question stringlengths 12 1.61k | answer stringlengths 0 2.23k | openr1_source stringclasses 8
values | id stringlengths 3 36 ⌀ | dataset stringclasses 2
values | choices sequencelengths 1 1 |
|---|---|---|---|---|---|
5. In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5, 5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$, such that for all integers $n$, $a_{n}=$ $b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$. The... | 5.2.
Divide the known sequence into groups as follows:
$$
\begin{array}{l}
(1),(3,3,3),(5,5,5,5,5), \cdots, \\
(\underbrace{2 k-1,2 k-1, \cdots, 2 k-1}_{2 k-1 \uparrow}),
\end{array}
$$
Let $a_{n}$ be in the $k$-th group, where $a_{n}=2 k-1$. Then we have
$$
\begin{array}{l}
1+3+5+\cdots+2 k-3+1 \\
\leqslant n0$, sol... | cn_contest | 4b9fa61e-9110-57b9-b459-577a14ab21fb | open-r1/OpenR1-Math-220k | [
""
] |
The situation of the roots of the equation $x^{2}-2\sqrt{2}x+2=0$ is ( ).
A: There is one real root
B: There are two distinct real roots
C: There are no real roots
D: There are two equal real roots | To analyze the situation of the roots of the given quadratic equation, we use the discriminant formula $\Delta = b^2 - 4ac$. For the equation $x^2 - 2\sqrt{2}x + 2 = 0$, we identify $a = 1$, $b = -2\sqrt{2}$, and $c = 2$.
Let's calculate the discriminant step by step:
\[
\begin{align*}
\Delta &= b^2 - 4ac \\
&= (-2\... | cn_k12 | f0ad8a51-464b-5530-a0a5-142c55dc6e7f | open-r1/OpenR1-Math-220k | [
""
] |
Factorize: $x^{2}-1=\_\_\_\_\_\_.$ | To factorize the expression $x^{2}-1$, we recognize it as a difference of squares. A difference of squares can be factored using the formula $a^{2}-b^{2}=(a+b)(a-b)$, where $a=x$ and $b=1$ in this case. Applying this formula, we get:
\[
x^{2}-1 = (x)^{2}-(1)^{2} = (x+1)(x-1)
\]
Therefore, the factorized form of $x^{2... | cn_k12 | 27503c94-0f6d-5ee6-9c61-7cc9ee321c9a | open-r1/OpenR1-Math-220k | [
""
] |
4. Masha chose a natural number $n$ and wrote down all natural numbers from 1 to 6 n on the board. Then, Masha halved half of these numbers, reduced a third of the numbers by a factor of three, and increased all the remaining numbers by a factor of six. Could the sum of all the resulting numbers match the sum of the or... | Solution: Yes, this could have happened. Let $n=2$, then the sum of all numbers from 1 to 12 is
$$
1+2+3+\ldots+12=78
$$
Suppose Masha halved the numbers 2, 3, 5, 9, 10, 11, reduced the numbers 4, 6, 8, 12 by a factor of three, and increased the numbers 1, 7 by a factor of six. Then the sum of the resulting numbers w... | olympiads | c8042012-4781-5e83-994a-e72d29647b2b | open-r1/OpenR1-Math-220k | [
""
] |
How many sets $M$ are there such that $M$ is a proper subset of $\{1, 2\}$? | By definition, a proper subset $M$ of a set $\{1, 2\}$ is a subset of $\{1, 2\}$ that is not equal to the entire set itself. Therefore, the set $M$ can be any of the following:
- The empty set $\emptyset$
- The set containing only the element $1$, which is $\{1\}$
- The set containing only the element $2$, which is $\{... | cn_k12 | 255e3b57-8d73-5b44-84ff-f193ad4afcfa | open-r1/OpenR1-Math-220k | [
""
] |
If the inequality \(mx^{2}+4mx-4 < 0\) always holds for any real number \(x\), then the range of values for the real number \(m\) is ______. | To solve the problem, consider the inequality \(mx^{2}+4mx-4 < 0\) which must be true for all real numbers \(x\).
1. If \(m=0\), the inequality reduces to \(-4 < 0\), which is true for any real number \(x\). Therefore, \(m=0\) is within the solution set.
2. When \(m \neq 0\), the quadratic function must be negative f... | cn_k12 | 0ad0bc6c-8faa-5569-84c7-ef57e125c17c | open-r1/OpenR1-Math-220k | [
""
] |
The War of $1812$ started with a declaration of war on Thursday, June $18$, $1812$. The peace treaty to end the war was signed $919$ days later, on December $24$, $1814$. On what day of the week was the treaty signed?
$\textbf{(A)}\ \text{Friday} \qquad \textbf{(B)}\ \text{Saturday} \qquad \textbf{(C)}\ \text{Sunday} ... | By: dragonfly
To find what day of the week it is in $919$ days, we have to divide $919$ by $7$ to see the remainder, and then add the remainder to the current day. We get that $\frac{919}{7}$ has a remainder of 2, so we increase the current day by $2$ to get $\boxed{\textbf{(B)}\ \text{Saturday}}$.
Note that the dates ... | amc_aime | a9532066-0983-59d9-8bce-510634721c27 | open-r1/OpenR1-Math-220k | [
""
] |
6. In a Cartesian coordinate system, a point whose both vertical and horizontal coordinates are integers is called an integer point. The number of integer points $(x, y)$ that satisfy the inequality $(|x|-1)^{2}+(|y|-1)^{2}<2$ is $\qquad$ . | 6. 16 Detailed Explanation: From $(|x|-1)^{2}+(|y|-1)^{2}<2$, we can deduce that $(|x|-1,|y|-1)$ are $(0,0),(0,1),(0,-1)$, $(1,0),(-1,0)$, thus yielding 16 pairs of $(x, y)$. | olympiads | 85f425d0-2b1e-538c-ac00-dac39f1abc04 | open-r1/OpenR1-Math-220k | [
""
] |
## Task 18
$a=9, b=628 ; \quad$ To the sixfold of $a$, add half of $b$. | $6 \cdot 9+628: 2=54+314=368$
$6 \cdot 9+628 \div 2=54+314=368$ | olympiads | 445a0438-ba7e-514a-958c-3f1f6b3323c7 | open-r1/OpenR1-Math-220k | [
""
] |
The graph of the function $f(x) = 2009 + a^x + \log_a(1-x)$ ($a>0$ and $a \neq 1$) passes through a fixed point. The coordinates of this fixed point are. | Since the logarithmic function $f(x) = \log_a x$ always passes through the fixed point $(1, 0)$,
Therefore, the graph of the function $f(x) = \log_a(1-x)$ ($a>0$, $a \neq 1$) always passes through the fixed point $(0, 0)$.
The graph of the exponential function $f(x) = a^x$ ($a>0$ and $a \neq 1$) always passes through... | cn_k12 | 92773e1c-a9c9-5b45-8fd3-b9f54d3ddaa4 | open-r1/OpenR1-Math-220k | [
""
] |
Given a random variable $X \sim B(n, 0.8)$, and $D(X) = 1.6$, then the value of $n$ is ( )
A: 8
B: 10
C: 12
D: 14 | Since the random variable $X \sim B(n, 0.8)$,
we have $D(X) = np(1-p) = n \times 0.8 \times (1-0.8) = 1.6$,
thus, $n = 10$.
Therefore, the correct choice is $\boxed{B}$. | cn_k12 | ee51a25e-af22-58bd-830f-496ad45f7189 | open-r1/OpenR1-Math-220k | [
""
] |
## 5. Music School
In a certain music school, each student plays exactly one instrument: piano, guitar, trumpet, or violin. Half of all students play the piano, and half of the half of all students play the guitar. The number of students who play the trumpet is equal to a quarter of the number of students who play the... | Solution.
The number of students can be represented by a diagram where the entire square represents all the students in the school.
Half of the students who play the piano are marked in red.
![](https://cdn.mathpix.com/cropped/2024_05_30_44689c7bad1ce05a818cg-3.jpg?height=227&width=227&top_left_y=2315&top_left_x=134... | olympiads | d776ae6a-a1d2-5228-b54d-164581ded449 | open-r1/OpenR1-Math-220k | [
""
] |
7. (2004 College Entrance Examination, Liaoning Province) A bag contains 10 identical balls, 5 of which are marked with the number 0, and 5 are marked with the number 1. If 5 balls are drawn from the bag, what is the probability that the sum of the numbers on the 5 balls drawn is less than 2 or greater than 3? | 7. $\frac{13}{63}$ The probability of the sum of the digits being exactly 2 or 3 is $\frac{2 C_{5}^{2} C_{5}^{3}}{C_{1}^{5}}=\frac{50}{63}$, so $P=\frac{13}{63}$. | olympiads | c75e2d5a-a534-5b51-af2a-42e2a50d87d1 | open-r1/OpenR1-Math-220k | [
""
] |
4. Let $x \in \mathbf{R}$. Then the maximum value of $y=\frac{\sin x}{2-\cos x}$ is | 4. $\frac{\sqrt{3}}{3}$.
As shown in Figure 3, let $A(\cos \theta, \sin \theta), B(2,0)$.
Then $k_{A B}=\frac{-\sin \theta}{2-\cos \theta}$.
Thus, $-k \in\left[-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right]$.
Therefore, $k_{\max }=\frac{\sqrt{3}}{3}$. | olympiads | 07434094-640e-5a59-8541-cbf0e5d64eef | open-r1/OpenR1-Math-220k | [
""
] |
The line $l_1$: $x+ay=2a+2$ is parallel to the line $l_2$: $ax+y=a+1$. Find the value of $a$. | To solve this, we first rewrite the equations of the lines $l_1$ and $l_2$ as follows:
For line $l_1$: $x+ay=2a+2$ can be rewritten as $x+ay-2a-2=0$.
For line $l_2$: $ax+y=a+1$ can be rewritten as $ax+y-a-1=0$.
Since lines $l_1$ and $l_2$ are parallel to each other, we have:
When $a \neq 0$ and $a \neq -1$, $\frac{... | cn_k12 | cd863f43-d1ae-5827-834d-6fc919cf69eb | open-r1/OpenR1-Math-220k | [
""
] |
9.2. Two spheres touch the plane of triangle $A B C$ at points $B$ and $C$ and are located on opposite sides of this plane. The sum of the radii of these spheres is 7, and the distance between their centers is 17. The center of a third sphere with radius 8 is at point $A$, and it touches each of the first two spheres e... | Answer. $2 \sqrt{15} \approx 7.75$. | olympiads | b02ee363-010a-565e-a9c1-b7c0bd7621f8 | open-r1/OpenR1-Math-220k | [
""
] |
An earthquake caused four buildings to collapse. Experts predicted that each following earthquake would have double the number of collapsing buildings as the previous one, since each one would make the foundations less stable. After three more earthquakes, how many buildings had collapsed including those from the first... | The second earthquake caused 2 * 4 = <<2*4=8>>8 buildings to collapse.
The third earthquake caused 2 * 8 = <<2*8=16>>16 buildings to collapse.
The fourth earthquake caused 16 * 2 = <<16*2=32>>32 buildings to collapse.
Including the first earthquake, the earthquakes caused 4 + 8 + 16 + 32 = <<4+8+16+32=60>>60 buildings ... | null | null | openai/gsm8k | [
""
] |
The derivative of the function $y=(x-2)^{2}$ at $x=1$ is equal to ( )
A: $-1$
B: $-2$
C: $-3$
D: $-4$ | **Analysis:** To find the derivative of the function, we simply apply the formula for derivatives. This question primarily tests the calculation of a function's derivative, and using the formula for derivatives is key to solving this problem. It's quite basic.
The derivative of the function is $y'=2x-4$,
$\therefore... | cn_k12 | eb83fbf6-3a7e-550d-8335-5a6829036358 | open-r1/OpenR1-Math-220k | [
""
] |
The ratio representing the age of Halima, Beckham, and Michelle is 4:3:7, respectively. If the total age for the three siblings is 126, calculate the age difference between Halima and Beckham. | The total ratio representing the age of the three siblings is 4+3+7 = <<4+3+7=14>>14
Halima's age is represented by the fraction 4/14, which means she is 4/14*126 = 36 years old.
Beckhams age, from the total age of the three, is 3/14*126 = <<3/14*126=27>>27
The difference in age between Halima and Beckham is 36-27 = <<... | null | null | openai/gsm8k | [
""
] |
16.2.13 * Find the smallest natural number such that when the unit digit of this number is moved to the first position, the resulting number is 5 times the original number. | Let the number we are looking for be $10a + b$, where $b \in \{1, 2, \cdots, 9\}$, and $a$ is an $n$-digit number, then $5(10a + b) = 10^n \cdot b + a$. Therefore,
$$
49 \cdot a = \underbrace{99 \cdots 95}_{n-1 \uparrow 9} \cdot 6,
$$
$49 \nmid b$, so $7 \mid 99 \cdots 95$. Among numbers of the form $99 \cdots 95$, the... | olympiads | e475d4eb-c0f4-5fb9-b2fa-853131d139e0 | open-r1/OpenR1-Math-220k | [
""
] |
Among the following propositions, the true proposition is:
A: The sum of two acute angles is an obtuse angle
B: The complement of an angle is less than the angle itself
C: Complementary angles are adjacent angles
D: In the same plane, two lines perpendicular to the same line are parallel | To evaluate each proposition:
**A: The sum of two acute angles is an obtuse angle.**
- An acute angle is less than $90^\circ$.
- The sum of two acute angles, each less than $90^\circ$, could be less than or equal to $180^\circ$ but not necessarily more than $90^\circ$.
- Therefore, proposition A is not necessarily tr... | cn_k12 | a3a047c1-26af-5a0a-9e8d-4f6b2bc9c75c | open-r1/OpenR1-Math-220k | [
""
] |
The equation of the tangent line to the curve $y=e^{2x+1}$ at its intersection point with the y-axis is. | Let $x=0$, we get $y=e$.
Differentiating, we get $y'=2e^{2x+1}$. Letting $x=0$, we find $y'=2e$.
Therefore, the equation of the tangent line to the curve $y=e^{2x+1}$ at its intersection point with the y-axis is $y-e=2ex$, which simplifies to $y=2ex+e$.
Hence, the answer is $\boxed{y=2ex+e}$. | cn_k12 | db430a52-34bd-526f-84e0-9339cf35c014 | open-r1/OpenR1-Math-220k | [
""
] |
In the diagram, wheel $B$ touches wheel $A$ and wheel $C$. Wheel $A$ has radius $35 \mathrm{~cm}$, wheel $B$ has radius $20 \mathrm{~cm}$, and wheel $C$ has radius $8 \mathrm{~cm}$. If wheel $A$ rotates, it causes wheel $B$ to rotate without slipping, which causes wheel $C$ to rotate without slipping. When wheel $A$ ro... | Since the three wheels touch and rotate without slipping, then the arc lengths through which each rotates will all be equal.
Since wheel $A$ has radius $35 \mathrm{~cm}$ and rotates through an angle of $72^{\circ}$, then the arc length through which it rotates is $\frac{72^{\circ}}{360^{\circ}}(2 \pi(35 \mathrm{~cm}))... | olympiads | a6ff197e-85c6-5564-8ab4-7974d9f72501 | open-r1/OpenR1-Math-220k | [
""
] |
Calculate the following expressions, simplify where possible:
1.25×3.2×0.8
14.5×10.1
2.8+3.5×2.8
4.8×56.7+0.567×520
8.7÷0.2÷0.5 | 1. 1.25×3.2×0.8
= 1.25 × 0.8 × 3.2
= 1 × 3.2
= $\boxed{3.2}$
2. 14.5×10.1
= 14.5 × (10 + 0.1)
= 14.5 × 10 + 14.5
= 145 + 14.5
= $\boxed{159.5}$
3. 2.8+3.5×2.8
= 2.8 × (1 + 3.5)
= 2.8 × 4.5
= $\boxed{12.6}$
4. 4.8×56.7+0.567×520
= 4.8 × 56.7 + 56.7 × 5.2
= (4.8 + 5.2) × 56.7
= 1... | cn_k12 | aca54852-e644-5d83-9154-5978bdb263d4 | open-r1/OpenR1-Math-220k | [
""
] |
# Task No. 6.1
## Condition:
A sheet of paper was folded like an accordion as shown in the figure, and then folded in half along the dotted line. After that, the entire resulting square stack was cut along the diagonal.
Now it is easy to count the resulting pieces. For convenience, they are highl... | olympiads | b24cc2a4-cf42-5f47-af72-e323a9763f22 | open-r1/OpenR1-Math-220k | [
""
] |
3. As shown in Figure 16, quadrilateral $ABCD$ is a square with side length $a$. A circular arc centered at $D$ with radius $DA$ intersects a semicircle with diameter $BC$ at another point $P$. Extending $AP$ intersects $BC$ at point $N$. Then $\frac{BN}{NC}=$ $\qquad$
(2004, National Junior High School Mathematics Lea... | Hint: The problem involves the basic figures of overlapping and hidden intersecting circles, common tangents, similarity, and projections. Extend $C P$ and $D A$ to meet at point $F$, and connect $B P$. Answer: $\frac{1}{2}$.
| cn_contest | 65b91948-d90c-5f0b-9950-0d2a3474a1ec | open-r1/OpenR1-Math-220k | [
""
] |
Given the function $f(x) = x^{-2m^2 + m + 3}$ where $m \in \mathbb{Z}$, it is known to be an even function and increases on the interval $(0, +\infty)$.
(1) Find the expression for $f(x)$;
(2) If $g(x) = \log_a[f(x) - ax]$ where $a > 0$ and $a \neq 1$ is an increasing function on the interval $[2, 3]$, find the range... | (1) Since $f(x)$ is an even function, $-2m^2 + m + 3$ must be even. Also, $-2m^2 + m + 3 > 0$, which implies $-1 0$ and $a \neq 1$.
Let $u(x) = x^2 - ax$ and $y = \log_a u$.
$\mathrm{(i)}$ When $a > 1$, $y = \log_a u$ is an increasing function, so we need $u(x) = x^2 - ax$ to be an increasing function on the interva... | cn_k12 | 4e9b35c1-62f8-5d7f-b03f-6ace40edb218 | open-r1/OpenR1-Math-220k | [
""
] |
Given that line $l$ is parallel to plane $α$, $P$ is a fixed point on line $l$, and point $B$ moves within plane $α$ such that $PB$ forms a $30^{\circ}$ angle with line $l$. What is the trajectory of point $B$?
A: Two lines
B: Ellipse
C: Hyperbola
D: Parabola | $P$ is a fixed point on line $l$, and there is a plane $α$ that is parallel to line $l$. Within plane $α$, point $B$ moves such that the connecting line $PB$ forms a $30^{\circ}$ angle with line $l$.
In three-dimensional space, all lines passing through $P$ and forming a $30^{\circ}$ angle with line $l$ constitute two... | cn_k12 | 10ab7f7d-4da4-5e01-b244-fbe610d2092d | open-r1/OpenR1-Math-220k | [
""
] |
If $m$ and $n$ are positive integers with $n>1$ such that $m^{n}=2^{25} \times 3^{40}$, then $m+n$ is
(A) 209962
(B) 1954
(C) 209957
(D) 6598
(E) 1049760
## Part C: Each correct answer is worth 8. | Since $m$ and $n$ are positive integers with $n>1$ and $m^{n}=2^{25} \times 3^{40}$, then 2 and 3 are prime factors of $m$ (since they are prime factors of $m^{n}$ ) and must be the only prime factors of $m$ (since if there were other prime factors of $m$, then there would be other prime factors of $m^{n}$ ).
Therefor... | olympiads | 9a77bd18-84b9-5c6f-8f91-dc6a1e25b7af | open-r1/OpenR1-Math-220k | [
""
] |
The solution to the equation $\frac{5}{x+1}-\frac{4}{x}=0$ is ______. | To solve the equation $\frac{5}{x+1}-\frac{4}{x}=0$, we proceed as follows:
Step 1: Start with the given equation
$$\frac{5}{x+1}-\frac{4}{x}=0$$
Step 2: Find a common denominator and combine the fractions
$$\frac{5x - 4(x+1)}{x(x+1)} = 0$$
Step 3: Simplify the numerator
$$\frac{5x - 4x - 4}{x(x+1)} = 0$$
$$\frac{x ... | cn_k12 | 796661b5-69fa-5109-802b-e17839ff2ea6 | open-r1/OpenR1-Math-220k | [
""
] |
## Task 1 - 260611
A square $A B C D$ with a side length of $8 \mathrm{~cm}$ has a square $E F G H$ with a side length of $2 \mathrm{~cm}$ inscribed in it, as shown in the image. $H G$ and $D C$ are parallel segments with a distance of $2 \mathrm{~cm}$ between them. $E H$ and $A D$ are parallel segments with a distanc... | a) The squares $A B C D$ and $E F G H$ have areas of $64 \mathrm{~cm}^{2}$ and $4 \mathrm{~cm}^{2}$, respectively, because $8 \cdot 8=64$ and $2 \cdot 2=4$. Therefore, the shaded area has an area of $60 \mathrm{~cm}^{2}$ because $64-4=60$.
b) The diagram shows one possible solution to the problem. | olympiads | 39e31a6e-a0ba-5a84-bf7c-b40b3419fbe4 | open-r1/OpenR1-Math-220k | [
""
] |
Example 1 Given $f(x)=\frac{a^{x}}{a^{x}+\sqrt{a}}$ (where $a$ is a constant greater than 0), find the value of $f\left(\frac{1}{1001}\right)+f\left(\frac{2}{1001}\right)+\cdots+$ $f\left(\frac{1000}{1001}\right)$. | Consider the value of $f(x)+f(1-x)$:
$$
\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 | [
""
] |
Real numbers $x$, $y$, and $z$ satisfy the inequalities
$0<x<1$, $-1<y<0$, and $1<z<2$.
Which of the following numbers is necessarily positive?
$\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z$ | Notice that $y+z$ must be positive because $|z|>|y|$. Therefore the answer is $\boxed{\textbf{(E) } y+z}$.
The other choices:
$\textbf{(A)}$ As $x$ grows closer to $0$, $x^2$ decreases and thus becomes less than $y$.
$\textbf{(B)}$ $x$ can be as small as possible ($x>0$), so $xz$ grows close to $0$ as $x$ approaches $0... | amc_aime | 8fa72092-f99b-593d-9bdb-427a75afbd6f | open-r1/OpenR1-Math-220k | [
""
] |
The difference in the number of pens that Alice and Clara each have matches their age difference. Alice has 60 pens and Clara has 2/5 times as many pens as Alice. If Alice's age is 20 and Clara is older than Alice, calculate Clara's age in 5 years to come. | If Alice has 60 pens, Clara 2/5*60=<<24=24>>24 pens.
The number of pens that Alice has more than Clara is 60-24=<<60-24=36>>36
Since Alice is 20, and the number of pens that Alice has more than Clara is their age difference, Clara is 20+36=<<20+36=56>>56 years old.
In five years, Clara's age will be 56+5=<<56+5=61>>61 ... | null | null | openai/gsm8k | [
""
] |
Given that the circumference of a sector is 40, find the maximum area of the sector. | Let's denote the radius of the sector as $r$ and the central angle in radians as $\theta$. The circumference $C$ of the sector includes the length of the arc and the two radii, so we have:
$$ C = r\theta + 2r = 40 $$
The area $A$ of the sector is given by:
$$ A = \frac{1}{2}r^2\theta $$
Now, we must express $\theta$ ... | cn_k12 | cf55b0ef-30a2-50b3-a06f-321dc9d8eb1d | open-r1/OpenR1-Math-220k | [
""
] |
Given that \\(a\\) and \\(b\\) are positive real numbers, and \\((a-b)^{2}=4(ab)^{3}\\), find the minimum value of \\( \dfrac {1}{a}+ \dfrac {1}{b}\\). | Given \\(a\\) and \\(b\\) are positive real numbers, and \\((a-b)^{2}=4(ab)^{3}\\),
Then: \\((a+b)^{2}=4(ab)^{3}+4ab\\),
Therefore: \\( \dfrac {1}{a}+ \dfrac {1}{b}= \dfrac {a+b}{ab}= \dfrac { \sqrt {4(ab)^{3}+4ab}}{ab}= \sqrt {4ab+ \dfrac {4}{ab}}\geqslant \sqrt {8}=2 \sqrt {2}\\).
Hence, the answer is: $\boxed{2 \... | cn_k12 | 86f17630-337f-58cc-b362-8ae9bedb02d9 | open-r1/OpenR1-Math-220k | [
""
] |
Let $p$ and $q$ be positive integers such that $\frac{5}{8}<\frac{p}{q}<\frac{7}{8}$. What is the smallest value of $p$ for which $p+q=2005$? | Given $q=2005-p$, we have
$$
\frac{5}{8}<\frac{p}{2005-p}<\frac{7}{8}
$$
from which it follows that
$$
5(2005-p)<8 p \quad \text { and } 8 p<7(2005-p)
$$
Therefore,
$$
\frac{5 \times 2005}{13}<p<\frac{7 \times 2005}{15} \Rightarrow 771.15<p<935.66
$$
Thus, 772 is the smallest value of $p$ that satisfies the condi... | olympiads | ec2f2fc4-da6d-55d6-bcbb-2526feab4ea8 | open-r1/OpenR1-Math-220k | [
""
] |
The total production value of a city increased continuously for two years. The growth rate for the first year was $p$, and the growth rate for the second year was $q$. What is the average annual growth rate of the city's total production value over these two years?
A: $\dfrac {p+q}{2}$
B: $\dfrac {(p+1)(q+1)}{2}$
C: $... | Let the average annual growth rate of the city's total production value over these two years be $x$.
Then, we have $(1+p)(1+q)=(1+x)^{2}$.
Solving for $x$, we get $x=\sqrt {(1+p)(1+q)}-1$.
Hence, the answer is $\boxed{D}$.
By setting the average annual growth rate as $x$, we can derive the equation $(1+p)(1+q)=(1+x... | cn_k12 | 6df06e43-803d-568e-9357-11a0fc84c107 | open-r1/OpenR1-Math-220k | [
""
] |
Let $y=(x-a)^2+(x-b)^2, a, b$ constants. For what value of $x$ is $y$ a minimum?
$\textbf{(A)}\ \frac{a+b}{2} \qquad \textbf{(B)}\ a+b \qquad \textbf{(C)}\ \sqrt{ab} \qquad \textbf{(D)}\ \sqrt{\frac{a^2+b^2}{2}}\qquad \textbf{(E)}\ \frac{a+b}{2ab}$ | Expanding the quadratic and collecting terms gives $y = 2x^2 - (2a + 2b)x + (a^2+b^2)$. For a quadratic of the form $y = Ax^2 + Bx + C$ with $A>0$, $y$ is minimized when $x = -\frac{B}{2A}$, which is the average of the roots.
Thus, the quadratic is minimized when $x = \frac{2a+2b}{2\cdot 2} = \frac{a+b}{2}$, which is ... | amc_aime | 77d4a65c-c0d8-5c09-a25a-4fab2c066ee0 | open-r1/OpenR1-Math-220k | [
""
] |
7.2. A city held three rounds of Go tournaments, with the same participants in each round. It is known that every 2 participants in the three rounds of the tournament won 1 time each, drew 1 time each. A person came in last place in the first two rounds. Question: What place did he get in the third round? | 7.2. Calculate the sum of scores for each participant in the three rounds of the competition, and find that they are all equal. Since someone scored less than any other participant in the first two rounds, he scored more than any other participant in the third round. Therefore, he came in first in the third round. | cn_contest | 0a6479c4-20dc-5fbb-86d0-e2ff04fb9f1f | open-r1/OpenR1-Math-220k | [
""
] |
1. Buratino left Papa Carlo's house and arrived at the Field of Wonders exactly at 22:00. If his walking speed had been 25% faster, he would have arrived at 21:30. At what time did he leave the house? | Answer: at 19:30.
Solution: If Buratino spent $t$ (hours) on his journey, then with the increased speed, he would have spent 1.25 times less, i.e., $\frac{4}{5} t$. Therefore, he would have saved $\frac{1}{5} t$, which amounted to 30 minutes. Thus, he spent 2.5 hours on the way to the Field of Wonders, and he left hom... | olympiads | e0e61de0-f6f0-59e0-82e0-e860e8c895d0 | open-r1/OpenR1-Math-220k | [
""
] |
In a certain measurement, the data of sample A are as follows: 52, 54, 54, 56, 56, 56, 55, 55, 55, 55. If the data of sample B are exactly the data of sample A each increased by 6, then the corresponding numerical characteristics of samples A and B that are the same are ( )
A: Mode
B: Mean
C: Median
D: Standard Deviat... | Let the data in sample A be $x_i$, then the data in sample B are $y_i = 6 + x_i$,
Thus, the mode, mean, and median in sample B differ from those in sample A by 6,
Only the standard deviation remains unchanged,
Therefore, the correct choice is: $\boxed{\text{D}}$
This conclusion can be drawn by considering the r... | cn_k12 | eb29185c-fa96-561d-8a9b-c445f3372e8f | open-r1/OpenR1-Math-220k | [
""
] |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=t-\sin t \\
y=2-\cos t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(t-\sin t)^{\prime}=1-\cos t$
$y_{t}^{\prime}=(2-\cos t)^{\prime}=\sin t$
## We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{\sin t}{1-\cos t}$
$\left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(\frac{\sin t}{1-\cos t}\right)^{\prime}=\frac{\cos t \cdot(1-\cos t)-\si... | olympiads | 4f4ec39f-e3cd-5207-b369-a5c3fbea17bd | open-r1/OpenR1-Math-220k | [
""
] |
In a dark box, there are 6 black balls and 4 white balls.
(1) Take out 3 balls in sequence without replacement. If the first ball taken out is white, find the probability of getting a black ball on the third draw;
(2) Take out 3 balls in sequence with replacement. If the first ball taken out is white, find the prob... | Let event A be "the first ball taken out is white, and the third ball taken out is black", B be "the second ball taken out is white", and C be "the third ball taken out is white",
then for (1) $$P(A)= \frac { C_{ 4 }^{ 1 }\cdot ( C_{ 6 }^{ 1 } C_{ 5 }^{ 1 }+ C_{ 3 }^{ 1 } C_{ 6 }^{ 1 })}{ C_{ 4 }^{ 1 } A_{ 9 }^{ 2 }}... | cn_k12 | 4af2f240-d774-5ac5-b24d-0b923c4bde2c | open-r1/OpenR1-Math-220k | [
""
] |
If the quadratic trinomial $x^{2}+kx+9$ is a perfect square trinomial, then the value of $k$ is:
A: $6$
B: $-6$
C: $\pm 6$
D: $\pm 3$ | To determine the value of $k$ for which the quadratic trinomial $x^{2}+kx+9$ is a perfect square trinomial, we compare it to the general form of a perfect square trinomial, which is $(x+a)^{2}=x^{2}+2ax+a^{2}$. Given that $9=3^{2}$, we can infer that $a=3$ in this case. Thus, the trinomial can be written as:
\[
x^{2}+... | cn_k12 | 43e6ef86-590c-547c-9506-2350f8915c5a | open-r1/OpenR1-Math-220k | [
""
] |
Let the sum of the first $n$ terms of an arithmetic sequence $\{a_n\}$ be $S_n$. If $1 \leq a_5 \leq 4$ and $2 \leq a_6 \leq 3$, then the range of values for $S_6$ is. | Given $a_5 = a_1 + 4d$ and $a_6 = a_1 + 5d$,
we have $1 \leq a_1 + 4d \leq 4$ and $2 \leq a_1 + 5d \leq 3$.
Thus, $-20 \leq -5a_1 - 20d \leq -5$ and $6 \leq 3a_1 + 15d \leq 9$.
Adding these inequalities, we get $-14 \leq -2a_1 - 5d \leq 4$.
Multiplying both sides by $-1$, we have $-4 \leq 2a_1 + 5d \leq 14$.
... | cn_k12 | c8a2a869-8a76-53a2-bae3-6fdb750b52cf | open-r1/OpenR1-Math-220k | [
""
] |
## Task 2
Sven is saving for a bicycle. It costs 330 marks. If he has saved six times as much money as he already has, he can buy the bike.
How much money has Sven already saved? | Sven has already saved 55 marks, because $6 \cdot 55=330$. | olympiads | 9be3a987-874b-5cc6-9e86-c5a00f7966c9 | open-r1/OpenR1-Math-220k | [
""
] |
If the value of the fraction $\frac{{a^2-4}}{{a-2}}$ is $0$, then the value of $a$ is ______. | Given the fraction $\frac{{a^2-4}}{{a-2}}$ equals $0$, we aim to find the value of $a$.
First, let's understand that for a fraction $\frac{N}{D}=0$, the numerator $N$ must be $0$ and the denominator $D$ must not be $0$ (since division by zero is undefined).
Given $\frac{{a^2-4}}{{a-2}} = 0$, we have two conditions to... | cn_k12 | ddd02dcc-ffc6-5309-b557-6dff8f28108a | open-r1/OpenR1-Math-220k | [
""
] |
9. Using 4 digit 4s and some addition, subtraction, multiplication, division signs, and parentheses, write five expressions whose values are $2$, $3$, $4$, $5$, and $6$ respectively. | Analysis: $4 \div 4+4 \div 4=2, \quad(4+4+4) \div 4=2, \quad 4+(4-4) \div 4=4, \quad(4 \times 4+4) \div 4=5, \quad 4+(4+4) \div 4=6$ | olympiads | 1ba84778-3203-57e4-9512-5888e3f9eee4 | open-r1/OpenR1-Math-220k | [
""
] |
Example 7 The equation $\frac{1}{5} \log _{2} x=\sin (5 \pi x)$ has how many real roots? | Dividing Just to determine the number of roots, the basic method is to use the graph of the function to solve by combining numerical and graphical analysis.
Solution The number of real roots of the original equation is the number of intersection points of the graphs of the two functions $y=\frac{1}{5} \log _{2} x$ and... | olympiads | 719ba01f-3b98-5d20-a383-cd5a791e0ebb | open-r1/OpenR1-Math-220k | [
""
] |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
x^{2}=(y-z)^{2}-3 \\
y^{2}=(z-x)^{2}-7 \\
z^{2}=(x-y)^{2}+21
\end{array}\right.
$$ | Answer: $(-1,-3,-5),(1,3,5)$.
Solution. Move the square of the difference in each equation to the left side and apply the formula for the difference of squares:
$$
\left\{\begin{array}{l}
(x-y+z)(x+y-z)=-3 \\
(y-z+x)(y+z-x)=-7 \\
(z-x+y)(z+x-y)=21
\end{array}\right.
$$
Let $X=-x+y+z, Y=x-y+z, Z=x+y-z$. Then
$$
\lef... | olympiads | 5ca2ae6b-75dd-5919-adca-191c04a68c0f | open-r1/OpenR1-Math-220k | [
""
] |
In right triangle $ABC$, $\angle C = 90^\circ$, $AC = 3$, and $BC = 4$. Then, $AB = $ ( )
A: 3
B: 4
C: 5
D: 6 | **Answer**: In right triangle $ABC$, with $AB$ as the hypotenuse, knowing $AC$ and $BC$ allows us to calculate $AB$.
Solution: In right triangle $ABC$, where $\angle C = 90^\circ$,
$AB$ is the hypotenuse,
thus $AB^2 = AC^2 + BC^2$,
since $AC = 3$ and $BC = 4$,
then $AB = 5$,
therefore, the correct choice is $\box... | cn_k12 | aec8e5bd-57a9-5d63-95bb-12c8a45c927f | open-r1/OpenR1-Math-220k | [
""
] |
Task 3. Determine the smallest four-digit number that is divisible by 9, if the product of its digits is equal to 180. | Solution. From $180=1 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5$ it follows that the digits which can give a product of 180 can be: $1,4,5,9$ or $1,5,6,6$ or $2,2,5,9$ or $3, 3,4,5$ or $2,3,5,6$. However, the number must be divisible by 9, which means the sum of its digits must be divisible by 9. From this, we conclude t... | olympiads | d458a612-51b8-5f9d-bd68-6482fa392d5e | open-r1/OpenR1-Math-220k | [
""
] |
The student council president team of a certain high school's senior year consists of 5 students, among which 3 students are from the same class, and the other two students are from two different classes. Now, two students are randomly selected to attend a meeting. The probability that the two selected students are fro... | Let's represent the 3 students from the same class as 1, 2, 3, and the other two students from different classes as A, B.
When randomly selecting two students to attend the meeting, there are a total of 10 possibilities: 12, 13, 1A, 1B, 23, 2A, 2B, 3A, 3B, AB.
Among these, the combinations where the two selected st... | cn_k12 | 81205499-5481-51c3-8567-431ef979a715 | open-r1/OpenR1-Math-220k | [
""
] |
The sum of the first $n$ terms of an arithmetic sequence $\left\{ a_n \right\}$ is denoted as $S_n$. Given $a_2=1$ and $a_3=3$, find $S_4$. | **Analysis**
This question mainly tests the formula and properties of the sum of an arithmetic sequence, focusing on computational skills. It is considered an easy question.
**Answer**
Since the common difference $d = a_3 - a_2 = 2$,
we have $a_1 = -1$.
Therefore, $S_4 = -1 \times 4 + \frac{4 \times 3}{2} \time... | cn_k12 | 8005ba32-e15f-5715-8274-a77e60ed3bc9 | open-r1/OpenR1-Math-220k | [
""
] |
Given that the sides opposite to the internal angles A, B, C of △ABC are a, b, c respectively, and sinA + $\sqrt {2}$sinB = 2sinC, b = 3, find the minimum value of cosC. | Given equation simplifies to a + $\sqrt {2}$b = 2c using the sine law. Squaring both sides, we get (a + $\sqrt {2}$b)2 = 4c2, which simplifies to a2 + 2$\sqrt {2}$ab + 2b2 = 4c2. Rearranging, we have 4a2 + 4b2 - 4c2 = 3a2 + 2b2 - 2$\sqrt {2}$ab. This simplifies to a2 + b2 - c2 = $\frac {3a^{2}+2b^{2}-2 \sqrt {2}ab}{4}$... | cn_k12 | 797a8e8e-b141-54d1-8f15-0a44f91b5ab5 | open-r1/OpenR1-Math-220k | [
""
] |
I3.1 If $\alpha+\beta=11, \alpha \beta=24$ and $\alpha>\beta$, find $\alpha$.
I3.2 If $\tan \theta=\frac{-\alpha}{15}, 90^{\circ}<\theta<180^{\circ}$ and $\sin \theta=\frac{b}{34}$, find $b$.
13.3 If $A$ is the area of a square inscribed in a circle of diameter $b$, find $A$.
I3.4 If $x^{2}+22 x+A \equiv(x+k)^{2}+d$, ... | $\alpha$ and $\beta$ are the roots of the equation $x^{2}-11 x+24=0$
$$
\begin{array}{l}
(x-3)(x-8)=0 \\
\because \alpha>\beta \therefore \alpha=8
\end{array}
$$
In the figure, $P=(8,-15)$
$r^{2}=8^{2}+(-15)^{2}$ (Pythagoras' theorem)
$$
\begin{array}{l}
r=17 \\
\sin \theta=\frac{8}{17}=\frac{16}{34} \\
b=16
\end{arra... | olympiads | b3ca1f07-14c0-50b6-a426-aabec5731eca | open-r1/OpenR1-Math-220k | [
""
] |
3. Let the function be
$$
f(x)=\frac{\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)+2 x^{2}+x}{2 x^{2}+\cos x}
$$
with the maximum value $M$ and the minimum value $m$. Then $M$ and $m$ satisfy ( ).
(A) $M+m=2$
(B) $M+m=4$
(C) $M-m=2$
(D) $M-m=4$ | 3.A.
$$
f(x)=\frac{\sin x+\cos x+2 x^{2}+x}{2 x^{2}+\cos x}=\frac{\sin x+x}{2 x^{2}+\cos x}+1 \text {. }
$$
Since $g(x)=\frac{\sin x+x}{2 x^{2}+\cos x}$ is an odd function, and $f(x)$ has a maximum value and a minimum value, therefore, $g(x)$ also has a maximum value $M'$ and a minimum value $m'$, and $M'+m'=0$.
Thus,... | cn_contest | bf7c70f1-5a15-5b15-98ce-ea9012da46d0 | open-r1/OpenR1-Math-220k | [
""
] |
1. Two cyclists set off simultaneously from point $A$ to point $B$. When the first cyclist had covered half the distance, the second cyclist had 24 km left to travel, and when the second cyclist had covered half the distance, the first cyclist had 15 km left to travel. Find the distance between points $A$ and $B$. | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the cyclists. Then $\frac{s}{2 v_{1}}=\frac{s-24}{v_{2}}$ and $\frac{s-15}{v_{1}}=\frac{s}{2 v_{2}}$. From this, $\frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24$; $s^{2}-52 s+480=0 ;... | olympiads | b1c2c3ab-f50d-5b07-aafc-6f801ceae501 | open-r1/OpenR1-Math-220k | [
""
] |
2. Simplify $(\mathrm{i}+1)^{2016}+(\mathrm{i}-1)^{2016}=$ | 2. $2^{1009}$.
From $(i+1)^{2}=2 i \Rightarrow(i+1)^{2016}=2^{1008}$.
Similarly, $(i-1)^{2016}=2^{1008}$.
Therefore, $(i+1)^{2016}+(i-1)^{2016}=2^{1009}$. | cn_contest | a6363ef1-5a77-504d-b582-d0b3862f5792 | open-r1/OpenR1-Math-220k | [
""
] |
The graph of the linear function $y=2x-3$ does not pass through which quadrant?
A: The first quadrant
B: The second quadrant
C: The third quadrant
D: The fourth quadrant | Since $k=2>0$, the function passes through the first and third quadrants.
Since $b=-3<0$, the function intersects with the negative half of the y-axis.
Therefore, the graph does not pass through the second quadrant.
Hence, the correct answer is $\boxed{\text{B}}$. | cn_k12 | a60ac443-0567-5592-9c46-badafff3f656 | open-r1/OpenR1-Math-220k | [
""
] |
89. Divide 40 chess pieces into 27 piles. The number of chess pieces in each pile is $1, 2$ or 3. If the number of piles with only 1 chess piece is twice the number of the remaining piles, then the number of piles with exactly 2 chess pieces is $\qquad$ piles. | Reference answer: 5 | olympiads | 8c2cd172-6535-5ec5-bd39-896b067860cc | open-r1/OpenR1-Math-220k | [
""
] |
Given that the lengths of all three sides of a triangle satisfy the equation $x^{2}-6x+8=0$, the perimeter of this triangle is ( )
A: $8$
B: $10$ or $8$
C: $10$
D: $6$ or $12$ or $10$ | From the equation $x^{2}-6x+8=0$, we get $x=2$ or $x=4$.
- When the three sides are $2$, $4$, $4$, the perimeter is $10$.
- When the three sides are $2$, $2$, $4$, it cannot form a triangle, so this case should be discarded.
- When all three sides are $2$, the perimeter is $6$.
- When all three sides are $4$, the peri... | cn_k12 | cac99eb5-6b5a-5c52-8f4d-e21e2977dccf | open-r1/OpenR1-Math-220k | [
""
] |
If $f(x)$ is a function defined on $\mathbb{R}$ with a period of $4$, and its expression on $[0,2]$ is given by
$$
f(x)=\begin{cases}
x(1-x), & 0\leqslant x\leqslant 1, \\
\cos \pi x, & 1 < x\leqslant 2,
\end{cases}
$$
then find the value of $f\left( f\left( \frac{29}{3} \right) \right)$. | **Analysis**
This problem examines the calculation of function values using the function's period, which is a basic question.
We use the period to approach the known interval.
**Solution**
Since the period of $f(x)$ is $4$, we have $f\left( \frac{29}{3} \right)=f\left( 8+ \frac{5}{3} \right)=f\left( \frac{5}{3} \ri... | cn_k12 | 6ddd01c4-e494-5107-97b6-893e26342ab2 | open-r1/OpenR1-Math-220k | [
""
] |
The value range of the function $f(x) = x^2 + 2x - 3$, where $x \in [1, 3]$, is ____. | By completing the square, we get $f(x) = (x + 1)^2 - 4$. The axis of symmetry of the function is the line $x = -1$.
Therefore, the function is monotonically increasing on the interval $[1, 3]$.
Since at $x = 1$, $f(1) = 0$; and at $x = 3$, $f(3) = 12$,
the value range of the function $f(x) = x^2 + 2x - 3$, where ... | cn_k12 | 911b7318-3a92-5229-bd84-83501c8a3250 | open-r1/OpenR1-Math-220k | [
""
] |
1. The number of real solutions to the equation $|x-2|+| x-3| = 1$ is ( ).
(A) 2.
(B) 3.
(C) 4.
(D) Infinitely many. | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | cn_contest | 0dd6408d-3eea-5924-8d9f-bedeaac7c2a1 | open-r1/OpenR1-Math-220k | [
""
] |
There are 20 cars lined up at a race. Each of the cars have 2 passengers and a driver, then when the cars go past the race’s halfway point, each car gains another passenger. The cars then continue racing, and the race finishes once every single car has passed the finish line. People who get into the cars have to stay i... | There are 2 passengers and a driver in each car, which means that in each car there are 2 passengers + 1 driver = <<2+1=3>>3 people.
This means there are a total of 20 cars * 3 people per car = <<20*3=60>>60 people in the car at the start of the race.
At the halfway point, there is an additional 20 cars * 1 additional ... | null | null | openai/gsm8k | [
""
] |
Given $(1-2x)^4 = a + a_1x + a_2x^2 + a_3x^3 + a_4x^4$, then the value of $(a + a_2 + a_4)^2 - (a_1 + a_3)^2$ is ( )
A: 1
B: 16
C: 41
D: 81 | Since $(1-2x)^4 = a + a_1x + a_2x^2 + a_3x^3 + a_4x^4$,
when $x = -1$, $3^4 = a - a_1 + a_2 - a_3 + a_4$ (1)
when $x = 1$, $1 = a + a_1 + a_2 + a_3 + a_4$ (2)
Adding (1) and (2), we get $82 = 2(a + a_2 + a_4)$
Therefore, $a + a_2 + a_4 = 41$
Subtracting (2) from (1), we get $a_1 + a_3 = 40$,
Therefore, $(a ... | cn_k12 | 930732ae-cd9c-5936-95b5-1e25e75f0f6a | open-r1/OpenR1-Math-220k | [
""
] |
Bend a 32 cm long wire into a rectangle with a length-to-width ratio of 5:3. The area of the rectangle is ( ). | Given that the perimeter of the rectangle is 32 cm and the length-to-width ratio is 5:3, we can denote the length as $5x$ and the width as $3x$. Therefore, the perimeter of the rectangle can be expressed as $2(5x + 3x) = 32$ cm. Simplifying this equation gives $16x = 32$, which means $x = 2$ cm. Thus, the length of the... | cn_k12 | d96d886b-91e5-5acf-aa53-95128ebb611b | open-r1/OpenR1-Math-220k | [
""
] |
B3. In triangle ABC, the angle $\alpha=58^{\circ}$ and the angle $\beta=84^{\circ}$. What is the measure of angle $x$ between the angle bisector of $\gamma$ and the altitude to side $c$? | B3. The angle $x$ can be calculated in two ways, namely $x=\frac{\gamma}{2}-$ $\left(90^{\circ}-\beta\right)=19^{\circ}-6^{\circ}=13^{\circ}$ or $x=90^{\circ}-\alpha-\frac{\gamma}{2}=32^{\circ}-19^{\circ}=$ $13^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-12.jpg?height=448&width=511&t... | olympiads | 110e91f4-71ba-5547-80e3-b32e764142a5 | open-r1/OpenR1-Math-220k | [
""
] |
3. Given real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=4$. Then the maximum value of $(2 x-y)^{2}+(2 y-z)^{2}+(2 z-x)^{2}$ is ( ).
(A) 12
(B) 20
(C) 28
(D) 36 | 3.C.
From the problem, we have
$$
\begin{array}{l}
(2 x-y)^{2}+(2 y-z)^{2}+(2 z-x)^{2} \\
\approx 20-4(x y+y z+z x) \\
=28-2(x+y+z)^{2} \leqslant 28 .
\end{array}
$$
When $x+y+z=0$, the equality holds.
Thus, the maximum value sought is 28. | cn_contest | 7a616d2a-284a-5ccc-8552-336fe71305c0 | open-r1/OpenR1-Math-220k | [
""
] |
If the complex number $z$ satisfies $(1+2i) \cdot z = 2+i$, where $i$ is the imaginary unit, then $|z| =$
A: $\dfrac{3}{5}$
B: $\dfrac{4}{5}$
C: $1$
D: $2$ | **Analysis:**
This problem examines the operation of complex numbers and the modulus of a complex number. According to the problem, we can find $z$ by using the operation of complex numbers, and then we can easily find the result.
Given $(1+2i) \cdot z = 2+i$,
we have $z= \dfrac{2+i}{1+2i}= \dfrac{(2+i)(1-2i)}... | cn_k12 | 773c8662-13fa-5e96-a2c6-afa808c14089 | open-r1/OpenR1-Math-220k | [
""
] |
On Saturdays and Sundays, Suzanne makes french toast for the family. She and her husband each have 1 whole slice and her daughters split 1 slice. She uses a thick sliced bread that comes 12 slices per loaf. Over 52 weeks, how many loaves of bread will she need to make french toast? | She and her husband have 1 slice each and her children split 1 slice so that's 1+1+.5+.5 = <<1+1+.5+.5=3>>3 slices
She makes 3 pieces of french toast on Saturday and Sunday for a total of 3*2 = <<3*2=6>>6 slices per weekend
If she uses 6 slices per weekend then over 52 weeks she will have used 6*52 = <<6*52=312>>312 sl... | null | null | openai/gsm8k | [
""
] |
Find the sum of all positive integers whose largest proper divisor is $55$. (A proper divisor of $n$ is a divisor that is strictly less than $n$.)
| 1. To find the sum of all positive integers whose largest proper divisor is $55$, we need to identify the numbers that have $55$ as their largest proper divisor. This means that these numbers must be of the form $55 \cdot k$, where $k$ is a prime number greater than $1$.
2. Let's list the prime numbers and check which... | aops_forum | de04173c-63cb-555e-bb91-81d25092398f | open-r1/OpenR1-Math-220k | [
""
] |
Point A (1, 7) lies on the terminal side of an acute angle $\alpha$, and acute angle $\beta$ satisfies $\sin\beta= \frac{\sqrt{5}}{5}$.
(1) Find the value of $\tan(\alpha+\beta)$.
(2) Find the value of $\alpha+2\beta$. | (1) Given that $\tan\alpha=7$ and $\tan\beta= \frac{1}{2}$,
$\therefore \tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}= \frac{7+ \frac{1}{2}}{1-7\times \frac{1}{2}}=\boxed{-3}$.
(2) Since $\tan(\alpha+2\beta)=\tan[\alpha+(\alpha+\beta)]= \frac{\tan(\alpha+\beta)+\tan\beta}{1-\tan(\alpha+\beta)... | cn_k12 | d1cb0acd-f04a-5000-b4f1-bf32aba30462 | open-r1/OpenR1-Math-220k | [
""
] |
15. (3 points) Xiao Lan added consecutive even numbers $2, 4, 6, 8, 10, 12, 14, 16, \cdots$ one by one, and got the result 2012. Upon checking, she found that she had missed adding one number. What is the number that was missed? $\qquad$ . | 【Analysis】Consider this sequence of even numbers as an arithmetic sequence with the first term being 2 and the common difference being 2. Let the total number of terms be $n$. According to the general term formula of an arithmetic sequence: $a_{n}=a_{1}+(n-1) d$ and the sum formula of the first $n$ terms: $S n=n\left(a... | olympiads | bfc077ec-15ac-5e21-b9cc-a680d4759276 | open-r1/OpenR1-Math-220k | [
""
] |
In the sequence ${a_n}$, $a_1=1$, $a_n=1-\frac{1}{a_{n-1}+1}$ ($n\geqslant 2$), find $a_3$. | Given $a_n=1-\frac{1}{a_{n-1}+1}=\frac{a_{n-1}}{a_{n-1}+1}$ ($n\geqslant 2$),
We have $\frac{1}{a_n}=\frac{a_{n-1}+1}{a_{n-1}}=1+\frac{1}{a_{n-1}}$ ($n\geqslant 2$),
Since $\frac{1}{a_1}=1$,
Thus, the sequence ${\frac{1}{a_n}}$ is an arithmetic sequence with first term and common difference both equal to $1$.
So, $... | cn_k12 | 341d4338-2631-5d7f-a586-0dded752b859 | open-r1/OpenR1-Math-220k | [
""
] |
Given the function $f(x) = 3x^2 + mx + 2$ is increasing on the interval $[1, +\infty)$, the range of $m$ is \_\_\_\_\_\_. | Since the function $f(x) = 3x^2 + mx + 2$ is increasing on the interval $[1, +\infty)$, its axis of symmetry is at $x = \frac{m}{-6}$.
Therefore, $\frac{m}{-6} \leq 1$, solving this yields $m \geq -6$.
Thus, the answer is: $\boxed{[-6, +\infty)}$.
From the problem, we know that the axis of symmetry of the quadratic ... | cn_k12 | 0f061ae2-d43b-566e-a951-b41a032d40b5 | open-r1/OpenR1-Math-220k | [
""
] |
A year ago, the total cost of buying a lawnmower was 2/5 times less than the cost it goes for now. If the cost was $1800 a year ago, calculate how much it would cost Mr. Lucian to buy 4 such lawnmowers. | If the price of a lawnmower was $1800 a year ago, 2/5 less than the cost right now, then it now costs 2/5*1800 = $720 more to buy one lawnmower.
The total cost of buying one lawnmower right now is $1800+$720 = $<<1800+720=2520>>2520
To purchase 4 such lawnmowers, Mr. Lucian will have to pay $2520*4 = $<<2520*4=10080>>1... | null | null | openai/gsm8k | [
""
] |
Let \\(A=\{x| \dfrac {1}{1-x}\geqslant 1\}\) and \\(B=\{x|x^{2}+2x-3 > 0\}\), then \\((\complement_{R}A)\cap B=\) \(\)
A: \([0,1)\)
B: \((-∞,-3)\)
C: \(\varnothing \)
D: \((-∞,-3)∪(1,+∞)\) | From the inequality in set \\(A\), we transform it to: \(\dfrac {1}{1-x}-1\geqslant 0\), which is equivalent to \(\dfrac {x}{1-x}\geqslant 0\),
After rearranging, we get: \(x(x-1)\leqslant 0\), and \(1-x\neq 0\),
Solving this gives: \(0\leqslant x 0\),
Solving this gives: \(x 1\), which means \(B=(-∞,-3)∪(1,+∞)\),... | cn_k12 | 669557f2-c863-5b48-9642-5e0d6e8b3f4d | open-r1/OpenR1-Math-220k | [
""
] |
316. $\left\{\begin{array}{l}(x+y)\left(x^{2}+y^{2}\right)=539200 \\ (x-y)\left(x^{2}-y^{2}\right)=78400 .\end{array}\right.$ | 316. Instruction. Dividing the 1st equation by the 2nd:
$$
\frac{x^{2}+y^{2}}{(1-y)^{2}}=\frac{337}{49}
$$
from which
$$
674 x y=288 x^{2}+288 y^{2}
$$
or
$$
98 x y=288(x-y)^{2}
$$
$$
\frac{(x-y)^{2}}{x y}=\frac{49}{144}=\frac{x}{y}+\frac{y}{x}-2
$$
$$
\frac{x}{y}+\frac{y}{x}=\frac{337}{144}
$$
let
$$
\frac{x}... | olympiads | ebaf3e8c-ee68-58a6-9614-090ffebba1fb | open-r1/OpenR1-Math-220k | [
""
] |
2. 2. 1 ^ Given the equation $\sin ^{2} x-(2 a+1) \cos x-a^{2}=0$ has real solutions for $x$. Then the set of real values for $a$ is ( ).
(A) $\left[-\frac{5}{4}, 1-\sqrt{2}\right]$
(B) $\left[-\frac{5}{4}, 1+\sqrt{2}\right]$
(C) $[1-\sqrt{2}, 1+\sqrt{2}]$
(D) $\left[-\frac{3}{2}, 1-\sqrt{2}\right]$ | Transform the equation into $\cos ^{2} x+(2 a+1) \cos x+a^{2}-1=0$.
Let $t=\cos x$, then the equation becomes
$$
t^{2}+(2 a+1) t+a^{2}-1=0 .
$$
Let $f(t)=t^{2}+(2 a+1) t+a^{2}-1, t \in[-1,1]$. By the problem, the real number $a$ should satisfy
$$
\left\{\begin{array}{l}
(2 a+1)^{2}-4\left(a^{2}-1\right) \geqslant 0, \... | olympiads | e8aa75e0-9038-526c-8079-40aef8cdb0ea | open-r1/OpenR1-Math-220k | [
""
] |
5. (20 points) Two mutually perpendicular rays, propagating in the same vertical plane, fall from air onto a horizontal glass surface.
The refractive index of the glass $n=1.5$. The angle of refraction for the first ray $\beta=25^{\circ}$. Determine the angle between the refracted rays. | Answer: $56^{\circ}$.
Solution. The law of refraction for the first ray $\frac{\sin \alpha}{\sin 25^{\circ}}=1.5$. Therefore, the angle of incidence of the first ray $\alpha=39.34^{\circ}$. The angle of incidence of the second ray $\beta=90^{\circ}-39.34^{\circ}=$ $50.66^{\circ}$. The law of refraction for the second ... | olympiads | 3344eef4-4b5b-5fc9-9a0a-9adf4e63b300 | open-r1/OpenR1-Math-220k | [
""
] |
Given that m is a real number, the complex number z = (2+i)m² - m(1-i) - (1+2i) (where i is the imaginary unit) corresponds to a point in the fourth quadrant of the complex plane. Determine the range of values for the real number m. | The complex number z can be rewritten as z = (2m² - m - 1) + (m² + m - 2)i. For z to be located in the fourth quadrant of the complex plane, the real part must be positive and the imaginary part must be negative. This leads to the following system of inequalities:
$$\begin{cases} 2m^{2}-m-1>0 \\ m^{2}+m-2<0 \end{cases... | cn_k12 | 95875c75-0c91-59ce-bc98-26a9a22b3875 | open-r1/OpenR1-Math-220k | [
""
] |
The solution set of the inequality system $\left\{\begin{array}{l}x-3＜2\\ 3x+1≥2x\end{array}\right.$ is ( )
A: $x \lt 5$
B: $1\leqslant x \lt 5$
C: $-1\leqslant x \lt 5$
D: $x\leqslant -1$ | To solve the given system of inequalities, we'll tackle each inequality step by step.
For the first inequality:
\[
\begin{align*}
x - 3 &< 2 \\
x &< 2 + 3 \\
x &< 5 \quad \text{(①)}
\end{align*}
\]
For the second inequality:
\[
\begin{align*}
3x + 1 &\geq 2x \\
3x - 2x &\geq -1 \\
x &\geq -1 \quad \text{(②)}
\end{ali... | cn_k12 | ebd5a0b1-e208-5fce-833d-48fd36277961 | open-r1/OpenR1-Math-220k | [
""
] |
Problem 7.6. The distance between cities A and B is an integer number of kilometers. Along the road between the cities, there is a signpost every kilometer: on one side is the distance to city A, and on the other side is the distance to city B. Slava walked from city A to city B. During his journey, Slava calculated th... | Answer: 39.
Solution. Suppose we are standing next to a sign with numbers $x$ and $y$ written on it. If $\text{GCD}(x, y) = d$, then $(x + y) \vdots d$, i.e., the distance between the cities is divisible by all the calculated GCDs.
Now suppose the distance between the cities (let's call it $S$) is divisible by some n... | olympiads | 880d9bcc-9ffd-528e-918d-17eecace5a00 | open-r1/OpenR1-Math-220k | [
""
] |
9.5. On the side $AC$ of triangle $ABC$, a point $M$ is taken. It turns out that $AM = BM + MC$ and $\angle BMA = \angle MBC + \angle BAC$. Find $\angle BMA$. | Answer. $\quad 60^{\circ}$. Solution. First, we will show that triangle $A B C$ is isosceles. Indeed, this follows from the condition $\angle B M A=\angle M B C+\angle B A C$ and the property of the exterior angle: $\angle B M A=\angle M B C+\angle B C A$. From these two equalities, we have $\angle B C A=\angle B A C$.... | olympiads | 3a510b0f-c2d6-5dd2-b1a3-7fae6431e8b1 | open-r1/OpenR1-Math-220k | [
""
] |
Given the standard equation of a hyperbola: $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 (a > 0, b > 0)$, and one of its asymptote equations is $y=3x$, what is the eccentricity of the hyperbola? | The standard equation of a hyperbola is given by: $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 (a > 0, b > 0)$. The equations of its asymptotes are given by $y = \pm \frac{b}{a}x$.
Comparing the given asymptote equation $y=3x$ with the general form $y = \pm \frac{b}{a}x$, we get:
$$\frac{b}{a} = 3$$
The eccentrici... | cn_k12 | bc3d77d2-0cf2-5b15-997a-c2173dec04ab | open-r1/OpenR1-Math-220k | [
""
] |
From the 8 vertices of the cube ABCD-A1B1C1D1, any 4 different vertices taken may form:
(1) The 4 vertices of a rectangle;
(2) The 4 vertices of a tetrahedron, each of whose faces is an equilateral triangle;
(3) The 4 vertices of a tetrahedron, each of whose faces is a right triangle;
(4) The 4 vertices of a tetrah... | The quadrilateral ABCD fits condition (1), the tetrahedron ACB1D1 fits condition (2), DB1C1D1 fits condition (3), and DA1C1D1 fits condition (4). Therefore, there are $\boxed{4}$ correct conclusions. | cn_k12 | eeafa3a3-db76-57b1-8272-7178bd55e9a8 | open-r1/OpenR1-Math-220k | [
""
] |
Tobias downloads $m$ apps. Each app costs $\$ 2.00$ plus $10 \%$ tax. He spends $\$ 52.80$ in total on these $m$ apps. What is the value of $m$ ?
(A) 20
(B) 22
(C) 18
(D) 24
(E) 26 | Since the tax rate is $10 \%$, then the tax on each $\$ 2.00$ app is $\$ 2.00 \times \frac{10}{100}=\$ 0.20$.
Therefore, including tax, each app costs $\$ 2.00+\$ 0.20=\$ 2.20$.
Since Tobias spends $\$ 52.80$ on apps, he downloads $\frac{\$ 52.80}{\$ 2.20}=24$ apps.
Therefore, $m=24$.
ANSWER: (D) | olympiads | 84eddb97-43d2-5877-ba41-643839b93f54 | open-r1/OpenR1-Math-220k | [
""
] |
5.1. Among the numbers exceeding 2013, find the smallest even number $N$ for which the fraction $\frac{15 N-7}{22 N-5}$ is reducible. | Answer: 2144. Solution: The presence of a common factor in the numbers $15 N-7$ and $22 N-5$ implies that the same factor is present in the number $(22 N-5)-(15 N-7)=7 N+2$, and subsequently in the numbers $(15 N-7)-2 \cdot(7 N+2)=N-11,(7 N+2)-7 \cdot(N-11)=79$. Since 79 is a prime number, the fraction is reducible by ... | olympiads | 893808da-e806-5c5e-87ac-a960a1771713 | open-r1/OpenR1-Math-220k | [
""
] |
Given vectors $\overrightarrow{a}=(\sin \theta, -2)$ and $\overrightarrow{b}=(\cos \theta, 1)$, if $\overrightarrow{a} \parallel \overrightarrow{b}$, then $\tan 2\theta=$ ______. | Since $\overrightarrow{a}=(\sin \theta,-2)$ and $\overrightarrow{b}=(\cos \theta,1)$, and $\overrightarrow{a} \parallel \overrightarrow{b}$,
$\therefore \sin \theta=-2\cos \theta$, $\therefore \tan \theta=-2$,
Thus, $\tan 2\theta= \dfrac {2\tan \theta}{1-\tan^2 \theta} = \dfrac {-4}{1-4}= \boxed{\dfrac {4}{3}}$.
Fro... | cn_k12 | e81b8ab8-e42c-53f6-b528-fe6afa07cb37 | open-r1/OpenR1-Math-220k | [
""
] |
8.2. The work was divided equally among the workers in the team. After the first day, they counted how many people had completed no less than 30 percent of their share; it turned out that 70 percent of all workers had done so. When they counted only those who had completed no less than 70 percent of their share, it tur... | Solution: Let's provide an example where the condition of the problem is met, but a third of the work is not done. Suppose there were 10 workers, and the work involves creating 300 identical parts. Then each worker needs to make 30 parts, and a third of the work consists of 100 parts. Suppose three workers made 21 part... | olympiads | ba652c0a-25d2-500e-b8ff-6539e0042e33 | open-r1/OpenR1-Math-220k | [
""
] |
3. Given $x=1-\sqrt{3}$. Then $x^{5}-2 x^{4}-2 x^{3}$ $+x^{2}-2 x-1$ is
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 1 (Hint: Original expression $=x^{3}\left(x^{2}-\right.$
$$
\left.2 x-2)+\left(x^{2}-2 x-2\right)+1\right)
$$ | cn_contest | d568b618-4541-56f4-a593-be8c1597007a | open-r1/OpenR1-Math-220k | [
""
] |
11. One Match (from 8th grade, 2 points). In the lottery ticket "6 out of 45," there are a total of 45 natural numbers. From these 45 numbers, a lottery participant must choose a combination that contains exactly 6 numbers. Two participants independently choose their combinations. What is the probability that their com... | Solution. Let's mark with some secret sign the six numbers chosen by the first participant. This results in 6 marked numbers and 39 unmarked numbers. We need to find the probability that the combination chosen by the second participant includes exactly 1 marked number and exactly 5 unmarked numbers. The probability of ... | olympiads | 2b28c8bd-8d30-5a57-986b-58057582107b | open-r1/OpenR1-Math-220k | [
""
] |
Given the function $f(x) = (x+1)^2e^x$, and let $k \in [-3, -1]$, for any $x_1, x_2 \in [k, k+2]$, the maximum value of $|f(x_1) - f(x_2)|$ is ( )
A: $4e^{-3}$
B: $4e$
C: $4e+e^{-3}$
D: $4e+1$ | To solve this, we first find the derivative of the function, which gives us $f'(x) = (x^2 + 4x + 3)e^x$.
By setting $f'(x) > 0$, we find that $x -1$; by setting $f'(x) < 0$, we find that $-3 < x < -1$.
Therefore, the function is increasing on the intervals $(-\infty, -3)$ and $(-1, +\infty)$, and decreasing on the... | cn_k12 | 2be7980e-4798-5632-859a-b528d4b6c2dd | open-r1/OpenR1-Math-220k | [
""
] |
17. Given in a circle $\odot O$ with radius 2, the side $AB=2\sqrt{3}$ of the inscribed $\triangle ABC$. Then the degree of $\angle C$ is ).
(A) $60^{\circ}$
(B) $30^{\circ}$
(C) $60^{\circ}$ or $120^{\circ}$
(D) $30^{\circ}$ or $150^{\circ}$ | 17. C.
From $\sin C=\frac{AB}{2R}=\frac{\sqrt{3}}{2}$, we get $\angle C=60^{\circ}$ or $120^{\circ}$. | cn_contest | 3c3a9a74-0b5c-5e2b-b14b-70d00cfa7f4c | open-r1/OpenR1-Math-220k | [
""
] |
If the inequality $x^2 - 4x \geq m$ holds for $x \in [3, 4)$, then ( )
A: $m \geq -3$
B: $-3 \leq m < 0$
C: $m \leq -3$
D: $m \geq -4$ | Since $x^2 - 4x = (x - 2)^2 - 4$, and $x \in [3, 4)$, when $x = 3$, the minimum value of $x^2 - 4x$ is $9 - 12 = -3$. Therefore, $m \leq -3$;
Thus, the correct choice is $\boxed{C}$.
**Analysis:** According to the problem, $m$ just needs to be less than or equal to the minimum value of $x^2 - 4x$. | cn_k12 | 0b2b132a-5086-5fff-8b56-1b2c93afe044 | open-r1/OpenR1-Math-220k | [
""
] |
Gunther just financed a John Deere tractor through the dealership. If his monthly payment is $150.00 a month, for 5 years, with no interest, how much did he finance the tractor for? | There are 12 months in 1 year and his loan is for 5 years so that's 12*5 = <<12*5=60>>60 months
His monthly payment is $150.00 and he pays this for 60 months so his loan for the tractor was 150*60 = $<<150*60=9000.00>>9,000.00
#### 9000 | null | null | openai/gsm8k | [
""
] |
Condition of the problem
Find the derivative.
$$
y=\frac{1}{24}\left(x^{2}+8\right) \sqrt{x^{2}-4}+\frac{x^{2}}{16} \arcsin \frac{2}{x}, x>0
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{24}\left(x^{2}+8\right) \sqrt{x^{2}-4}+\frac{x^{2}}{16} \arcsin \frac{2}{x}\right)^{\prime}= \\
& =\frac{1}{24} \cdot 2 x \cdot \sqrt{x^{2}-4}+\frac{1}{24}\left(x^{2}+8\right) \cdot \frac{1}{2 \sqrt{x^{2}-4}} \cdot 2 x+\frac{2 x}{16} \cdot \arcsin \frac{2}{x}+... | olympiads | df4f8552-e64a-5168-a68f-2561adbeabef | open-r1/OpenR1-Math-220k | [
""
] |
End of preview. Expand in Data Studio
README.md exists but content is empty.
- Downloads last month
- 27