plane $\alpha$ is parallel to plane $\beta$? ( )
A: There are infinitely many lines on $\alpha$ parallel to $\beta$
B: Line $a$ is parallel to $\alpha$, and $a$ is parallel to $\beta$
C: Line $b$ is parallel to $\alpha$, and plane $\alpha$ is parallel to plane $\beta$
D: N... | For option A, if there are infinitely many lines in $\alpha$ that are all parallel to $\beta$, it is possible that plane $\alpha$ and plane $\beta$ are parallel, but it is also possible that they intersect, hence option A is incorrect.
For option B, if line $a$ is parallel to $\alpha$, and $a$ is also parallel to $\be... | cn_k12 | a303a406-d064-51a1-9a10-257d18aceb79 | open-r1/OpenR1-Math-220k | [
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] |
$\left.\begin{array}{l}\text { [Inclusion-Exclusion Principle]} \\ {[\quad \text { Word Problems (Miscellaneous). }}\end{array}\right]$
In the garden, Anya and Vitya had 2006 rose bushes. Vitya watered half of all the bushes, and Anya watered half of all the bushes. It turned out that exactly three bushes, the most be... | Vitya watered 1003 bushes, of which 1000 he watered alone, and three - together with Anya. Similarly, Anya watered 1003 bushes, of which 1000 she watered alone, and three - with Vitya. Therefore, together they watered $1000+1000+3=2003$ bushes. Thus, $2006-2003=3$ rose bushes remained unwatered.
## Answer
3 bushes. | olympiads | 16626c61-e67e-5baa-b356-e3b49a2cd062 | open-r1/OpenR1-Math-220k | [
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] |
6. Find the smallest natural number $N$ such that the number $99 N$ consists only of threes.
ANSWER: 3367. | Solution. The number 33N must consist of all ones. A number is divisible by 33 if it is divisible by 3 and by 11. A number consisting of all ones is divisible by 3 if the number of ones is a multiple of 3, and it is divisible by 11 if the number of ones is a multiple of 2. The smallest such number is 111111, so 33N = 1... | olympiads | 5b8eba88-9522-5c14-90d9-7d05a6cd8d94 | open-r1/OpenR1-Math-220k | [
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] |
12. (5 points) Before New Year's Day, Xiaofang made greeting cards for her five classmates. When putting the cards into envelopes, she made a mistake, and none of the five classmates received the card Xiaofang made for them; instead, they received cards Xiaofang made for others. In total, there are $\qquad$ possible sc... | 【Analysis】(1) When all five greeting cards are given incorrectly, and there are no mutual wrongs between any two, then student No. 1 has four wrong ways. For example, if student No. 1 gets card No. 2, then student No. 2 cannot get card No. 1 or card No. 2, leaving only 3 wrong ways. Student No. 3, besides not getting c... | olympiads | a42f4bef-9b81-5837-9653-9d9e93610737 | open-r1/OpenR1-Math-220k | [
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] |
96. The school has newly purchased 50 office desks, which need to be transported back to the campus. It is known that a large cart can transport 6 desks at a cost of 10 yuan per cart; a small cart can transport 4 desks at a cost of 8 yuan per cart. To transport all the desks, the minimum transportation cost is $\qquad$... | Answer: 86 | olympiads | a99834c7-8456-588c-9e46-d713f861b865 | open-r1/OpenR1-Math-220k | [
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] |
2. If the graph of the function $y=x^{2}+a x+a$ given in Figure 1 intersects the $x$-axis at only one point, then $a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 2. D.
From the function $y=x^{2}+a x+a$ having only one zero, we know
$$
\begin{array}{l}
\Delta=a^{2}-4 \times 1 \times a=0 \\
\Rightarrow a(a-4)=0 .
\end{array}
$$
Thus, $a=0$ or 4.
But from the axis of symmetry $-\frac{a}{2} \neq 0$, we get $a \neq 0$.
Therefore, $a=4$. | cn_contest | a8874ac4-de16-50c6-b8f3-7db91fc39a32 | open-r1/OpenR1-Math-220k | [
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] |
[ Arithmetic operations. Numerical identities ]
Find the sum of the digits in the decimal representation of the number $4^{12} \cdot 5^{21}$.
# | Let's transform: $4^{12} \cdot 5^{21}=2^{3} \cdot 10^{21}=80 \ldots 0$.
## Answer
8. | olympiads | 0c9bb0bf-6b15-5fb2-beb3-616e81375b95 | open-r1/OpenR1-Math-220k | [
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] |
1. Kelvin the Frog is going to roll three fair ten-sided dice with faces labelled $0,1,2, \ldots, 9$. First he rolls two dice, and finds the sum of the two rolls. Then he rolls the third die. What is the probability that the sum of the first two rolls equals the third roll? | Answer: $\square$
First, there are $10^{3}=1000$ triples $(a, b, c)$. Now, we should count how many of these triples satisfy $a+b=c$. If $c=0$, we get 1 triple $(0,0,0)$. If $c=1$, we get two triples $(1,0,1)$ and $(0,1,1)$. Continuing, this gives that the total number of triples is $1+2+\cdots+10=55$. Therefore, our f... | olympiads | 3911c757-e952-5968-8140-9137696c9f3e | open-r1/OpenR1-Math-220k | [
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] |
If an even function $f(x)$ defined on $\mathbb{R}$ is a decreasing function on $[0, +\infty)$, then we have ( )
A: $f(3) < f(-2) < f(1)$
B: $f(1) < f(-2) < f(3)$
C: $f(-2) < f(1) < f(3)$
D: $f(3) < f(1) < f(-2)$ | Since the function $f(x)$ defined on $\mathbb{R}$ is a decreasing function on $[0, +\infty)$,
it follows that $f(3) < f(2) < f(1)$,
Since the function is an even function,
it follows that $f(3) < f(-2) < f(1)$,
Therefore, the correct choice is: $\boxed{\text{A}}$.
**Analysis:** By utilizing the monotonicity a... | cn_k12 | ab007d4f-6aa1-51fa-ae09-31d2957658e2 | open-r1/OpenR1-Math-220k | [
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] |
Bucky earns money each weekend catching and selling fish. He wants to save up for a new video game, which costs $60. Last weekend he earned $35. He can earn $5 from trout and $4 from blue-gill. He caught five fish this Sunday. If 60% were trout, and the rest were blue-gill, how much more does he need to save before he ... | He is $25 short for the game because 60 - 35 = <<60-35=25>>25
He caught 3 trout because 5 x .6 = <<5*.6=3>>3
He caught 2 blue-gill because 5 - 3 = <<5-3=2>>2
He earned $15 from the trout because 3 x 5 = <<3*5=15>>15
He earned $8 from the blue-gill because 2 x 4 = <<2*4=8>>8
He earned $23 total because 15 + 8 = <<15+8=2... | null | null | openai/gsm8k | [
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] |
Given $f(x) = \sin x + 2x$, where $x \in \mathbb{R}$, and $f(1-a) + f(2a) < 0$, then the range of $a$ is. | Since $f(x) = \sin x + 2x$, for $x \in \mathbb{R}$, and $f(-x) = \sin(-x) + 2(-x) = -\sin x - 2x = -f(x)$,
it follows that the function is an odd function.
Furthermore, $f'(x) = \cos x + 2 > 0$, which means the function is increasing.
Therefore, $f(1-a) + f(2a) < 0$ can be rewritten as $f(1-a) < -f(2a) = f(-2a)$,... | cn_k12 | 91544c36-7d9e-5fcb-95bd-33a25d476909 | open-r1/OpenR1-Math-220k | [
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] |
An isosceles right triangle with legs of length $8$ is partitioned into $16$ congruent triangles as shown. The shaded area is
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$ | Solution 1
Because the smaller triangles are congruent, the shaded area take $\frac{10}{16}$ of the largest triangles area, which is $\frac{8 \times 8}{2}=32$, so the shaded area is $\frac{10}{16} \times 32= \boxed{\text{(B)}\ 20}$.
Solution 2
Each of the triangle has side length of $\frac{1}{4} \times 8=2$, so the a... | amc_aime | 6791b888-bdc3-5528-aebd-95b8f72b4ef0 | open-r1/OpenR1-Math-220k | [
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Example 6 Given that $f(x)$ is an $n(n>0)$ degree polynomial of $x$, and for any real number $x$, it satisfies:
$$
8 f\left(x^{3}\right)-x^{6} f(2 x)-2 f\left(x^{2}\right)+12=0
$$
Find $f(x)$ | Let the highest degree term of $f(x)$ be $a_{n} x^{n}\left(a_{n} \neq 0\right)$, then the highest degree terms of $8 f\left(x^{3}\right), x^{6} f(2 x), 2 f\left(x^{2}\right)$ are $8 a_{n} \cdot x^{3 n}, 2^{n} a_{n} x^{n \cdot 6}, 2 a_{n} x^{2 n}$, respectively. Since $2 n<3 n$, by (1) and the Polynomial Identity Theore... | olympiads | b82e0b09-7f72-5048-bbdc-d1a850ceb549 | open-r1/OpenR1-Math-220k | [
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] |
Melanie does her weekly shopping at the farmer's market. She starts with an 8-ounce wheel of brie cheese. Next is a 1 pound loaf of bread. She grabs a pound of tomatoes and 2 pounds of zucchini. After that, she grabs 1 1/2 pounds of chicken breasts and treats herself to 8 ounces of fresh raspberries and 8 ounces of... | She bought 8ounces of cheese, 8-ounce raspberries, 8 ounces of blueberries for a total of 8*3 = <<8*3=24>>24 ounces
16 ounces are in 1 pound and she has 24 ounces of food so she has 24/16 = <<24/16=1.5>>1.5 pounds of food
Those items are 1.5 pounds and she bought 1 pound of bread, 1 pound of tomatoes, 2 pounds of zucch... | null | null | openai/gsm8k | [
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] |
1. It is known that the sequence of numbers $a_{1}, a_{2}, \ldots$, is an arithmetic progression, and the sequence of numbers $a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4} \ldots$, is a geometric progression. It is known that $a_{1}=1$. Find $a_{2017}$. | Answer: $a_{2017}=1$;
Trunov K.V.
## Solution:
Since the sequence of numbers $a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4} \ldots$, is a geometric progression, then $\left(a_{n} a_{n+1}\right)^{2}=\left(a_{n-1} a_{n}\right)\left(a_{n+1} a_{n+2}\right)$ for $n \geq 2$. From this, we obtain that $a_{n} a_{n+1}=a_{n-1} a_{n+2... | olympiads | a735bbe0-9f60-5713-9bc6-d58ea36c5f9b | open-r1/OpenR1-Math-220k | [
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] |
Which of the following statements is correct? ( )
A: $-2$ is not an algebraic expression
B: $-a$ represents a negative number
C: The coefficient of $$\frac {3ac}{4}$$ is 3
D: $x+1$ is an algebraic expression | Solution: A, $-2$ is an algebraic expression, so this option is incorrect;
B, $-a$ is not necessarily a negative number, so this option is incorrect;
C, The coefficient of $$\frac {3ac}{4}$$ is $$\frac {3}{4}$$, so this option is incorrect;
D, $x+1$ is an algebraic expression, so this option is correct.
Therefo... | cn_k12 | fa88a443-197c-5330-9bd7-513527f89abf | open-r1/OpenR1-Math-220k | [
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] |
Given that $a=-6$, it is a condition for the line $l\_1$: $ax+(1-a)y-3=0$ to be perpendicular to the line $l\_2$: $(a-1)x+2(a+3)y-2=0$ ($\quad$).
A: Sufficient and unnecessary condition
B: Necessary and insufficient condition
C: Necessary and sufficient condition
D: Neither necessary nor sufficient condition | First, we need to find the values of $a$ for which the lines $l\_1$ and $l\_2$ are perpendicular. We can do this by equating the negative reciprocals of their slopes:
$$-\frac{a}{1-a} = -\frac{1}{2}\cdot\frac{a-1}{a+3}$$
Solving this equation, we get $a=1$ or $a=-6$.
Now let's examine each case:
1. When $a=1$, the ... | cn_k12 | c9d78d4e-f7d0-57a9-bac4-6834884158e2 | open-r1/OpenR1-Math-220k | [
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] |
Given a sequence \\(\{a_n\}\) where \\(a_1=1\\) and \\(a_{n+1}=2a_n+4(n\in \mathbb{N}^*)\\), find the general formula for \\(a_n=\\) . | **Analysis**
This problem examines the derivation of a sequence's general formula from its recursive relation. The key is to construct a new sequence based on the recursive relation of the given sequence to find its general formula. This is a basic question.
**Solution**
Given the sequence \\(\{a_n\}\) where \\(a_... | cn_k12 | d09eb002-d292-51f8-829a-77c44843c9c8 | open-r1/OpenR1-Math-220k | [
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] |
$S=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+\ldots+\frac{1}{1991}}$, Find: The integer part of $S$. | 10.【Solution】 $\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+\cdots+\frac{1}{1991}12 \times \frac{1}{1991}=\frac{12}{1991}$
$\therefore \mathrm{S}>165$ and $\mathrm{s}<\frac{1991}{12}=165 \frac{11}{12}$
Thus, the integer part of S is 165 | olympiads | 953ad59d-e8f5-5996-a1ed-f8f359b30728 | open-r1/OpenR1-Math-220k | [
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] |
Let $a$ be a given rational number. Write down the measurements of the sides of all "rational" right triangles (where the lengths of all sides are rational) for which the length of one of the legs is $a$.
If $a$ is an integer, find all "integer" - i.e., Pythagorean - triangles among them! | Let $b$ be the other leg, $c$ the measure of the hypotenuse, then we seek the solution of $c^{2}-b^{2}=a^{2}$ with rational $b$ and $c$. $a^{2}=(c+b) \cdot(c-b)$, here $c-b=\frac{p}{q}$ must be rational. Let this fraction be in its simplest form, that is, $p$ and $q$ are relatively prime numbers. From here, $c+b=\frac{... | olympiads | 9f1bde40-e9f6-541c-b2d7-c56d3eaea3f1 | open-r1/OpenR1-Math-220k | [
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] |
In a right triangle, the lengths of the two sides are $6$ and $8$. The length of the third side is ( )
A: $10$
B: $2\sqrt{7}$
C: $2\sqrt{5}$
D: $10$ or $2\sqrt{7}$ | To solve for the length of the third side in a right triangle where the lengths of the two sides are $6$ and $8$, we consider two cases based on the Pythagorean theorem, $a^2 + b^2 = c^2$, where $c$ is the length of the hypotenuse.
**Case 1:** When $8$ is the hypotenuse, and $6$ is one of the legs, we let the length o... | cn_k12 | 88b4f861-b330-5551-9da4-c3ca77e01f4b | open-r1/OpenR1-Math-220k | [
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] |
Given a circle (x-a)2+y2=9 (a>5) with a point M on it such that |OM|=2|MQ| (O is the origin) holds true, Q(2,0), the range of values for the real number a is _____. | Let M(x, y). From |OM|=2|MQ|, we get $$\sqrt {x^{2}+y^{2}}$$=2$$\sqrt {(x-2)^{2}+y^{2}}$$, which simplifies to x2+y2-$$\frac {16}{3}$$x+$$\frac {16}{3}$$=0. The center of this circle is ($$\frac {8}{3}$$,0) with a radius of $$\frac {4}{3}$$.
The problem is transformed into finding the intersection of the circle (x-a)2... | cn_k12 | fc13a0af-96b7-5e06-a842-8da445ebf7d7 | open-r1/OpenR1-Math-220k | [
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] |
2. Let there be a rectangle $A B C D$. The perpendicular bisector of the diagonal $\overline{A C}$ intersects the side $\overline{A B}$ at point $E$, and the side $\overline{C D}$ at point $F$ such that the triangle $E B C$ is isosceles. Determine the measure of the angle $\measuredangle D F E$. | First method:
Since triangle $E B C$ is an isosceles right triangle, it follows that $|\measuredangle E C B|=|\measuredangle B E C|=45^{\circ}$.
Since $A B C D$ is a rectangle, it follows ... | olympiads | eb433cab-1f5c-598b-a6a4-5b4cf90d5312 | open-r1/OpenR1-Math-220k | [
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] |
4. In $\triangle A B C$, if $\angle A=2 \angle B$, side $b=4, c=5$, then side $a$ equals ( ).
(A) 6
(B) 7
(C) $3 \sqrt{5}$
(D) 5 | 4. A.
As shown in the figure, draw the angle bisector $AD$ of $\angle A$. By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
CD=\frac{4}{9} a, BD=\frac{5}{9} a . \\
\chi \angle CDA=2 \angle DAB=
\end{array}
$$
$\angle A$,
$$
\begin{array}{l}
\therefore \triangle ADC \sim \triangle ABC . \\
\text { Hence } \fra... | cn_contest | e0941be6-b78d-5419-86a0-509a6e082cb2 | open-r1/OpenR1-Math-220k | [
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] |
4. Solve the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
x^{3}+y^{5}=1
\end{array}\right.
$$ | Answer: $(0 ; 1),(1 ; 0)$.
Solution. Subtract the first equation from the second:
$$
x^{2}(x-1)+y^{2}\left(y^{3}-1\right)=0 .
$$
From the first equation of the system, it follows that $x \leqslant 1$ and $y \leqslant 1$. Therefore, $x^{2}(x-1) \leqslant 0$ and $y^{2}\left(y^{3}-1\right) \leqslant 0$. The sum of two ... | olympiads | 279d1a44-bcfa-5550-8088-f4ee42d870cf | open-r1/OpenR1-Math-220k | [
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## SUBJECT II
Solve the following logarithmic equation: $x^{\log _{3}(x-1)}+2 \cdot(x-1)^{\log _{3} x}=3 \cdot x^{2}$ | ## SUBJECT II
Solve the following logarithmic equation: $x^{\log _{3}(x-1)}+2 \cdot(x-1)^{\log _{3} x}=3 \cdot x^{2}$
Selected by Prof. Bara Lajos from "The Most Beautiful Math Problems" by Dan and Vlad Sachelarie.
Grading Rubric
Initial conditions are set: $x>0, x>1, x \neq 1, x-1 \neq 1 \Rightarrow x>1, x \neq 2$... | olympiads | 6c1bd64a-f72d-5a49-b696-f0feafb6529c | open-r1/OpenR1-Math-220k | [
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] |
12. There are 2021 balls in a crate. The balls are numbered from 1 to 2021 . Erica works out the digit sum for each ball. For example, the digit sum of 2021 is 5, since $2+0+2+1=5$.
Erica notes that balls with equal digit sums have the same colour and balls with different digit sums have different colours.
How many dif... | SOLUTION
028
The largest possible digit sum is that of 1999 which is 28 . The smallest is that of 1 , which is 1 . Each of the $9 \mathrm{~s}$ in 1999 can be replaced by any of $0,1, \ldots, 8$. So all digit sums between 1 and 28 can be achieved. Therefore there are 28 different digit sums and colours. | olympiads | 66f1dd59-e362-5de1-b6c5-ec01e500d0eb | open-r1/OpenR1-Math-220k | [
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] |
The front view of a cylinder is a rectangle with an area of 6. Its lateral area is ( )
A: $8\pi$
B: $6\pi$
C: $4\pi$
D: $3\pi$ | Let the height of the cylinder be $h$.
Since the front view of the cylinder is a rectangle with an area of 6, the diameter of the base circle of the cylinder is $\frac{6}{h}$.
Therefore, the lateral area of this cylinder is $S = \pi \cdot \frac{6}{h} \cdot h = 6\pi$.
Hence, the correct option is: $\boxed{B}$.
Le... | cn_k12 | b710abf2-53ac-5622-9400-9552efdcfedd | open-r1/OpenR1-Math-220k | [
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8. Three girls and a boy must sit around a table with five chairs, numbered from 1 to 5. To decide their seat, each of the four draws at random one of five slips of paper (numbered from 1 to 5). What is the probability that the empty chair will be between two girls?
(A) $3 / 5$
(B) $2 / 5$
(C) $2 / 3$
(D) $3 / 4$
(E) $... | (8) The correct answer is $(E)$.
Considering only the gender of the 4 people (male or female), the total number of possible configurations is equal to 20: that is, 5 (corresponding to choosing one of the 5 chairs to leave empty) multiplied by 4 (the number of possible choices for the boy's seat, once the empty chair i... | olympiads | f40c1092-2f9e-59e0-9584-bd87a02d52bf | open-r1/OpenR1-Math-220k | [
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Problem 8.6. Vasya thought of three natural numbers with a sum of 1003. Calculating their product, Vasya noticed that it ends with $N$ zeros. What is the maximum value that
$N$ can take? | Answer: 7.
Solution. The product of the three thought-of numbers could end with 7 zeros, for example, if these were the numbers $625, 250, 128$. Indeed, $625+250+128=1003$ and
$$
625 \cdot 250 \cdot 128=5^{4} \cdot\left(2 \cdot 5^{3}\right) \cdot 2^{7}=2^{8} \cdot 5^{7}=2 \cdot 10^{7}=20000000
$$
Suppose there exist... | olympiads | 7255d5e9-fcd2-5ebe-a592-b2640fda3583 | open-r1/OpenR1-Math-220k | [
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] |
The vertices of triangle ABC correspond to the complex numbers $Z_1$, $Z_2$, and $Z_3$, respectively. The complex number $Z$ satisfies $|Z-Z_1|=|Z-Z_2|=|Z-Z_3|$. The corresponding point of $Z$ in triangle ABC is ( )
A: Circumcenter
B: Incenter
C: Centroid
D: Orthocenter | Given $|Z-Z_1|=|Z-Z_2|=|Z-Z_3|$,
this means the distance from $Z$ to the three vertices is equal,
therefore, $Z$ is the center of the circumscribed circle of the triangle,
hence, the correct choice is $\boxed{\text{A}}$.
**Analysis:** Based on the equal distances from $Z$ to the three vertices of the triangle,... | cn_k12 | 85975bb0-b453-5f8b-bc70-70d4ae1ad172 | open-r1/OpenR1-Math-220k | [
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9.1. Let the numbers $x, y, u, v$ be distinct and satisfy the relation $\frac{x+u}{x+v}=\frac{y+v}{y+u}$. Find all possible values of the sum $x+y+u+v$. | Answer. $x+y+u+v=0$.
Solution. Let's bring the difference between the left and right parts of the expression from the condition, equal to zero, to a common denominator and factor the numerator: $\frac{x+u}{x+v}-\frac{y+v}{y+u}=\frac{x u+y u+u^{2}-y v-x v-v^{2}}{(x+v)(y+u)}=\frac{(x+y)(u-v)+(u+v)(u-v)}{(x+v)(y+u)}=\fra... | olympiads | bd06fc75-d5a3-5e85-afa2-17396101d8be | open-r1/OpenR1-Math-220k | [
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] |
The value of $0.3+0.03$ is
(A) 0.303
(B) 0.6
(C) 3.3
(D) 0.33
(E) 0.06 | Evaluating, $0.3+0.03=0.33$.
ANSwer: (D) | olympiads | f1c62e7f-9e81-5cc6-8a86-eabd348f6fd4 | open-r1/OpenR1-Math-220k | [
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] |
Problem 9.2. The least common multiple of four pairwise distinct numbers is 165. What is the maximum value that the sum of these numbers can take? | Answer: 268.
Solution. Since 165 is the least common multiple of four numbers, these numbers are divisors of 165. To maximize the sum of these numbers, it is sufficient to take the four largest divisors of 165. If one of them is the number 165 itself, then the LCM will definitely be equal to it.
Then the maximum sum ... | olympiads | f10cfb1a-5ee9-566b-a457-2a2182c8c69c | open-r1/OpenR1-Math-220k | [
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7. Person A and Person B independently and repeatedly roll a fair die until the first 6 appears. The probability that the number of rolls by A and B differs by no more than 1 is $\qquad$ | 7. $\frac{8}{33}$.
Let the number of throws made by A and B be $\xi$ and $\eta$, respectively. Then the required probability is
$$
\begin{array}{l}
\sum_{i=1}^{\infty}(P(\xi=\eta=i)+P(\xi=i, \eta=i+1)+ \\
P(\xi=i+1, \eta=i)) .
\end{array}
$$
By independence, the required probability is
$$
\begin{array}{l}
\sum_{i=1}^... | olympiads | bf5457c8-2c4e-52c2-b486-6b83aa3966db | open-r1/OpenR1-Math-220k | [
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30. The figure below shows two circles with centres $\mathrm{A}$ and $\mathrm{B}$, and a line $\mathrm{L}$ which is a tangent to the circles at $X$ and $Y$. Suppose that $X Y=40 \mathrm{~cm}, A B=41 \mathrm{~cm}$ and the area of the quadrilateral $\mathrm{ABYX}$ is $300 \mathrm{~cm}^{2}$. If $a$ and $b$ denote the area... | 30. Answer: 16
Let the radii of the circles with centres $\mathrm{A}$ and $\mathrm{B}$ be $x \mathrm{~cm}$ and $y \mathrm{~cm}$ respectively, and let $\mathrm{C}$ be the point on the line segment $\mathrm{BY}$ such that $\mathrm{AC}$ is parallel to $\mathrm{XY}$. Then $\mathrm{AC}=40 \mathrm{~cm}$ and $\mathrm{BC}=(y-... | olympiads | 83e1f019-54b4-5413-b288-3905d4ee52b6 | open-r1/OpenR1-Math-220k | [
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] |
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