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2. In quadrilateral $ABCD$, $AC$ bisects $\angle BCD$, $\angle CAD=20^{\circ}, \angle CBD=40^{\circ}$. Then $\angle ABD=(\quad)$.
(A) $50^{\circ}$
(B) $60^{\circ}$
(C) $70^{\circ}$
(D) $80^{\circ}$ | $$
\begin{array}{l}
\angle C A E = \angle C B E \\
= 40^{\circ}, \\
\angle D A E \\
= \angle C A E - \angle C A D \\
= 20^{\circ} = \angle C A D, \\
\angle A E B = \angle A C B = \angle A C D . \\
\text { Also, } A D = A D, \text { so } \\
\triangle A D E \cong \triangle A D C \\
\Rightarrow A E = A C \Rightarrow \ove... | cn_contest | bdcc701f-6f5e-53de-a486-fd8df3b200e4 | open-r1/OpenR1-Math-220k | [
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] |
(5 points) Among the following propositions:
(1) Two planes parallel to the same line are parallel;
(2) Two planes parallel to the same plane are parallel;
(3) Two lines perpendicular to the same line are parallel;
(4) Two lines perpendicular to the same plane are parallel.
The number of correct propositions ... | B
Key point: Relationship between planes.
Topic: Reading comprehension.
Analysis: (1) Two planes parallel to the same line can either be parallel or intersect; (2) According to the parallel postulate, two planes parallel to the same plane are parallel; (3) Two lines perpendicular to the same line can either be pa... | cn_k12 | 5c0290eb-7d41-54a0-adc2-aa1103181dc1 | open-r1/OpenR1-Math-220k | [
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] |
Given $x^{a}=2$ and $x^{b}=9$, find $x^{3a-b}$. | Given $x^{a}=2$ and $x^{b}=9$, we aim to find $x^{3a-b}$.
Starting with the given equations:
1. $x^{a}=2$
2. $x^{b}=9$
We can manipulate these equations to find $x^{3a-b}$ as follows:
- First, we raise both sides of the first equation to the power of 3, which gives us $(x^{a})^{3} = 2^{3}$.
- This simplifies to $x^{3... | cn_k12 | 78d8d910-153f-52c9-937e-df673c621b9f | open-r1/OpenR1-Math-220k | [
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] |
Given the sets $A=\left\{x|x < -2\text{ or }3 < x < 4\right\},B=\left\{x|{x}^{2}-2x-15\leqslant 0\right\}$. Find:
$(1)A\cap B$;
$(2)$ If $C=\left\{x|x\geqslant a\right\}$, and $B\cap C=B$, find the range of $a$. | This problem involves the operation of set intersections, the solution of quadratic inequalities, and the determination of parameter ranges using set relationships, which is a basic problem.
$(1)$ First, solve the quadratic inequality to find set $B$. The quadratic inequality ${x}^{2}-2x-15\leqslant 0$ can be factored... | cn_k12 | b1546af9-bc15-5138-b411-9fd6d10dddf0 | open-r1/OpenR1-Math-220k | [
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] |
One barnyard owl makes 5 hoot sounds per minute. If 5 less than 20 hoots per minute are heard coming out of the barn, how many Barnyard owls would this noise be coming from? | 5 less than 20 hoots per minute is 20-5=<<20-5=15>>15 hoots per minute.
If one barnyard owl makes 5 hoot sounds per minute, then 15 hoots per minute are coming from 15/5=<<15/5=3>>3 barnyard owls.
#### 3 | null | null | openai/gsm8k | [
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] |
Task 1. Represent in the form of an irreducible fraction:
$$
\frac{5+10}{15}+\frac{20+25}{30}+\ldots+\frac{80+85}{90}
$$ | Answer: $\frac{191}{20}$.
## Solution.
$$
\begin{gathered}
\quad \frac{5+10}{15}+\frac{20+25}{30}+\ldots+\frac{80+85}{90}=\frac{1+2}{3}+\frac{4+5}{6}+\ldots+\frac{16+17}{18}= \\
=\frac{(3-2)+(3-1)}{3}+\frac{(6-2)+(6-1)}{6}+\ldots+\frac{(18-2)+(18-1)}{18}= \\
=2+\frac{-3}{3}+2+\frac{-3}{6}+2+\frac{-3}{9}+\ldots+2+\fra... | olympiads | 2174ff55-9e36-568e-8a68-822050cf0366 | open-r1/OpenR1-Math-220k | [
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] |
Three. (20 points) In the pyramid $S-ABC$, $SA=4, SB \geqslant 7, SC \geqslant 9, AB=5, BC \leqslant 6, AC \leqslant 8$. Try to find the maximum volume of the pyramid $S-ABC$. | Three, let the pyramid $S-ABC$ satisfy the given conditions:
$$
\begin{array}{c}
S A \leqslant 4, S B \geqslant 7, S C \geqslant 9, \\
A B=5, B C \leqslant 6, A C \leqslant 8 .
\end{array}
$$
We estimate the volume of the pyramid using $\triangle A B S$ as the base.
According to the cosine rule, we have
$$
\cos \angle... | cn_contest | 7d4af377-e0d2-5e7d-956b-55b9715a0817 | open-r1/OpenR1-Math-220k | [
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] |
The focus of the parabola $y^{2}=16x$ is $F$, and its directrix is $l$. Let $P$ be a point on the parabola, and $PA\perp l$ with $A$ being the foot of the perpendicular. If the slope of line $AF$ is $-1$, then the value of $|PF|$ is ______. | Since the equation of the parabola is $y^{2}=16x$,
the focus $F$ is at $(4,0)$, and the equation of the directrix $l$ is $x=-4$.
Since the slope of line $AF$ is $-1$, the equation of line $AF$ is $y=-(x-4)$.
By solving $\begin{cases} x=-4 \\ y=-(x-4) \end{cases}$, we find the coordinates of point $A$ to be $(-4,8... | cn_k12 | 80900dca-a0a0-5941-b6f3-2a91343da2a7 | open-r1/OpenR1-Math-220k | [
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] |
13. Given that $a$ and $b$ are constants, and $a \neq 0$, the function $f(x)=\frac{x}{a x+3}$ satisfies $f(2)=1$, and the value of $x$ that makes $f(x)=x$ is unique. (1) Find the values of $a$ and $b$; (2) Let $x_{1}=1, x_{n}=$ $f\left(x_{n-1}\right)(n=2,3, \cdots)$, find the general term formula of the sequence $\left... | 13. (1) From $f(2)=1$ we get $2a+b=2$, and the equation $f(x)=x$ has only one solution, i.e., $ax^2+(b-1)x=0$. It has only one solution, which must mean $b-1=0$, so $b=1, a=\frac{1}{2} \cdot f(x)=\frac{2x}{x+2}$;
(2) $x_{n}=\frac{2x_{n-1}}{x_{n-1}+2}$, thus $\frac{1}{x_{n}}=\frac{1}{x_{n-1}}+\frac{1}{2}$, hence $\left\... | olympiads | 9d5e795c-92fd-552c-9fe1-15a064e70e71 | open-r1/OpenR1-Math-220k | [
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] |
Example 1 Given $f(x)=\frac{a^{x}}{a^{x}+\sqrt{a}}$ (where $a$ is a constant greater than 0), find the value of $f\left(\frac{1}{1001}\right)+f\left(\frac{2}{1001}\right)+\cdots+$ $f\left(\frac{1000}{1001}\right)$. | Consider the value of $f(x)+f(1-x)$:
$$
\begin{array}{l}
f(x)+f(1-x)=\frac{a^{x}}{a^{x}+\sqrt{a}}+\frac{a^{1-x}}{a^{1-x}+\sqrt{a}}=\frac{a^{x}}{a^{x}+\sqrt{a}}+\frac{a}{a+\sqrt{a} \cdot a^{x}}=\frac{a^{x}}{a^{x}+\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{a}+a^{x}}=1 \\
\text { Therefore } \quad f\left(\frac{1}{1001}\right)+f\left... | olympiads | b337de95-200a-5074-9093-c760a11fc08b | open-r1/OpenR1-Math-220k | [
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] |
Given $a+b+c=1$, $a^2+b^2+c^2=2$, and $a^3+b^3+c^3=3$, find the value of $a^4+b^4+c^4$. | **Solution**: Start with $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$,
which means $1=2+2(ab+bc+ac)$,
thus $ab+bc+ac=-\frac{1}{2}$,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$,
which gives $3-3abc=2+\frac{1}{2}$,
thus $abc=\frac{1}{6}$;
Since $(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)$,
we have $\frac{1}{... | cn_k12 | 541e44a1-e2f2-54bf-a3f4-58a055ed7313 | open-r1/OpenR1-Math-220k | [
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] |
26. In the diagram below, $A$ and $B(20,0)$ lie on the $x$-axis and $C(0,30)$ lies on the $y$-axis such that $\angle A C B=90^{\circ}$. A rectangle $D E F G$ is inscribed in triangle $A B C$. Given that the area of triangle $C G F$ is 351 , calculate the area of the rectangle $D E F G$. | 26. Answer: 468 .
Note that $A O=\frac{30^{2}}{20}=45$. Then the area of $\triangle A B C$ is $\frac{(20+45) \times 30}{2}=975$.
Let the height of $\triangle C G F$ be $h$. Then
$$
\left(\frac{h}{30}\right)^{2}=\frac{351}{975}=\left(\frac{3}{5}\right)^{2} \Rightarrow \frac{h}{30-h}=\frac{3}{2}
$$
Note that the rectang... | olympiads | 037bed16-2b18-54bc-8bae-c9952adc1cab | open-r1/OpenR1-Math-220k | [
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] |
7. Highway (from 7th grade, 3 points). A highway running from west to east intersects with $n$ equal roads, numbered from 1 to $n$ in order. Cars travel on these roads from south to north and from north to south. The probability that a car will approach the highway from each of these roads is $\frac{1}{n}$. Similarly, ... | Answer: $\frac{2 k n-2 k^{2}+2 k-1}{n^{2}}$.
Solution. The specified event will occur in one of three cases. | olympiads | 07e7ce45-f8fc-5816-ac84-32ba5e8b2780 | open-r1/OpenR1-Math-220k | [
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] |
1. Ivan and Petr are running in different directions on circular tracks with a common center, and initially, they are at the minimum distance from each other. Ivan completes one full circle every 20 seconds, while Petr completes one full circle every 28 seconds. After what least amount of time will they be at the maxim... | Answer: $35 / 6$ seconds.
Solution. Ivan and Petr will be at the minimum distance from each other at the starting points after the LCM $(20,28)=140$ seconds. During this time, Ivan will complete 7 laps, and Petr will complete 5 laps relative to the starting point. Consider this movement in a reference frame where Petr... | olympiads | 720e7d85-4e9e-5a8f-b149-6308f9ded35b | open-r1/OpenR1-Math-220k | [
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] |
7. Given $\boldsymbol{A} \boldsymbol{B}=(k, 1), \boldsymbol{A} \boldsymbol{C}=(2,3)$. Then the value of $k$ that makes $\triangle A B C$ a right triangle is ( ).
(A) $\frac{3}{2}$
(B) $1-\sqrt{2}$
(C) $1-\sqrt{3}$
(D) $-\sqrt{5}$ | 7.C.
If $\angle B A C=90^{\circ}$, then $A B \cdot A C=2 k+3=0$.
Solving for $k$ gives $k=-\frac{3}{2}$.
If $\angle A B C=90^{\circ}$, then
$$
C B \cdot A B=(A B-A C) \cdot A B=k^{2}-2 k-2=0 \text {. }
$$
Solving for $k$ gives $k=1 \pm \sqrt{3}$.
If $\angle A C B=90^{\circ}$, then
$$
C B \cdot A C=(A B-A C) \cdot A C... | cn_contest | 4f0a28e6-16d6-55e9-9e73-3bb1664a1396 | open-r1/OpenR1-Math-220k | [
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] |
$$
\begin{array}{l}
\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{199}-\frac{1}{200}}{\frac{1}{201^{2}-1^{2}}+\frac{1}{202^{2}-2^{2}}+\cdots+\frac{1}{300^{2}-100^{2}}} \\
= \\
\end{array}
$$ | 3. 400 .
Original expression
$$
\begin{array}{l}
=\frac{\left(1+\frac{1}{2}+\cdots+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{200}\right)}{\frac{1}{202 \times 200}+\frac{1}{204 \times 200}+\cdots+\frac{1}{400 \times 200}} \\
=\frac{\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}}{\frac{1}{400... | cn_contest | c88e4070-cef4-5d28-8e96-414480cd535d | open-r1/OpenR1-Math-220k | [
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] |
3. As shown in Figure 2, in the equilateral $\triangle A B C$, side $A B$ is tangent to $\odot O$ at point $H$, and sides $B C$ and $C A$ intersect $\odot O$ at points $D, E, F, G$. Given that $A G=2, G F=6, F C$ $=1$. Then $D E=$ $\qquad$ | 3. $\sqrt{21}$.
By the secant-tangent theorem, we know $A H^{2}=A G \cdot A F=16$, which means $A H=4$.
Also, $A C=A G+G F+F C=9$, so, $A B=9$.
Therefore, $B H=5$.
Let $B D=x, C E=y$.
By the secant-tangent theorem, we know $B H^{2}=B D \cdot B E$, that is,
$$
25=x(9-y) \text {. }
$$
Also, $C E \cdot C D=C F \cdot C G... | cn_contest | e5290083-8b1c-5b5f-83c4-a558dc9918ae | open-r1/OpenR1-Math-220k | [
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] |
In the x-y plane of the Cartesian coordinate system, the center of a hyperbola is at the origin and its focus is on the y-axis. One of its asymptotes passes through the point (-3, 1). Determine the eccentricity of the hyperbola. | Since the center of the hyperbola is at the origin and its focus is on the y-axis, let the equation of the hyperbola be: $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$
Given that one of its asymptotes passes through the point (-3, 1), we have $$\frac{b}{a} = 3$$, which implies $$b^2 = 9a^2 = c^2 - a^2$$.
Solving for the e... | cn_k12 | e549f357-fd40-52ae-8de3-d87be568b8db | open-r1/OpenR1-Math-220k | [
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] |
The value of $\frac{4 \times 4+4}{2 \times 2-2}$ is
(A) 2
(B) 6
(C) 10
(D) 12
(E) 18 | The value of $\frac{4 \times 4+4}{2 \times 2-2}$ is
(A) 2
(B) 6
(C) 10
(D) 12
(E) 18
Solution
$\frac{4 \times 4+4}{2 \times 2-2}=\frac{16+4}{4-2}=\frac{20}{2}=10$
ANSWER: (C) | olympiads | bbf79f1f-7b32-5806-a937-063936a970a1 | open-r1/OpenR1-Math-220k | [
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] |
1. Calculate:
$$
2 \times\left(1+\frac{-1}{2}\right) \times\left[1+\frac{(-1)^{2}}{3}\right] \times\left[1+\frac{(-1)^{3}}{4}\right] \times \cdots \times\left[1+\frac{(-1)^{2019}}{2020}\right]=
$$
$\qquad$ | $1$ | olympiads | 4cffd246-4f6f-5f13-a34c-98f9f6e048a7 | open-r1/OpenR1-Math-220k | [
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] |
Task 10. (12 points)
The Ivanov family owns an apartment with necessary property worth 3 million rubles, a car that is currently valued at 900 thousand rubles on the market, and savings, part of which, amounting to 300 thousand rubles, is placed in a bank deposit, part is invested in securities worth 200 thousand rubl... | # Answer: 2300000
## Comment
Solution: equity (net worth) = value of assets - value of liabilities. Value of assets $=3000000+900000+300000+200000+100000=$ 4500000 rubles. Value of liabilities $=1500000+500000+200000=2200000$ rubles. Net worth $=4500000-2200000=2300000$ rubles
## MOSCOW FINANCIAL LITERACY OLYMPIAD ... | olympiads | 857b029c-78d7-5f09-9fac-67f463a25a9c | open-r1/OpenR1-Math-220k | [
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] |
6. In the Cartesian coordinate system, $O$ is the origin, points $A(3, a) 、 B(3, b)$ make $\angle A O B=45^{\circ}$, where $a 、 b$ are integers, and $a>b$. Then the number of pairs $(a, b)$ that satisfy the condition is. | 6.6.
Let $\angle A O X=\alpha, \angle B O X=\beta$. Then $\tan \alpha=\frac{a}{3}, \tan \beta=\frac{b}{3}$.
Given $a>b$, we have
$$
\begin{array}{l}
1=\tan 45^{\circ}=\tan (\alpha-\beta) \\
=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta}=\frac{3(a-b)}{9+a b} .
\end{array}
$$
Rearranging gives $(a+3)(b... | cn_contest | 80dbb906-83f4-582b-ae4e-c9b8c64f4ebe | open-r1/OpenR1-Math-220k | [
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] |
In a particular game, each of $4$ players rolls a standard $6{ }$-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the... | Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a $5$ is just the probability that the highest roll in the first round is $5$.
Let $P(x)$ indicate the probability that event $x$ occurs.
We find that $P(\text{No one rolls a 6})-P(\text{No... | amc_aime | 18eb8e08-e5ea-5ff5-8e8a-2e461699bc1e | open-r1/OpenR1-Math-220k | [
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] |
Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$. A fence is located at the horizontal line $y = 0$. On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point... | Clearly Freddy's $x$-coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$-coordinate. Observe that $E(24)=0$, and \[E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}\] for all $y$ such that $1\le y\le 23$. Also note that $E(0)=1+\frac{2E(0)+E(1)... | amc_aime | c567ec81-610d-5837-b349-3593ad703ef2 | open-r1/OpenR1-Math-220k | [
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1. Evaluate the expression: $$125^{ \frac {2}{3}}+( \frac {1}{2})^{-2}-( \frac {1}{27})^{- \frac {1}{3}}+100^{ \frac {1}{2}}+ \frac {lg3+ \frac {1}{4}lg9-lg \sqrt {3}}{lg81-lg27}$$
2. Simplify the expression: $$\frac {a^{ \frac {3}{2}}-1}{a+a^{ \frac {1}{2}}+1}- \frac {a+a^{ \frac {1}{2}}}{a^{ \frac {1}{2}}+1}+ \frac {... | 1. The original expression equals to $$25+4-3+10+ \frac {lg3+lg \sqrt {3}-lg \sqrt {3}}{lg3}=37$$. Therefore, the final answer is $\boxed{37}$.
2. The original expression equals to $$\frac {(a^{ \frac {1}{2}}-1)\cdot (a+a^{ \frac {1}{2}}+1)}{a+a^{ \frac {1}{2}}+1}-$$ $$\frac {a^{ \frac {3}{2}}-a+a-a^{ \frac {1}{2}}-a^... | cn_k12 | 207fdf64-ce59-5254-a09c-bd8914087f25 | open-r1/OpenR1-Math-220k | [
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] |
In acute triangle $\triangle ABC$, the sides opposite angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, and it is given that $2\cos^2 \left(\frac{B+C}{2}\right)+\sin 2A=1$.
(Ⅰ) Find $A$;
(Ⅱ) Given $a=2\sqrt{3}-2$ and the area of $\triangle ABC$ is $2$, find the value of $b+c$. | Solution:
(Ⅰ) In acute triangle $\triangle ABC$, from $2\cos^2 \left(\frac{B+C}{2}\right)+\sin 2A=1$, we can derive $\cos(B+C)+\sin 2A=0$,
which implies $\sin 2A=\cos A$, or $2\sin A\cos A=\cos A$. Solving this gives $\sin A=\frac{1}{2}$, thus $A=\boxed{\frac{\pi}{6}}$.
(Ⅱ) Given $a=2\sqrt{3}-2$ and the area of ... | cn_k12 | 5aea3c18-8bd7-582e-ba29-f1f0ba3adb00 | open-r1/OpenR1-Math-220k | [
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] |
4. There are three mathematics courses: Algebra, Geometry, and Number Theory. If no two of these courses can be scheduled consecutively, then a student has $(\quad)$ different ways to arrange the 6 periods in one day.
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24 | 4. E.
If no two courses can be scheduled consecutively, the three math classes can be arranged in periods $1, 3, 5$; periods $1, 3, 6$; periods $1, 4, 6$; and periods $2, 4, 6$, for a total of four scenarios. Each scenario has
$$
3!=3 \times 2 \times 1=6 \text{ (ways). }
$$
Therefore, there are a total of $4 \times 3... | olympiads | 9532c03f-b46d-5771-aca2-c2f6f9ad5585 | open-r1/OpenR1-Math-220k | [
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] |
A certain high school has a total of 2000 students. Using stratified sampling, a sample of 100 students is drawn from students across three grades. If 30 students are drawn from both the first and second grades, then the number of students in the third grade is $\boxed{800}$. | Given that a total of 100 students are sampled from three grades, and 30 students are drawn from both the first and second grades, we can calculate the number of students sampled from the third grade as follows:
The total number of students sampled from the first and second grades is $30 + 30 = 60$ students.
Therefor... | cn_k12 | 956d591a-d3d5-5426-8710-61421c249c84 | open-r1/OpenR1-Math-220k | [
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] |
Given that $F\_1$ and $F\_2$ are the left and right foci of the hyperbola $C$: $\frac{x^2}{9} - \frac{y^2}{27} = 1$, and point $A$ is on $C$. Point $M$ has coordinates $(2, 0)$, and $AM$ is the bisector of $\angle F\_1AF\_2$. Find the length of $|AF\_2|$. | Let's assume point $A$ is on the right branch of the hyperbola.
Since $AM$ is the bisector of $\angle F\_1AF\_2$, by the angle bisector theorem, we have:
$\frac{|AF\_1|}{|AF\_2|} = \frac{|F\_1M|}{|MF\_2|} = \frac{8}{4} = 2$
Also, by the definition of a hyperbola, we know that:
$|AF\_1| - |AF\_2| = 2a = 6$
Solving ... | cn_k12 | be85457f-dd02-57fe-bea6-161d1ef89cbe | open-r1/OpenR1-Math-220k | [
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] |
# Problem No. 6 (10 points)
The density of a body is defined as the ratio of its mass to the volume it occupies. A homogeneous cube with a volume of \( V = 8 \, \text{m}^3 \) is given. As a result of heating, each of its edges increased by 4 mm. By what percentage did the density of this cube change?
Answer: decrease... | # Solution and evaluation criteria:
Volume of the cube: $v=a^{3}$, where $a$ is the length of the edge, therefore:
$a=2 \partial m=200 mm$.
Final edge length: $a_{\kappa}=204$ mm.
Thus, the final volume: $V_{\kappa}=a_{K}^{3}=2.04^{3}=8.489664 \partial \mu^{3} \approx 1.06 V$.
Therefore, the density:
$\rho_{K}=\f... | olympiads | 6ee0c17f-9a3a-5699-9ea8-b4004f41e901 | open-r1/OpenR1-Math-220k | [
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] |
Evaluate $\frac{9+3 \times 3}{3}$. | Evaluating, $\frac{9+3 \times 3}{3}=\frac{9+9}{3}=\frac{18}{3}=6$. | olympiads | 10dd1cfb-3bae-53a4-a876-487414ffa0b1 | open-r1/OpenR1-Math-220k | [
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] |
(2) If $f(g(x))=\sin 2 x, g(x)=\tan \frac{x}{2}(0<x<\pi)$, then $f\left(\frac{\sqrt{2}}{2}\right)=$ $\qquad$ . | (2) $\frac{4 \sqrt{2}}{9}$ Hint:
$$
\begin{aligned}
f(g(x)) & =\sin 2 x=2 \sin x \cos x \\
& =2 \cdot \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \cdot \frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}},
\end{aligned}
$$
Therefore,
$$
f\left(\frac{\sqrt{2}}{2}\right)=2 \times \frac{\sqrt{2}}{1+\frac{1}{2}}... | olympiads | 9c80c35c-638e-546e-94be-75c627d19f23 | open-r1/OpenR1-Math-220k | [
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1. The solution set of the inequality $\arccos x<\arctan x$ is
$\qquad$ | $$
\text { II. 1. }\left\{x \left\lvert\, \sqrt{\frac{\sqrt{5}-1}{2}}0$, both sides of the inequality are acute angles. The original inequality is equivalent to
$\operatorname{tan}(\arccos x)\frac{\sqrt{5}-1}{2}$ or $x^{2}\sqrt{\frac{\sqrt{5}-1}{2}}$.
Also, since $x \in(0,1]$,
$$
\therefore \sqrt{\frac{\sqrt{5}-1}{2}}<... | cn_contest | fa60bfa8-b71c-5446-b73c-39945d319807 | open-r1/OpenR1-Math-220k | [
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] |
(5) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=0, a_{n+1}=a_{n}+1+2 \sqrt{1+a_{n}}(n=$ $1,2, \cdots)$, then $a_{2009}=(\quad)$.
(A) 4036080
(B) 4036078
(C) 4036082
(D) 4036099 | (5) A Hint: From the given, we have
$$
a_{n+1}+1=a_{n}+1+2 \sqrt{1+a_{n}}+1=\left(\sqrt{a_{n}+1}+1\right)^{2} .
$$
Since $a_{n+1}>0$, we have
$$
\sqrt{a_{n+1}+1}=\sqrt{a_{n}+1}+1,
$$
thus the sequence $\left\{\sqrt{a_{n}+1}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1, $\sqrt{a_{n}... | olympiads | 4ec59c11-e47f-56da-9564-ba16e26e0ded | open-r1/OpenR1-Math-220k | [
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] |
Which of the following statements is correct?
A: All integers are positive numbers
B: Integers, $0$, and fractions are collectively referred to as rational numbers
C: $0$ is the smallest rational number
D: $-1$ is the largest negative integer | To determine which statement is correct, let's analyze each option step by step:
- **Option A: All integers are positive numbers.**
- Integers include three types of numbers:
- Positive integers (e.g., 1, 2, 3, ...)
- Negative integers (e.g., -1, -2, -3, ...)
- Zero (0)
- Since integers also include ne... | cn_k12 | 149a14ea-4e77-53ee-badb-e85465c97dd4 | open-r1/OpenR1-Math-220k | [
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] |
Given a triangle ABC, let the lengths of the sides opposite to angles A, B, C be a, b, c, respectively. If a, b, c satisfy $a^2 + c^2 - b^2 = \sqrt{3}ac$,
(1) find angle B;
(2) if b = 2, c = $2\sqrt{3}$, find the area of triangle ABC. | (1) In triangle ABC, we have
$$\cos B = \frac{a^2 + c^2 - b^2}{2ac}.$$
Given that $a^2 + c^2 - b^2 = \sqrt{3}ac$, we can substitute this into the cosine formula:
$$\cos B = \frac{\sqrt{3}ac}{2ac} = \frac{\sqrt{3}}{2}.$$
Thus, $$B = \frac{\pi}{6}.$$
(2) From part (1), we have $B = \frac{\pi}{6}$. Since b = 2 and c = $2\... | cn_k12 | 01fcc46c-fad5-55fb-a3dc-27953605c08f | open-r1/OpenR1-Math-220k | [
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] |
Given the set $M=\{x\in \mathbb{Z} \,|\, |x|<5\}$, which of the following statements is correct? ( )
A: $2.5\in M$
B: $0\subseteq M$
C: $\{0\}\in M$
D: $\{0\}\subseteq M$ | From the set $M=\{x\in \mathbb{Z} \,|\, |x|<5\}=\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$, by using the symbols for the relationship between elements and sets, and between sets and sets, we can obtain the correct result.
Therefore, the correct answer is $\boxed{\text{D}}$. | cn_k12 | 34799155-6732-5952-9632-a6b52a587856 | open-r1/OpenR1-Math-220k | [
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] |
17. Three composite numbers $A, B, C$ are pairwise coprime, and $A \times B \times C=11011 \times 28$, then the maximum value of $A+B+C$ is . $\qquad$ | Answer: 1626 | olympiads | de85e1bc-7240-567f-8207-311a18c79e59 | open-r1/OpenR1-Math-220k | [
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] |
Given that the radius of the base of a certain cone is $1$ and its height is $2\sqrt{2}$, what is the surface area of the cone?
A: $2\pi$
B: $3\pi$
C: $4\pi$
D: $5\pi$ | To find the surface area of the cone, we first need to calculate its slant height, which can be found using the Pythagorean theorem in the right triangle formed by the radius, height, and slant height of the cone.
Given:
- Radius $r = 1$
- Height $h = 2\sqrt{2}$
The slant height $l$ can be calculated as follows:
\[l... | cn_k12 | e1c34975-59a6-5470-8f1c-3c452d9c81a7 | open-r1/OpenR1-Math-220k | [
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] |
Given \\(m \in \mathbb{R}\\), the complex number \\(z = (1+i)m^{2} - mi - 1 - 2i\\) where \\(i\\) is the imaginary unit.
(Ⅰ) For what value of the real number \\(m\\) is the complex number \\(z\\) purely imaginary;
(Ⅱ) If the complex number \\(z\\) corresponds to a point in the third quadrant on the complex plane, ... | Solution: The complex number \\(z = (1+i)m^{2} - mi - 1 - 2i = m^{2} - 1 + (m^{2} - m - 2)i\\).
(Ⅰ) From \\( \begin{cases} m^{2} - 1 = 0 \\ m^{2} - m - 2 \neq 0 \end{cases} \\), we find that when \\(m = 1\\), the complex number \\(z\\) is purely imaginary;
(Ⅱ) From \\( \begin{cases} m^{2} - 1 < 0 \\ m^{2} - m - 2 <... | cn_k12 | c0becabc-8442-59cb-8889-db0e8c8fe965 | open-r1/OpenR1-Math-220k | [
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] |
12. As shown in Figure 3, $\triangle A B C$ is an equilateral triangle, point $C$ is on the side $D E$ of rectangle $A B D E$, the inradius of $\triangle A B C$ is 1. Then the diameter of the circumcircle of rectangle $A B D E$ is $\qquad$ | 12. $\sqrt{21}$.
Let point $O$ be the center of the incircle of $\triangle ABC$, and $P$ be the point where the incircle touches side $AB$. Connect $OB$ and $OP$. Then $OB=2$, $BP=\sqrt{3}$, $AB=2\sqrt{3}$.
Thus, $PC=3$.
Since quadrilateral $ABDE$ is a rectangle and $CP \perp AB$, we have $AE=3$.
By the Pythagorean t... | cn_contest | adaeb866-3202-5237-b998-959f08d38970 | open-r1/OpenR1-Math-220k | [
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] |
2. Usain runs one lap around the school stadium at a constant speed, while photographers Arina and Marina are positioned around the track. For the first 4 seconds after the start, Usain was closer to Arina, then for 21 seconds he was closer to Marina, and then until the finish, he was closer to Arina again. How long do... | # Solution:
It is not hard to see that regardless of Arina and Marina's positions, the entire circle of the school stadium is divided into two equal parts - half of the circle is closer to Arina and the other half is closer to Marina (this is half of the shorter arc between Arina and Marina and half of the longer arc ... | olympiads | 7b9519ff-6ccb-5922-85de-4eb9b7302831 | open-r1/OpenR1-Math-220k | [
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] |
Determine the range of the real number $a$ such that the inequality $|x+ \frac{1}{x}|\geqslant |a-2|+\sin y$ holds for all non-zero real numbers $x$ and for all $y$. | Since $x+ \frac{1}{x}$ belongs to the interval $(-\infty,-2]\cup[2,+\infty)$,
it follows that $|x+ \frac{1}{x}|$ belongs to the interval $[2,+\infty)$, with its minimum value being $2$.
Furthermore, the maximum value of $\sin y$ is $1$,
thus, for the inequality $|x+ \frac{1}{x}|\geqslant |a-2|+\sin y$ to always hold... | cn_k12 | 559dca69-7b59-5414-b645-d95bf1fbe050 | open-r1/OpenR1-Math-220k | [
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] |
4.
On a coordinate line, 16 points are marked and numbered from left to right. The coordinate of any point, except for the extreme points, is equal to half the sum of the coordinates of the two adjacent points. Find the coordinate of the fifth point if the first point has a coordinate of 2 and the sixteenth point has... | # Solution
Solution. Let $a, b$ and $c$ be the coordinates of three consecutive points (from left to right). Then $b=\frac{a+c}{2}$, which means the second point is the midpoint of the segment with endpoints at the neighboring points. This condition holds for any triple of consecutive points, meaning the distances bet... | olympiads | 488729ba-6e43-5952-be14-347b361552a9 | open-r1/OpenR1-Math-220k | [
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] |
In triangle $\triangle ABC$, the lengths of the sides opposite to angles A, B, and C are a, b, and c respectively. a and c are the first term $a_1$ and the second term $a_2$ of the geometric sequence $\{a_n\}$, respectively. The solution set to the inequality $-x^2+6x-8>0$ is $\{x|a<x<c\}$. Find the general term formul... | First, we start by solving the inequality $-x^2+6x-8>0$. This can be rewritten as $(x-4)(x-2)<0$.
From this factorization, we can determine that the solution to the inequality is $2<x<4$. Thus, we can conclude that $a=2$ and $c=4$.
Since a and c are successive terms of a geometric sequence, the common ratio $r$ can ... | cn_k12 | 467786ba-c72c-5aff-8bf3-aac08a1f3cf6 | open-r1/OpenR1-Math-220k | [
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] |
Billy is breeding mice for an experiment. He starts with 8 mice, who each have 6 pups. When the pups grow up, all the mice have another 6 pups. Then each adult mouse eats 2 of their pups due to the stress of overcrowding. How many mice are left? | First find the number of pups in the first generation: 8 mice * 6 pups/mouse = <<8*6=48>>48 pups
Then add the number of adult mice to find the total number of mice: 48 pups + 8 mice = <<48+8=56>>56 mice
Then find the number of pups per mouse that survive from the second generation: 6 pups/mouse - 2 pups/mouse = <<6-2=4... | null | null | openai/gsm8k | [
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] |
2. Indicate all values of $n$ for which the sum of $n$ consecutive terms of the arithmetic progression $25,22,19, \ldots$, starting from the first, is not less than 66. | Solution: For the given arithmetic progression, we have: $a_{1}=25, d=-3$. The inequality $S_{n} \geq 66$ must be satisfied, i.e., $\frac{2 a_{1}+d(n-1)}{2} \cdot n \geq 66$. Substituting $a_{1}$ and $d$ into the last inequality, we get
$$
\frac{50-3(n-1)}{2} \cdot n \geq 66 \Leftrightarrow 53 n-3 n^{2} \geq 132 \Left... | olympiads | 7acef38e-d524-5fb9-b752-01d0a80032ef | open-r1/OpenR1-Math-220k | [
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] |
Given a line segment $AB=6$, point $C$ lies on the line $AB$, and $BC=4$. Find the length of $AC$. | To solve the problem, we consider two cases based on the position of point $C$ relative to the line segment $AB$.
**Case 1:** Point $C$ lies on the line segment $AB$.
In this case, we can find the length of $AC$ by subtracting the length of $BC$ from the length of $AB$. This is because the lengths of segments $AC$ an... | cn_k12 | 4fa139ca-ccc2-56a9-9a44-a95667bdb9ec | open-r1/OpenR1-Math-220k | [
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] |
Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?
$\textbf{(A)} \ 96 \qqua... | There are two possibilities regarding the parents.
1) Both are in the same store. In this case, we can treat them both as a single bunny, and they can go in any of the 4 stores. The 3 baby bunnies can go in any of the remaining 3 stores. There are $4 \cdot 3^3 = 108$ combinations.
2) The two are in different stores. I... | amc_aime | 80e07682-a825-5f69-9e21-aaca2180132b | open-r1/OpenR1-Math-220k | [
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] |
24. Let real numbers $a, b, c$ satisfy $a^{2}+2 b^{2}+3 c^{2}=\frac{3}{2}$, prove: $3^{-a}+9^{-b}+27^{-c} \geqslant 1$. (First China Southeast Mathematical Olympiad) | 24. By the Cauchy-Schwarz inequality, $(a+2 b+3 c)^{2} \leqslant(1+2+3)\left(a^{2}+2 b^{2}+3 c^{2}\right)=9$, then $a+2 b+3 c \leqslant 3$.
Therefore,
$$3^{-a}+9^{-b}+27^{-c} \geqslant 3 \cdot \sqrt[3]{3^{-(a+2 b+3 c)}}=3 \cdot \sqrt[3]{3^{-3}}=1$$ | inequalities | 3f82c68d-5068-525f-bd94-71a88c2817ca | open-r1/OpenR1-Math-220k | [
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] |
Problem 4. 13 children sat at a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their right neighbor: "The majority of us are boys." The latter said to their right neighbor... | # Answer: 7.
Solution. It is clear that there were both boys and girls at the table. Let's see how the children were seated. After a group of boys sitting next to each other comes a group of girls, then boys again, then girls, and so on (a group can consist of just one person). Groups of boys and girls alternate, so t... | olympiads | 01846d40-6943-570a-a205-3c5d18eca69e | open-r1/OpenR1-Math-220k | [
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] |
8. Let the function $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$, where $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ are non-negative integers. Given that $f(1)=4$, $f(5)=152$, then $f(6)=$ $\qquad$ | \[
\left\{\begin{array}{l}
a_{0}+a_{1}+a_{2}+a_{3}=4, \\
a_{0}+5 a_{1}+25 a_{2}+125 a_{3}=152 .
\end{array}\right.
\]
Obviously, \(a_{3}=1\), otherwise
\[
\left\{\begin{array}{l}
a_{0}+a_{1}+a_{2}=4, \\
a_{0}+5 a_{1}+25 a_{2}=152
\end{array}\right.
\]
\(\Rightarrow a_{0}+5 a_{1}+25(4-a_{0}-a_{1})=100-24 a_{0}-20 a_{1}... | olympiads | 1d75f76e-742e-5a54-8aa9-9a367a63e7cc | open-r1/OpenR1-Math-220k | [
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] |
Given that $a,b$ are real numbers, then "$a+b \leqslant 2$" is a condition for "$a \leqslant 1$ and $b \leqslant 1$" to be
A: Sufficient but not necessary
B: Necessary but not sufficient
C: Necessary and sufficient
D: Neither sufficient nor necessary | **Analysis**
This question tests the judgment of sufficient and necessary conditions. The key to solving the problem is to master the concepts of sufficient conditions, necessary conditions, and necessary and sufficient conditions.
The solution can be derived from the concepts of sufficient conditions, necessary cond... | cn_k12 | 59206051-1276-581c-aaa4-cab8a8addda3 | open-r1/OpenR1-Math-220k | [
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] |
Proposition $p$: The equation $x^2 + mx + 1 = 0$ has two distinct positive real roots. Proposition $q$: The equation $4x^2 + 4(m+2)x + 1 = 0$ has no real roots. If "proposition $p$ or proposition $q$" is true, find the range of values for $m$. | Since "proposition $p$ or proposition $q$" is true,
then proposition $p$ is true, or proposition $q$ is true.
When proposition $p$ is true,
we have $\begin{cases} \Delta = m^2 - 4 > 0 \\ x_1 + x_2 = -m > 0 \\ x_1x_2 = 1 > 0 \end{cases}$,
which gives $m < -2$;
When proposition $q$ is true,
we have $\Delta = 16(m... | cn_k12 | 6714f5fb-91d5-5b66-9554-ec2fb8a30fda | open-r1/OpenR1-Math-220k | [
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] |
## SUBIECTUL I
(4p) a) Calculaţi: $x=\sqrt{\left(\frac{12}{11}+\frac{13}{22}+\frac{14}{33}+\ldots+\frac{110}{1089}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{99}\right)}$;
(3p)b)Calculaţi: $y=\left(\frac{1}{1 \bullet 4}+\frac{1}{2 \bullet 6}+\frac{1}{3 \bullet 8}+\ldots+\frac{1}{49 \bullet 100}\right)$;
| ## SUBIECTUL I
a) $11=11 \cdot 1 ; 22=11 \cdot 2, \ldots, 1089=11 \cdot 99 \quad$ (1p); $\frac{12}{11}-1=\frac{1}{11}, \frac{13}{22}-\frac{1}{2}=\frac{1}{11}, \ldots, \frac{110}{1089}-\frac{1}{99}=\frac{1}{11}$ (2p); $\sqrt{\frac{1}{11} \cdot 99}=3$ (1p);
b) $y=\frac{1}{2}\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+... | olympiads | f399754f-b074-5c68-9def-d78ff4bef113 | open-r1/OpenR1-Math-220k | [
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] |
What does not occur during prophase I of meiosis?
A: Chromosome replication
B: Chromosome shortening and thickening
C: Pairing of homologous chromosomes
D: Crossing over between homologous chromosomes | **Step 1: Understanding the Problem**
The problem asks us to identify the event that does not occur during prophase I of meiosis.
**Step 2: Analyzing the Options**
- Option A: Chromosome replication occurs during the interphase before meiosis I, not during prophase I. So, this could be the correct answer.
- Option B: ... | cn_k12 | 8f3cceb8-7a69-5363-8948-d078c048d722 | open-r1/OpenR1-Math-220k | [
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] |
1. Let $x, y, z$ be positive real numbers, satisfying $x y+z=(x+z)(y+z)$.
Then the maximum value of $x y z$ is $\qquad$ | $-1 . \frac{1}{27}$.
From the given condition, we have $z=z(x+y+z)$.
Then $x+y+z=1$.
Thus, $x y z \leqslant\left(\frac{x+y+z}{3}\right)^{3}=\frac{1}{27}$.
When $x=y=z=\frac{1}{3}$, the equality holds.
Therefore, the maximum value of $x y z$ is $\frac{1}{27}$. | cn_contest | 4dc81000-23d6-5511-9745-23f6a0f4a5a4 | open-r1/OpenR1-Math-220k | [
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] |
## Task 2
Calculate the difference between the product and the sum of the numbers 9 and 2. | $9 \cdot 2=18,9+2=11,18-11=7$. The difference is 7.
### 11.4 5th Olympiad 1967
### 11.4.1 1st Round 1967, Class 2 | olympiads | a45af2af-69d1-5ce4-914e-916e5a1f29ca | open-r1/OpenR1-Math-220k | [
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] |
Given vectors $\overrightarrow{a}=(1,2)$, $\overrightarrow{b}=(x,1)$, and $(\overrightarrow{a}+2\overrightarrow{b})\parallel(2\overrightarrow{a}-\overrightarrow{b})$, find the real number $x=$ ( ).
A: $10$
B: $5$
C: $\dfrac{1}{2}$
D: $-10$ | **Analysis:** This problem primarily tests your understanding of coordinate operations of plane vectors and the necessary and sufficient conditions for the collinearity of plane vectors. First, use scalar multiplication and vector addition to calculate the coordinates of the two collinear vectors. Then, solve the equat... | cn_k12 | e0da47bc-e3d6-5400-a2ba-b9916f4d1d54 | open-r1/OpenR1-Math-220k | [
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] |
The solution set of the inequality $\log_{2}(2x-4) \leq 1$ is. | The inequality $\log_{2}(2x-4) \leq 1$
is equivalent to $\log_{2}(2x-4) \leq \log_{2}2$, which implies $0 < 2x-4 \leq 2$,
Solving this, we get $2 < x \leq 3$,
Therefore, the solution set of the inequality $\log_{2}(2x-4) \leq 1$ is $(2, 3]$
Hence, the answer is $\boxed{(2, 3]}$. | cn_k12 | 43e78bc5-fcf5-58d9-8f0d-c4826662314c | open-r1/OpenR1-Math-220k | [
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] |
Chinese mathematician Yang Hui of the Southern Song Dynasty proposed a problem in the year $1275$: "A rectangular field has an area of $864$ square steps. The width is $12$ steps less than the length. What are the width and length?" If we denote the width as $x$ steps, which equation correctly represents the problem?
... | To solve the problem, we start by denoting the width of the rectangular field as $x$ steps. According to the problem statement, the length is $12$ steps more than the width, which means the length can be represented as $x + 12$ steps.
Given that the area of a rectangle is calculated by multiplying its width by its len... | cn_k12 | 6fede62c-65c1-5800-93b1-79f5121ed9e4 | open-r1/OpenR1-Math-220k | [
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] |
Find the equation of the tangent line to the function $y=x^3-2x+1$ at the point $(1,0)$. | To find the equation of the tangent line to a curve at a given point, we need to find the derivative of the function at that point (which gives the slope of the tangent line) and then use the point-slope form of a linear equation.
1. **Find the derivative of the function**: The derivative of $y=x^3-2x+1$ is $y'=3x^2-2... | cn_k12 | 62589cad-7d41-52b0-9b1e-a64689db44c2 | open-r1/OpenR1-Math-220k | [
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] |
Convert the quadratic equation $2(x+2)^2+(x+3)(x-2)=-11$ into general form ( ).
A: $x^2+3x+4=0$
B: $3x^2+9x+12=0$
C: $3x^2+8x+13=0$
D: $3x^2+9x+13=0$ | Solution: Expand the original equation to get
$2x^2+8x+8+x^2+x-6=-11$.
Rearrange and combine terms to get
$3x^2+9x+13=0$.
Hence, the answer is: $\boxed{D}$.
The general form of a quadratic equation is $ax^2+bx+c=0$ ($a$, $b$, and $c$ are constants and $a\neq 0$). Using this, we can transform the original equation into... | cn_k12 | ea5f6e95-d675-587b-bb84-1e26ad9371b8 | open-r1/OpenR1-Math-220k | [
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] |
Task B-2.7. Today is Valentine's Day, and Valentino wants to go out with his girlfriend Lorna. With his 120 kuna, he can buy several roses and two lemonades. The price of one rose is equal to the price of one lemonade. Since they have a math test tomorrow, they will postpone the outing until Saturday. On Saturday, the ... | Solution. Let $x$ be the number of roses, and $y$ be the price of one rose on Valentine's Day. From the conditions of the problem, we have
$$
\begin{aligned}
x y+2 y & =120 \\
x(y-7)+78 & =126
\end{aligned}
$$
We solve the system
$$
\left\{\begin{array}{l}
x y+2 y=120 \\
x y-7 x=48
\end{array}\right.
$$
If we expre... | olympiads | 2ffec7ac-2640-5b35-aa6f-3584d0c957d0 | open-r1/OpenR1-Math-220k | [
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] |
## 165. Math Puzzle $2 / 79$
A motorcycle battery ( $6 \mathrm{~V}$ ) was discharged through a lamp ( $6 \mathrm{~V}, 0.5 \mathrm{~W}$ ). The lamp was on for a total of 48 hours.
How many Ah was the battery charged?
Explanation: Ah is the abbreviation for Ampere-hour, a unit of electrical charge. | When discharging the motorcycle battery, the current $I$ flows for 48 hours,
$$
I=\frac{P}{U}=\frac{0.5 \mathrm{~W}}{6 \mathrm{~V}}=\frac{1}{12} \mathrm{~A}
$$
This corresponds to a charge $Q$ of 4 Ah, because
$$
Q=I \cdot t=\frac{1}{12} \mathrm{~A} \cdot 48 \mathrm{~h}=4 \mathrm{Ah}
$$ | olympiads | 7eec110e-db9c-5fe4-94fb-d09124e12f41 | open-r1/OpenR1-Math-220k | [
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] |
If $x > 1$, then the function $f(x) = x + \frac{4}{x - 1}$ has:
A. Minimum value $5$
B. Maximum value $5$
C. Minimum value $-5$
D. Maximum value $-5$ | Given that $x > 1$, we can rewrite the function $f(x) = x + \frac{4}{x - 1}$ as:
$$f(x) = x - 1 + \frac{4}{x - 1} + 1$$
Applying the inequality of arithmetic and geometric means (AM-GM inequality) to the first two terms, we get:
$$x - 1 + \frac{4}{x - 1} \geq 2\sqrt{(x - 1) \cdot \frac{4}{x - 1}} = 4$$
Equality occ... | cn_k12 | 0b1263e7-6b7e-532e-8274-f2238a0b0e0a | open-r1/OpenR1-Math-220k | [
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] |
15th Putnam 1955 Problem A4 n vertices are taken on a circle and all possible chords are drawn. No three chords are concurrent (except at a vertex). How many points of intersection are there (excluding vertices)? Solution | : nC4. Easy. Every 4 vertices correspond to a unique point of intersection (the intersection of the two diagonals defined by the 4 points). [Actually, I got this the hard way. Take one of the vertices. Consider the diagonals from it. The diagonal to the next vertex but one has one vertex on one side and (n - 3) on the ... | olympiads | 98f24a62-46e3-5a3a-b3ef-85d1f7d3a57d | open-r1/OpenR1-Math-220k | [
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] |
In a basketball game, Cyrus made exactly eighty percent of the shots he attempted. He attempted twenty shots. How many times did he miss the shots? | Cyrus made 20 x 80/100 = <<20*80/100=16>>16 shots.
So, he missed 20 - 16 = <<20-16=4>>4 shots.
#### 4 | null | null | openai/gsm8k | [
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] |
## Task 4 - 320624
A rectangular children's room is $4 \mathrm{~m}$ and $40 \mathrm{~cm}$ long and $3 \mathrm{~m}$ and $30 \mathrm{~cm}$ wide. It has exactly one door, which is $90 \mathrm{~cm}$ wide. Thomas wants to install a new baseboard around the walls of this room. He calculates the required total length of the ... | Thomas calculates, taking into account the length $\mathrm{a}=4.40 \mathrm{~m}$, the width $\mathrm{b}=3.30 \mathrm{~m}$ of the room, and the door width $\mathrm{t}=0.90 \mathrm{~m}$, the required total length $2 a+2 b-t=8.80 \mathrm{~m}+6.60 \mathrm{~m}-0.90 \mathrm{~m}=$ $14.50 \mathrm{~m}$.
For this, $5 \cdot 14.50... | olympiads | 13fdaa55-f16f-50df-b189-e086ad52ddb0 | open-r1/OpenR1-Math-220k | [
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] |
3. (5 points) The average of four numbers is 30. If one of them is changed to 50, the average becomes 40. The original number was $\qquad$ | 【Solution】Solve: $50-(40 \times 4-30 \times 4)$
$$
\begin{array}{l}
=50-(160-120) \\
=50-40 \\
=10
\end{array}
$$
Answer: The original number is 10.
Therefore, the answer is: 10. | olympiads | 519e8a48-59bb-5f83-b0b4-a00db856e650 | open-r1/OpenR1-Math-220k | [
""
] |
Example 7. Find $\lim _{x \rightarrow 0} \frac{\sin ^{2} \frac{x}{3}}{x^{2}}$. | Solution. Taking into account formula (1.67), based on the properties of limits, we obtain
$$
\lim _{x \rightarrow 0} \frac{\sin ^{2} \frac{x}{3}}{x^{2}}=\lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{3}}{x}\right)^{2}=\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x}{3}}{x}\right]^{2}=\left(\frac{1}{3}\right)^{2}=\... | olympiads | 65da2308-b99c-5ca7-b0bb-250a201f1daf | open-r1/OpenR1-Math-220k | [
""
] |
In how many ways can we select two disjoint subsets from a set of $n$ elements. | We denote $A$ and $B$ as two disjoint sets. For each element, it can either belong to $A$, $B$, or neither $A$ nor $B$. Therefore, there are $3^{n}$ possible configurations. | olympiads | db8bfa7a-ac93-5489-8025-4b2f41974c1b | open-r1/OpenR1-Math-220k | [
""
] |
3. Given $M=2^{7} t=3^{5} s$, where $t$ is odd, and $s$ cannot be divided by 3. Then the sum of all divisors of $M$ in the form of $2^{P} 3^{q}$ is $\qquad$ . | 3. The sum of all divisors of $M$ is
$$
\begin{array}{l}
\left(1+2+\cdots+2^{7}\right)\left(1+3+\cdots+3^{5}\right) \\
=92820 .
\end{array}
$$ | cn_contest | 5118c70f-f544-5df6-be36-8e8c430b56db | open-r1/OpenR1-Math-220k | [
""
] |
4. (7 points) On the board, 46 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 46 minutes? | Answer: 1035.
Solution: Let's represent 46 units as points on a plane. Each time we combine numbers, we will connect the points of one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line... | olympiads | 2e33aead-d4b7-5058-9a0d-8c2262539916 | open-r1/OpenR1-Math-220k | [
""
] |
1. Calculate:
$$
2 \times\left(1+\frac{-1}{2}\right) \times\left[1+\frac{(-1)^{2}}{3}\right] \times\left[1+\frac{(-1)^{3}}{4}\right] \times \cdots \times\left[1+\frac{(-1)^{2019}}{2020}\right]=
$$
$\qquad$ | $1$ | olympiads | 4cffd246-4f6f-5f13-a34c-98f9f6e048a7 | open-r1/OpenR1-Math-220k | [
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] |
7.169. $\left(16^{\sin x}\right)^{\cos x}+\frac{6}{4^{\sin ^{2}\left(x-\frac{\pi}{4}\right)}}-4=0$. | Solution.
Transform the denominator of the second term of the equation:
$4^{\sin ^{2}\left(x-\frac{\pi}{4}\right)}=4^{\left(\sin x \cos \frac{\pi}{4}-\cos x \sin \frac{\pi}{4}\right)^{2}}=4^{\left(\frac{\sqrt{2}}{2}\right)^{2}\left(\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x\right)}=4^{\frac{1}{2}(1-\sin 2 x)}=$
$=4^{\fr... | olympiads | b9495b3f-ddd8-5b81-ba5d-d1b36340c5ec | open-r1/OpenR1-Math-220k | [
""
] |
Given that the function $f(x)=5\cos(wx+\varphi)$ satisfies $f\left(\frac{\pi}{3}+x\right)=f\left(\frac{\pi}{3}-x\right)$ for any real number $x$, and the function $g(x)=4\sin(wx+\varphi)+1$, then $g\left(\frac{\pi}{3}\right)=$ ()
A: $1$
B: $5$
C: $-3$
D: $0$ | From the given condition, we have $f\left(\frac{\pi}{3}\right)=\pm5$, which implies $\sin\left(w\cdot\frac{\pi}{3}+\varphi\right)=0$.
Since $g(x)=4\sin(wx+\varphi)+1$,
we have $g\left(\frac{\pi}{3}\right)=1$.
Therefore, the correct choice is $\boxed{A}$.
The condition $f\left(\frac{\pi}{3}\right)=\pm5$ and $\sin\le... | cn_k12 | fd8d90b9-b10a-56e5-a167-35cc8a3d3c74 | open-r1/OpenR1-Math-220k | [
""
] |
27. 8 letters, $A, B, C, D, E, F, G, H$ in all permutations, find the number of permutations where only 4 elements are not in their original positions. | 27. 8 elements taken 4 at a time, there are $\mathrm{C}_{8}^{4}$ ways to choose, 4 elements staying in their original positions have only one way to be chosen, the other 4 elements not being in their original positions have $D_{4}$ ways, in total there are $\mathrm{C}_{8}^{4} \cdot D_{4}$ ways. | olympiads | 0efa323d-e800-58f3-9886-eae4420de371 | open-r1/OpenR1-Math-220k | [
""
] |
4. $[\mathbf{2 0}]$ Thomas and Michael are just two people in a large pool of well qualified candidates for appointment to a problem writing committee for a prestigious college math contest. It is 40 times more likely that both will serve if the size of the committee is increased from its traditional 3 members to a who... | Answer: 16. Suppose there are $k$ candidates. Then the probability that both serve on a 3 membered committee is $(k-2) /\binom{k}{3}$, and the odds that both serve on an $n$ membered committee are $\binom{k-2}{n-2} /\binom{k}{n}$. The ratio of the latter to the former is
$$
\frac{\binom{k}{3}\binom{k-2}{n-2}}{(k-2)\bin... | olympiads | b12b29b2-edd2-5c23-95a7-8325759527a3 | open-r1/OpenR1-Math-220k | [
""
] |
3B. Let $z$ and $w$ be complex numbers such that $|z|=|w|=|z-w|$. Calculate the value of $\left(\frac{Z}{w}\right)^{300}$. | Solution. If $x=\frac{z}{w}$, then
$$
|x|=\left|\frac{Z}{w}\right|=\frac{|z|}{|w|}=1,|x-1|=\left|\frac{Z}{w}-1\right|=\frac{|z-w|}{|w|}=1
$$
From
$$
1=|x-1|^{2}=(x-1)(\overline{x-1})
$$
we get $x \bar{x}-x-\bar{x}+1=1$. Since $1=|x|^{2}=x \bar{x}$, it follows that $x-1+\bar{x}=0$. If we multiply by $x(x \neq 0)$, w... | olympiads | 087c701d-f137-5064-a7a2-d359da6d58f5 | open-r1/OpenR1-Math-220k | [
""
] |
Given points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, $C(x_{3}, y_{3})$ are three points on the graph of the inverse proportion function $y=-\frac{2}{x}$, and $x_{1} \lt x_{2} \lt 0 \lt x_{3}$, then the relationship between $y_{1}$, $y_{2}$, $y_{3}$ is ( )
A: $y_{1} \lt y_{2} \lt y_{3}$
B: $y_{3} \lt y_{2} \lt y_{1}$
C... | Given the inverse proportion function $y = -\frac{2}{x}$, we know that the constant of proportionality $k = -2$. This means that as $x$ increases or decreases, $y$ will move in the opposite direction because the constant of proportionality is negative.
Given that $x_{1} 0$, since $x_{3}$ is positive and the function... | cn_k12 | e2030fbf-f89e-5f84-82a5-2fab4f97b72d | open-r1/OpenR1-Math-220k | [
""
] |
3. $n$ is a positive integer, and it is defined that $n!=1 \times 2 \times \cdots \times n$ (for example, $1!=1,2!=1 \times 2$), then
$$
1!\times 1+2!\times 2+3!\times 3+\cdots+250!\times 250
$$
when divided by 2008, the remainder is ( ).
(A) 1
(B) 2
(C) 2007
(D) 2008 | 3.C.
Notice
$$
\begin{array}{l}
n! \times n = n! \times [(n+1)-1] \\
= n! \times (n+1) - n! = (n+1)! - n!.
\end{array}
$$
Then the original expression $=(2!-1!)+(3!-2!)+$
$$
\begin{aligned}
& (4!-3!)+\cdots+(251!-250!) \\
= & 251!-1=(251!-2008)+2007 .
\end{aligned}
$$
Since $2008 = 251 \times 8$, we have,
$$
2008 \t... | cn_contest | 332303c3-ec89-593c-8add-1f57fee3e94f | open-r1/OpenR1-Math-220k | [
""
] |
Determine the coefficient of the term containing $x^3$ in the expansion of ${(1+2x)}^{5}$. (The result should be represented as a number.) | **Analysis**
This problem primarily tests the application of the general term formula in binomial theorem, which is a basic question type.
We can utilize the general term formula, $T_{r+1}=C_{n}^{r}a^{n-r}b^{r}$, to solve the problem. By setting the exponent of $x$ in the general term to $3$, we can find which term c... | cn_k12 | ae8f7f04-84e3-5c09-b7dc-9354ea85a3e2 | open-r1/OpenR1-Math-220k | [
""
] |
Problem. For a sequence $a_{1}<a_{2}<\cdots<a_{n}$ of integers, a pair $\left(a_{i}, a_{j}\right)$ with $1 \leq i<j \leq n$ is called interesting if there exists a pair $\left(a_{k}, a_{l}\right)$ of integers with $1 \leq k<l \leq n$ such that
$$
\frac{a_{l}-a_{k}}{a_{j}-a_{i}}=2
$$
For each $n \geq 3$, find the larg... | Consider the numbers $a_{i}=2^{i}$ for $2 \leq i \leq n$ and $a_{1}=0$, and choose any pair $(i, j)$ with $1 \leq i < j \leq n$. If $(i, j)$ is interesting, then $a_{j}-a_{i} \leq a_{n}-a_{1}$ (otherwise, if $a_{j}-a_{i} > a_{n}-a_{1}$ we would have $a_{l}-a_{k} > a_{n}-a_{1}$, which is not possible.)
Finally, for any... | olympiads_ref | NaN | open-r1/OpenR1-Math-220k | [
""
] |
Given the function $f(x)=2x-\sqrt{x-1}$, determine the range of $f(x)$. | Let's substitute $t=\sqrt{x-1}$, where $t\geq 0$,
Then, $x=1+t^2$;
After substitution, we have $g(t)=2(t^2+1)-t=2t^2-t+2$;
The parabola $g(t)$ opens upwards and has its axis of symmetry at $t=\frac{1}{4}$, which is within $(0, +\infty)$;
Thus, the minimum value of $g(t)$ is $g(\frac{1}{4})=\frac{15}{8}$;
Therefore, the... | cn_k12 | 0cba1b51-bdfc-59cb-bfb2-320f559bce3a | open-r1/OpenR1-Math-220k | [
""
] |
In each row of the table, the sum of the first two numbers equals the third number. Also, in each column of the table, the sum of the first two numbers equals the third number. What is the sum of the nine numbers in the
| $m$ | 4 | $m+4$ |
| :---: | :---: | :---: |
| 8 | $n$ | $8+n$ |
| $m+8$ | $4+n$ | 6 |
table?
(A)... | Solution 1
Looking at the third row of the table, $(m+8)+(4+n)=6$ or $m+n+12=6$ or $m+n=-6$. The sum of the nine numbers in the table is
$m+4+m+4+8+n+8+n+m+8+4+n+6=3(m+n)+42=3(-6)+42=24$
## Solution 2
Try setting $m=0$.
Then the table becomes
| 0 | 4 | 4 |
| :---: | :---: | :---: |
| 8 | $n$ | $8+n$ |
| 8 | $4+n$... | olympiads | a48c1669-9ab7-5e39-a1f7-4cb07548f006 | open-r1/OpenR1-Math-220k | [
""
] |
4 students each sign up for one of the school's sports teams: soccer, basketball, or table tennis. Each person can only sign up for one team. The number of different ways they can sign up is ( )
A: $3^4$
B: $4^3$
C: 24
D: 12 | The four students sign up for table tennis, basketball, or soccer teams, with each person limited to one sport.
Each person has 3 ways to sign up;
According to the principle of counting by steps, there are a total of $3 \times 3 \times 3 \times 3 = 3^4$ different ways to sign up;
Therefore, the correct option is ... | cn_k12 | 4e765756-e9dc-5451-a2eb-296e4848fa94 | open-r1/OpenR1-Math-220k | [
""
] |
Let's determine the prime numbers $p$ and $q$ (both positive) if
$$
p+p^{2}+p^{4}-q-q^{2}-q^{4}=83805
$$ | Solution. One of $p$ and $q$ is even, the other is odd, because otherwise the value of the sixth-degree expression could not be odd. It is also clear that $p>q$, so only $q=2$ is possible. Then $p+p^{2}+p^{4}=83$827. If we increase the value of $p$, then $p+p^{2}+p^{4}$ also increases, so if there is a solution, there ... | olympiads | 46a2e620-7a4b-5acb-9ffe-5f43b34d129d | open-r1/OpenR1-Math-220k | [
""
] |
16. If $\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}$, where $x, y$, $z$ are three different positive numbers, then $\frac{x}{y}=(\quad$.
(A) $\frac{1}{2}$.
(B) $\frac{3}{5}$.
(C) $\frac{2}{3}$.
(D) $\frac{5}{3}$.
(E) 2. | 16
E | cn_contest | 6ba13ff9-d9ad-5b8b-9986-25f536dc7c16 | open-r1/OpenR1-Math-220k | [
""
] |
Given the equation $\frac{x^2}{2+m} + \frac{y^2}{1-m} = 1$ represents an ellipse with foci on the x-axis, the range of the real number $m$ is __________. | To ensure that the given equation $\frac{x^2}{2+m} + \frac{y^2}{1-m} = 1$ represents an ellipse with foci on the x-axis, the denominators of both fractions must be positive and the denominator associated with $x^2$ must be larger than that associated with $y^2$. This leads to the following three conditions:
1. $2 + m >... | cn_k12 | 4f5f80f1-b058-59d6-8040-14ab9c83cb30 | open-r1/OpenR1-Math-220k | [
""
] |
The graph of the function $f(x) = a^{x-1} + 4$ (where $a>0$ and $a \neq 1$) passes through a fixed point. Find the coordinates of this fixed point. | To find the fixed point through which the graph of the function passes, regardless of the value of $a$, we set the exponent to zero. This corresponds to the case where $x - 1 = 0$ or $x = 1$.
Let's substitute $x = 1$ back into the function:
\[
\begin{align*}
f(1) &= a^{1-1} + 4 \\
&= a^0 + 4 \\
&= 1 + 4 \\
&= 5
\end{... | cn_k12 | 58ce07fd-7b2a-57ab-90cb-4ec63ad718cf | open-r1/OpenR1-Math-220k | [
""
] |
The motion equation of a particle is $S=2t^2+3$ (displacement unit: meters, time unit: seconds). What is the instantaneous velocity of the particle at $t=2$ seconds, in meters per second? | Given $S=2t^2+3$, we find the derivative with respect to time, $s' = 4t$.
Therefore, the instantaneous velocity of the particle at $t=2$ seconds is $s'|_{t=2} = 4 \times 2 = 8$ (meters/second).
Hence, the answer is $\boxed{8}$. | cn_k12 | 063b6b76-9fa8-55f0-8025-98a8a18c5180 | open-r1/OpenR1-Math-220k | [
""
] |
If $ \dfrac{y}{x} = \dfrac{3}{4} $, then the value of $ \dfrac{x+y}{x} $ is ( ).
A: $1$
B: $ \dfrac{4}{7} $
C: $ \dfrac{5}{4} $
D: $ \dfrac{7}{4} $ | **Analysis**
This question tests knowledge related to the properties of proportions. First, transform the expression in question, and then substitute the values to solve it.
**Solution**
Given $ \dfrac{y}{x} = \dfrac{3}{4} $,
Therefore, the original expression $ = 1 + \dfrac{y}{x} = 1 + \dfrac{3}{4} = \dfrac{7}{4} ... | cn_k12 | 32db8cb5-2129-5e12-bfa8-8e758e5ab2b1 | open-r1/OpenR1-Math-220k | [
""
] |
Let $a,b,c$ be positive real numbers. Find all real solutions $(x,y,z)$ of the system:
\[ ax+by=(x-y)^{2}
\\ by+cz=(y-z)^{2}
\\ cz+ax=(z-x)^{2}\] | \( (0, 0, c) \).
The final answer is \( \boxed{ (0, 0, 0), (0, 0, c), (a, 0, 0), (0, b, 0) } \). | aops_forum | 6d9ffb53-4602-5285-8faa-8ac19cc999f0 | open-r1/OpenR1-Math-220k | [
""
] |
A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$, where $m$ and $n$ a... | The probability of three heads and five tails is $\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\binom {8}{3}p^5(1-p)^3$.
\begin{align*} 25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ 25(1-p)^2&=p^2 \\ 25p^2-50p+25&=p^2 \\ 24p^2-50p+25&=0 \\ p&=\frac {5}{6}\end{align*}
Therefore,... | amc_aime | a1f7c67b-37da-5dea-a977-77c657eedfc2 | open-r1/OpenR1-Math-220k | [
""
] |
4B. Calculate the sum
$$
\frac{1}{2 \sqrt{1}+1 \sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\frac{1}{4 \sqrt{3}+3 \sqrt{4}}+\ldots+\frac{1}{1999 \sqrt{1998}+1998 \sqrt{1999}}
$$ | Solution. For every natural number $n$ it holds that
$$
\begin{array}{r}
\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}=\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}} \cdot \frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{(n+1) \sqrt{n}-n \sqrt{n+1}} \\
=\frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{n(n+1)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
\end{array}
$... | olympiads | b499b02a-0b5a-570a-8c5a-8c98dd0a4570 | open-r1/OpenR1-Math-220k | [
""
] |
3. Let $M_{i}\left(x_{i}, y_{i}\right)(i=1,2,3,4)$ be four points on the parabola $y=a x^{2}$ $+b x+c(a \neq 0)$. When $M_{1}, M_{2}, M_{3}, M_{4}$ are concyclic, $x_{1}+x_{2}+x_{3}+x_{4}=$ $\qquad$ . | 3. $-\frac{2 b}{a}$.
$M_{2}, M_{2}, M_{3}, M_{6}$ are concyclic. Let's assume the equation of this circle is $x^{2} + y^{2} + D x + E y + F = 0$. Combining this with $y = a x^{2} + b x + c$ and eliminating $y$, we get $a^{2} x^{4} + 2 a b x^{3} + (2 a c + E a + 1) x^{2} + (2 b c + D + E b) x + c^{2} + E c + F = 0$. It ... | cn_contest | f8246e81-0829-5c6f-8e25-5262cd3658b2 | open-r1/OpenR1-Math-220k | [
""
] |
Among the following conditions, the one that can determine the congruence of two triangles is ( )
A: Two isosceles triangles with one obtuse angle equal
B: Two right-angled triangles containing a 60° angle
C: Two isosceles triangles with sides of length 3 and 6
D: Two right-angled triangles with one corresponding side... | **Answer**: A, Two isosceles triangles with one obtuse angle equal, does not necessarily mean the side lengths are equal, hence this option is incorrect;
B, Two right-angled triangles containing a 60° angle, does not necessarily mean the side lengths are equal, hence this option is incorrect;
C, Two isosceles trian... | cn_k12 | 5a53f453-44b4-5d77-abbd-085e3aa16d9e | open-r1/OpenR1-Math-220k | [
""
] |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\ldots+\frac{n-1}{n^{2}}\right)$ | Solution
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\ldots+\frac{n-1}{n^{2}}\right)=\lim _{n \rightarrow \infty}\left(\frac{1+2+3+\ldots+(n-1)}{n^{2}}\right)=$
$=\lim _{n \rightarrow \infty} \frac{1}{n^{2}}\left(\frac{n(n-1)}{2}\right)=\lim _{n \rightarrow \infty} \frac{n-1}{2 ... | olympiads | ec493a8c-8691-5f72-b2ee-6b390f754239 | open-r1/OpenR1-Math-220k | [
""
] |
Given $f(x) = \sin x (\cos x + 1)$, then $f'(x)$ equals to ( )
A: $\cos 2x - \cos x$
B: $\cos 2x - \sin x$
C: $\cos 2x + \cos x$
D: $\cos 2x + \cos x$ | Since $f(x) = \sin x (\cos x + 1)$,
then $f'(x) = (\sin x)'(\cos x + 1) + \sin x (\cos x + 1)' = \cos x (\cos x + 1) + \sin x (-\sin x) = \cos^2 x - \sin^2 x + \cos x = \cos 2x + \cos x$.
Therefore, the correct option is $\boxed{C}$. | cn_k12 | e96859a8-160e-5f0f-a744-e1f84c50bb2e | open-r1/OpenR1-Math-220k | [
""
] |
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