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Which of the following statements about the classical probability model is correct ( )?
① There are only a finite number of elementary events that can occur in the experiment;
② The probability of each event occurring is equal;
③ The probability of each elementary event occurring is equal;
④ If there are a total of $n$... | Analyzing option ①, from the definition of a random experiment, we know that a finite number of elementary events can occur, hence ① is correct.
Considering option ②, the definition of a random experiment does not imply that each event (combination of elementary events) has an equal chance of occurring, therefore ② is... | cn_k12 | b3f3bbb7-4a61-523c-b4a5-4f424bb68ce3 | open-r1/OpenR1-Math-220k | [
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20. [11] There exist unique nonnegative integers $A, B$ between 0 and 9 , inclusive, such that
$$
(1001 \cdot A+110 \cdot B)^{2}=57,108,249 \text {. }
$$
Find $10 \cdot A+B$. | Answer: 75 We only need to bound for $A B 00$; in other words, $A B^{2} \leq 5710$ but $(A B+1)^{2} \geq 5710$. A quick check gives $A B=75$. (Lots of ways to get this...) | olympiads | 78f53e6f-ab87-506a-bee7-df0eedde2529 | open-r1/OpenR1-Math-220k | [
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Find the equation of the circle with center at the point (2, -1) that is tangent to the line $x+y=7$. | First, convert the equation of the line $x+y=7$ into the form $x+y-7=0$.
The radius $r$ of the circle can be calculated using the distance from the center of the circle to the line, which is $\frac {|2-1-7|}{ \sqrt {2}} = 3 \sqrt {2}$.
Therefore, the equation of the circle is $(x-2)^2+(y+1)^2=18$.
Hence, the answer ... | cn_k12 | f59c5b5f-bc8e-5d86-9601-d5b8e693b4be | open-r1/OpenR1-Math-220k | [
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The graph of the function $y=\sin x-\sqrt{3}\cos x$ is translated $m(m>0)$ units to the left, and the resulting graph is symmetric about the $y$-axis. What can be the value of $m$?
A: $\frac{5\pi}{6}$
B: $\frac{2\pi}{3}$
C: $\frac{\pi}{3}$
D: $\frac{\pi}{6}$ | First, let's rewrite the given function.
$y=\sin x-\sqrt{3}\cos x=2(\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x)=2\sin(x-\frac{\pi}{3})$.
After translating the graph $m(m>0)$ units to the left, the function becomes $y=2\sin(x+m-\frac{\pi}{3})$.
Since the resulting graph is symmetric about the $y$-axis, it must be true... | cn_k12 | 04532472-1608-52b1-a8b2-db379f999f9d | open-r1/OpenR1-Math-220k | [
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The positional relationship between the line $4x-3y=0$ and the circle $x^{2}+y^{2}-18x-45=0$ is ( ).
A: Intersect.
B: Separate.
C: Tangent.
D: Uncertain. | **Analysis**
This question examines the general equation of a circle and the positional relationship between a line and a circle. It is not difficult and is considered a basic problem.
**Solution**
Given the circle $x^{2}+y^{2}-18x-45=0$, we can find the center of the circle is $(9,0)$, and the radius is $\sqrt{126}... | cn_k12 | 0c7985fe-0c2c-5c51-9ccf-b7b6ec9bccbf | open-r1/OpenR1-Math-220k | [
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3. The numbers $1,2,3, \ldots, 29,30$ were written in a row in a random order, and partial sums were calculated: the first sum $S_{1}$ equals the first number, the second sum $S_{2}$ equals the sum of the first and second numbers, $S_{3}$ equals the sum of the first, second, and third numbers, and so on. The last sum $... | Answer: 23.
Solution: Evaluation: adding an odd number changes the parity of the sum, there are 15 odd numbers, so the parity of the sums changes at least 14 times. Therefore, there will be at least 7 even sums, and thus no more than 23 odd sums.
Implementation: arrange the numbers as follows: 1, then all even number... | olympiads | d0a3a2a8-4b1e-5923-974c-0ede5dc27a50 | open-r1/OpenR1-Math-220k | [
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Find the intersection point $M$ of the lines $l_{1}$: $3x+4y-5=0$ and $l_{2}$: $2x-3y+8=0$, and determine the equations of the lines that satisfy the following conditions:
- $(1)$ Parallel to the line $2x+y+7=0$.
- $(2)$ Perpendicular to the line $2x+y+7=0$. | To find the intersection point $M$ of the lines $l_{1}$ and $l_{2}$, we solve the system of equations given by $l_{1}$ and $l_{2}$:
1. Solve the system $\left\{\begin{array}{l}3x+4y-5=0\\ 2x-3y+8=0\end{array}\right.$:
- Multiply the first equation by $3$ and the second by $4$ to eliminate $y$:
\[
\begin{align*}
... | cn_k12 | 94905a66-c722-5c86-8c83-0c0b71afdb88 | open-r1/OpenR1-Math-220k | [
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] |
Bogdanov I.I.
In a certain country, there are 100 cities (consider the cities as points on a plane). In a directory, for each pair of cities, there is a record of the distance between them (a total of 4950 records).
a) One record has been erased. Can it always be uniquely restored from the remaining ones?
b) Suppose... | a) Let 98 points lie on a straight line $l$, and two points $A$ and $B$ lie outside it. If the distance between $A$ and $B$ is unknown, it cannot be restored: by replacing point $B$ with $B^{\prime}$, symmetric to $B$ relative to $l$, the other distances will not change.
b) By induction, we will show that for $n \geq 4... | olympiads | f1041779-d479-58fb-b7b9-65e57f318f35 | open-r1/OpenR1-Math-220k | [
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Which of the following equations is calculated correctly?
A: $\sqrt{2}+\sqrt{3}=\sqrt{5}$
B: $\sqrt{3}\times \sqrt{2}=\sqrt{6}$
C: $\sqrt{9}-\sqrt{3}=\sqrt{3}$
D: $\sqrt{12}\div \sqrt{3}=4$ | To solve this problem, let's evaluate each option step by step:
**Option A: $\sqrt{2}+\sqrt{3}=\sqrt{5}$**
- We know that the addition of square roots, $\sqrt{a} + \sqrt{b}$, does not simplify to $\sqrt{a+b}$ unless $a$ and $b$ are under the same radical and can be combined directly, which is not the case here. There... | cn_k12 | ab615f59-b68e-5ef2-9292-ce1fb7354652 | open-r1/OpenR1-Math-220k | [
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Given the quadratic function $y=x^{2}-4x+3$.
$(1)$ Find the coordinates of the vertex $C$ of this function and explain how to translate it to obtain $y=x^{2}$;
$(2)$ Find the coordinates of the points where the graph of the function intersects the $x$-axis. | ### Step-by-Step Solution
#### Part 1: Finding the Vertex $C$
Given the quadratic function $y = x^2 - 4x + 3$, we can complete the square to find the vertex form of the quadratic function.
1. Start with the given function:
$$y = x^2 - 4x + 3$$
2. Rewrite the quadratic and linear terms:
$$y = (x^2 - 4x) + 3$$
... | cn_k12 | 7a5d3f4a-a535-5d5b-be0a-7eafb2f1dec0 | open-r1/OpenR1-Math-220k | [
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In the plane, three vectors are given: $\overrightarrow{a} = (3, 2)$, $\overrightarrow{b} = (-1, 2)$, $\overrightarrow{c} = (4, 1)$.
(1) Find $3\overrightarrow{a} + \overrightarrow{b} - 2\overrightarrow{c}$;
(2) If $(\overrightarrow{a} + k\overrightarrow{c}) \parallel (2\overrightarrow{b} - \overrightarrow{a})$, find... | Solution:
(1) $3\overrightarrow{a} + \overrightarrow{b} - 2\overrightarrow{c} = 3(3, 2) + (-1, 2) - 2(4, 1) = (9, 6) + (-1, 2) - (8, 2) = (9-1-8, 6+2-2) = \boxed{(0, 6)}$.
(2) $\overrightarrow{a} + k\overrightarrow{c} = (3+4k, 2+k)$, $2\overrightarrow{b} - \overrightarrow{a} = (-5, 2)$.
Since $(\overrightarrow{a} + ... | cn_k12 | a760c97a-b524-518d-b2be-2a61f169ddf7 | open-r1/OpenR1-Math-220k | [
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Observe the following equations: \\(1^{3}=1\\), \\(1^{3}+2^{3}=9\\), \\(1^{3}+2^{3}+3^{3}=36\\), \\(1^{3}+2^{3}+3^{3}+4^{3}=100\\), ... Based on these equations, make a general conclusion. For \\(n \in \mathbb{N}^*\\), \\(1^{3}+2^{3}+3^{3}+...+n^{3}= \) \_\_\_\_\_\_. | First, let's analyze the given equations:
\\(1^{3}=1\\)
\\(1^{3}+2^{3}=9=(1+2)^{2}\\)
\\(1^{3}+2^{3}+3^{3}=36=(1+2+3)^{2}\\)
\\(1^{3}+2^{3}+3^{3}+4^{3}=100=(1+2+3+4)^{2}\\)
From these equations, we can observe that the left side represents the sum of cubes of consecutive natural numbers, while the right side is the sq... | cn_k12 | 63c71115-068c-5d36-a032-827b20fedb57 | open-r1/OpenR1-Math-220k | [
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Example 3 If $x, y, z$ are all positive real numbers, find the maximum value of $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)}$. (2003 Singapore Mathematical Olympiad) | Solve: By the generalization of Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
(1+5 x)(4 x+3 y)(5 y+6 z)(z+18) \geqslant \\
(\sqrt[4]{1 \cdot 4 x \cdot 5 y \cdot z}+\sqrt[4]{5 x \cdot 3 y \cdot 6 z \cdot 18})^{4}=5120 x y z
\end{array}$$
Therefore, the maximum value of $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+... | inequalities | 3d6300a6-f915-5311-aac2-50e7e18db325 | open-r1/OpenR1-Math-220k | [
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In boxes labeled $0$, $1$, $2$, $\dots$, we place integers according to the following rules:
$\bullet$ If $p$ is a prime number, we place it in box $1$.
$\bullet$ If $a$ is placed in box $m_a$ and $b$ is placed in box $m_b$, then $ab$ is placed in the box labeled $am_b+bm_a$.
Find all positive integers $n$ that are ... | 1. **Base Case:**
- If \( p \) is a prime number, it is placed in box 1. This is given by the problem statement.
2. **Inductive Step:**
- Assume that for \( p_1p_2\cdots p_k \), where \( p_1, p_2, \ldots, p_k \) are primes, the number is placed in box \( \left(p_1p_2\cdots p_k\right) \left(\frac{1}{p_1} + \frac{... | aops_forum | 5bf7f8f5-8bf9-5a1a-95f6-b13a8293bc20 | open-r1/OpenR1-Math-220k | [
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] |
The age of a lioness in a park is twice the age of a hyena in the same park. The lioness is 12 years old. If the babies of the two animals are half the age of their mothers, calculate the sum of the age of the babies in five years. | If the lioness is 12 years old, the hyena, half her age, is 12/2 = <<12/2=6>>6 years old.
The baby of the lioness is 12/2 = <<12/2=6>>6 years old, half the mother's age.
Five years from now, the age of the lioness's cub will be 6+5 = <<6+5=11>>11
The baby of the hyena is 6/2 = <<6/2=3>>3 years old, which is half the mo... | null | null | openai/gsm8k | [
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] |
Given the complex number $\dfrac{a+i}{b-i}=2-i$ where $a, b$ are real numbers, then the point corresponding to the complex number $a+bi$ in the complex plane is located in ( )
A: First quadrant
B: Second quadrant
C: Third quadrant
D: Fourth quadrant | To solve $\dfrac{a+i}{b-i}=2-i$, where $a, b$ are real numbers, first multiply both sides of the equation by the complex conjugate of the denominator $(b+i)$ to simplify the left side:
\[
a+i=(2-i)(b-i)
\]
Expanding the right side, we get:
\[
2b-bi-2i+i^2 = 2b-(b+2)i-1 \quad \text{(since $i^2=-1$)}
\]
For two compl... | cn_k12 | 94c2b577-4123-5963-88ec-ca43a40f96af | open-r1/OpenR1-Math-220k | [
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4. In the diagram, $\angle A F C=90^{\circ}, D$ is on $A C$, $\angle E D C=90^{\circ}, C F=21, A F=20$, and $E D=6$. Determine the total area of quadrilateral $A F C E$. | 4. The area of quadrilateral $A F C E$ equals the sum of the areas of $\triangle A F C$ and $\triangle A C E$.
Since $\triangle A F C$ is right-angled at $F$, then its area equals $\frac{1}{2}(A F)(C F)=\frac{1}{2}(20)(21)=210$.
Since $\triangle A F C$ is right-angled at $F$, then we can use the Pythagorean Theorem to ... | olympiads | 19dd11a6-2025-539a-9580-c73b7ce3691b | open-r1/OpenR1-Math-220k | [
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Given that $a$, $b$, $c$ are the opposite sides of angles $A$, $B$, $C$ in $\triangle ABC$, and $D$ is the midpoint of side $AC$, with $a=BD=4\sqrt{7}$ and $c=12$. $(1)$ Find the value of $b$; $(2)$ If the angle bisector of $\angle BAC$ intersects $BC$ at point $E$, find the length of segment $AE$. | ### Step-by-Step Solution
#### Part (1): Finding the value of $b$
Given that $a = BD = 4\sqrt{7}$ and $c = 12$, and $D$ is the midpoint of side $AC$, we let $b = 2x$, which implies $AD = DC = x$.
- In $\triangle ABD$, we use the Law of Cosines to find $\cos \angle ADB$:
\[
\cos \angle ADB = \frac{x^2 + 112 - 144... | cn_k12 | 25558858-c126-5032-a757-73b554a23b0c | open-r1/OpenR1-Math-220k | [
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A rectangle with sides $a$ and $b$ has an area of $24$ and a diagonal of length $11$. Find the perimeter of this rectangle. | 1. **Given Information:**
- The area of the rectangle is \(24\), which implies:
\[
ab = 24
\]
- The length of the diagonal is \(11\), which implies:
\[
a^2 + b^2 = 11^2 = 121
\]
2. **Finding \(a + b\):**
- We use the identity for the square of a sum:
\[
(a + b)^2 = a^2 ... | aops_forum | b0df0230-e26b-59fa-9959-aeb45c522509 | open-r1/OpenR1-Math-220k | [
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June has $500 for buying school supplies for the new school year. She buys four maths books at $20 each, six more science books than maths books at $10 each, and twice as many art books as maths books at $20 each. If she also bought music books, how much money did she spend on music books? | The total cost of maths books is 4*20 = $<<4*20=80>>80$.
She bought six more science books than maths books which totals 6+4 = <<6+4=10>>10 books.
If each science book cost her $10 she spent 10*10 = $<<10*10=100>>100 on science books.
There were twice as many art books as maths books which total 2*4 = <<2*4=8>>8
The to... | null | null | openai/gsm8k | [
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8.3. On 10 cards, different natural numbers are written, their sum is 60. Masha laid out the cards in ascending order. What can the sum of the numbers on the two middle cards be? Find all possible answers. | Answer: 11,12
Solution. If the smallest possible natural numbers, i.e., the card numbers (from 1 to 10), were written on the cards, then their sum would be equal to $1+2+. .+10=55$. But the actual sum is 5 more. This means that some numbers are greater than their numbers. If the number on the fifth card is greater tha... | olympiads | 31673fa4-2afe-52d7-9c5e-b77483ebbb81 | open-r1/OpenR1-Math-220k | [
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Given an even function $f(x)$ is monotonically decreasing on $(-\infty,0]$, let $a=f(\log_{2}3)$, $b=f(\log_{4}5)$, and $c=f(2^{\frac{3}{2}})$. Then, $a$, $b$, and $c$ satisfy which of the following?
A: $a < b < c$
B: $b < a < c$
C: $c < a < b$
D: $c < b < a$ | Since the even function $f(x)$ is monotonically decreasing on $(-\infty,0]$, it follows that $f(x)$ is monotonically increasing on $(0,+\infty)$.
Given that $2 > \log_{2}3=\log_{4}9 > \log_{4}5$ and $2^{\frac{3}{2}} > 2$, we have $f(\log_{4}5) < f(\log_{2}3) < f(2^{\frac{3}{2}})$.
Therefore, $b < a < c$.
Hence, t... | cn_k12 | b6ff51ea-580e-53d6-a4ad-191b1ddd959c | open-r1/OpenR1-Math-220k | [
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] |
Given the space vectors $\overrightarrow{a}=(0,1,1)$ and $\overrightarrow{b}=(1,0,1)$, the angle between vector $\overrightarrow{a}$ and $\overrightarrow{b}$ is ( )
A: $60^{\circ}$
B: $120^{\circ}$
C: $30^{\circ}$
D: $150^{\circ}$ | **Analysis**
This question examines the application of spatial vectors, specifically the calculation of the angle between two spatial vectors, which involves the use of the dot product of vectors.
**Solution**
Given: $\overrightarrow{a}=(0,1,1), \overrightarrow{b}=(1,0,1)$,
Therefore, $|\overrightarrow{a}|= \sqrt{2... | cn_k12 | c8e51b0a-d994-5e6b-a932-02600dfcfd16 | open-r1/OpenR1-Math-220k | [
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11. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M, N$ are points on the diagonals $A C$ and $B F$ respectively. And $A M=F N$, if $A B=1$, then the range of values for $M N$ is $\qquad$ | 11. $\left[\frac{\sqrt{3}}{2}, 1\right]$
As shown in the figure, draw $M P \perp A B$ at $P$, connect $P N$. According to the problem, we can get $P N \perp A B$,
so $\angle M P N$ is the dihedral angle of the plane $D-A B$
$-E$, that is, $\angle M P N=120^{\circ}$,
Let $\Lambda M=F N=x$, then $M P=\frac{\sqrt{2}}{2} ... | olympiads | 9e7e4328-7094-579c-8925-e9473485e7aa | open-r1/OpenR1-Math-220k | [
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Given the quadratic equation in $x$: $x^{2}+2(k-1)x+k^{2}-1=0$, which has two unequal real roots.
1. Find the range of values for the real number $k$.
2. Is it possible for $0$ to be a root of the equation? If so, find the other root; if not, explain why. | 1. The discriminant $\Delta$ of the quadratic equation is given by:
$$\begin{align}
\Delta &= [2(k-1)]^{2}-4(k^{2}-1) \\
&= 4k^{2}-8k+4-4k^{2}+4 \\
&= -8k+8
\end{align}$$
Since the equation has two unequal real roots, the discriminant must be positive:
$$-8k+8 > 0$$
Solving for $k$, we find that:
$$k < 1$$
Thus, the ra... | cn_k12 | 5093ff8c-5d91-586d-9d20-088b40dd50d3 | open-r1/OpenR1-Math-220k | [
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Problem 2. 2-1. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-100)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it as the answer. | Solution. Knowledge of arithmetic progression is required. It turns out to be a quadratic function
$f(x)=51 x^{2}-2(2+4+6+\ldots+100) x+(2^{2}+4^{2}+6^{2}+\ldots+100^{2})=51 x^{2}-2 \cdot 50 \cdot 51 x+4 \cdot(1^{2}+2^{2}+3^{2}+\ldots+50^{2})$.
The minimum is achieved at the point $x_{0}=50$. In this case,
$f(50)=50... | olympiads | d0c323a8-253b-588e-8f78-dbd2c92b36e6 | open-r1/OpenR1-Math-220k | [
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] |
# Problem 7. (4 points)
The lengths of the sides of a trapezoid are 2, 10, 10, and 20. Find the length of the segment connecting the midpoints of the diagonals.
(It is known that these points are distinct). If there are multiple possible answers, list them in ascending order, separated by commas or semicolons. | # Problem 7. (4 points)
The lengths of the sides of a trapezoid are 2, 10, 10, and 20. Find the length of the segment connecting the midpoints of the diagonals.
(It is known that these points are distinct). If there are multiple possible answers, list them in ascending order, separated by commas or semicolons.
Answe... | olympiads | 6cf3c939-9c6b-5fe3-8be5-a1c976e802c9 | open-r1/OpenR1-Math-220k | [
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25. The sequence $\left(x_{n}\right)$ is defined by the conditions: $x_{1}=1$, and for each natural number $n$, the number $x_{n+1}$ is the largest number that can be obtained by rearranging the digits of the number $x_{n}+1$. Find the smallest $n$ for which the decimal representation of the number $x_{n}$ has exactly ... | 25. Answer: the smallest such number $n$ is $18298225=9(1+2+3+\ldots+2016)+1$.
To understand how this sequence is constructed, let's write out the first few terms. Initially, the sequence contains single-digit numbers:
$$
1, \quad 2, \quad 3, \quad \ldots, \quad 9 .
$$
Next come the two-digit numbers:
![](https://c... | olympiads | 4a038cf5-3da0-5b76-8661-a1c7aebe7b79 | open-r1/OpenR1-Math-220k | [
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Given a quadratic function $f(x)$ satisfies $f(x+1)-f(x)=2x-1$, and $f(0)=3$.
(1) Find the expression for the function $f(x)$.
(2) If the function $y=f(\log_{3}x+m)$, where $x \in \left[\frac{1}{3}, 3\right]$, has a minimum value of 3, find the value of the real number $m$. | Solution:
(1) Let $f(x)=ax^2+bx+c$, then $f(x+1)=a(x+1)^2+b(x+1)+c$,
Since $f(x+1)-f(x)=2x-1$,
we have $a=1$, $b=-2$, $c=3$,
thus $f(x)=x^2-2x+3$;
(2) Let $t=\log_{3}x+m$, then $t \in [m-1, m+1]$,
thus $y=f(\log_{3}x+m)=f(t)=t^2-2t+3=(t-1)^2+2$,
When $1 \leq m-1 \Leftrightarrow m \geq 2$, then $f(m-1)=3 \... | cn_k12 | 686b612f-a777-5493-a476-dbf72b784537 | open-r1/OpenR1-Math-220k | [
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3. Given an integer $n \geqslant 5$. Find the smallest integer $m$, such that there exist two sets of integers $A$ and $B$, satisfying the following conditions:
(1) $|A|=n,|B|=m$, and $A \subseteq B$;
(2) For any two distinct elements $x, y$ in set $B$, $x+y \in B$ if and only if $x, y \in A$.
（Zhenhua Niu, problem con... | 3. First, take an integer $N \geqslant 2 n$, and consider
$A=\{N+1, N+2, \cdots, N+n\}$,
$B=A \cup\{2 N+3,2 N+4, \cdots, 2 N+2 n-1\}$,
Condition (1) is clearly satisfied.
On one hand, for any $x, y \in A, x \neq y$, we have
$$
2 N+3 \leqslant x+y \leqslant 2 N+2 n-1 \text {. }
$$
Thus, $x+y \in B$.
On the other hand, ... | olympiads | 06b04226-85d9-5256-a0cc-ffdec8b46660 | open-r1/OpenR1-Math-220k | [
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] |
1.34. On side $AC$ of triangle $ABC$, a point $E$ is taken. Through point $E$, a line $DE$ parallel to side $BC$ and a line $EF$ parallel to side $AB$ are drawn ($D$ and $F$ are points on these sides, respectively). Prove that $S_{BDEF}=2 \sqrt{S_{ADE} \cdot S_{EFC}}$. | 1.34. $S_{B D E F} / 2 S_{A D E}=S_{B D E} / S_{A D E}=D B / A D=E F / A D=\sqrt{S_{E F C} / S_{A D E}}$. Therefore $S_{B D E F}=2 \sqrt{S_{A D E} \cdot S_{E F C}}$. | olympiads | e50a4e1e-31b7-5819-a24d-a2b5544e7dd4 | open-r1/OpenR1-Math-220k | [
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] |
As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE$?
$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C... | By angle subtraction, we have $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.$ Note that $\triangle DEF$ is isosceles, so $\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.$ Finally, we get $\angle AFE = 180^\circ - \angle DFE = \boxed{\textbf{(D) }170}$ degrees.
~MRENTHUSIASM ~[Aops-g5-gethsemanea... | amc_aime | c4d425cd-96ce-5dbb-80e2-ed92762bdc03 | open-r1/OpenR1-Math-220k | [
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] |
B2. The chord is 3 cm away from the center of a circle with a diameter of $6 \sqrt{2} \mathrm{~cm}$. Calculate the length of the chord, which is the intersection of the circle and the line. Draw a sketch.
(6 points) | B2. Let $S$ be the center of the circle, and $A$ and $B$ the endpoints of the chord $t$. The segments $S A$ and $S B$ are radii of the circle. Therefore, triangle $A B S$ is isosceles. The height from $S$ to the base of triangle $A B S$ bisects the chord $t$ and is 3 cm long (the distance from $S$ to the line). Using t... | olympiads | d021445f-ed03-545c-9cbd-054bca2b2eb8 | open-r1/OpenR1-Math-220k | [
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The statement "For I From 2 To 20" indicates that the loop body is executed \_\_\_\_\_ times. | **Analysis:**
According to the problem, the statement "For I From 2 To 20" means that $i$ goes from 2 to 20. Therefore, the number of times the loop body is executed is $20 - 2 + 1 = 19$. Hence, the answer is 19.
**Key Point:** Program Statements
Therefore, the loop body is executed $\boxed{19}$ times. | cn_k12 | cf30c8e7-1d29-5d9f-b4ac-08b4cfbc4e3b | open-r1/OpenR1-Math-220k | [
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4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=10 \\
\operatorname{LCM}(a ; b ; c)=2^{17} \cdot 5^{16}
\end{array}\right.
$$ | Answer: 8640.
Solution. Let $a=2^{\alpha_{1}} \cdot 5^{\alpha_{2}}, b=2^{\beta_{1}} \cdot 5^{\beta_{2}}, c=2^{\gamma_{1}} \cdot 5^{\gamma_{2}}$ (the numbers $a$, $b$, $c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=2^{... | olympiads | d794158c-e72f-50c2-beea-7d1b084d09da | open-r1/OpenR1-Math-220k | [
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14. The perimeter of a rectangle is 20 decimeters. If it is cut along the line connecting the midpoints of the longer sides, it is divided into two identical smaller rectangles, and the sum of their perimeters is 6 decimeters more than the original perimeter. The area of the original rectangle is $\qquad$ square decime... | answer: 21 | olympiads | 787ce05d-abcb-59f2-a33d-dbc1107bbe77 | open-r1/OpenR1-Math-220k | [
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## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2 \pi}(\cos x)^{\frac{\operatorname{ctg} 2 x}{\sin 3 x}}$ | ## Solution
$\lim _{x \rightarrow 2 \pi}(\cos x)^{\frac{\operatorname{ctg} 2 x}{\sin 3 x}}=\lim _{x \rightarrow 2 \pi}\left(e^{\ln (\cos x)}\right)^{\frac{\operatorname{ctg} 2 x}{\sin 3 x}}=$
$=\lim _{x \rightarrow 2 \pi} e^{\frac{\operatorname{ctg} 2 x}{\sin 3 x} \cdot \ln (\cos x)}=\exp \left\{\lim _{x \rightarrow ... | olympiads | a34e39fc-36a2-5742-9f6a-7bfa71f2772d | open-r1/OpenR1-Math-220k | [
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Among the following points, which can serve as the center of symmetry for the function $y=\tan x$? ( )
A: $\left(\frac{\pi}{4}, 0\right)$
B: $\left(\frac{\pi}{4}, 1\right)$
C: $\left(-\frac{\pi}{4}, 0\right)$
D: $\left(\frac{\pi}{2}, 0\right)$ | The function $y=\tan x$ is periodic with period $\pi$, and its graph is symmetric about the line $x = \frac{k\pi}{2}$ for $k \in \mathbb{Z}$, which is a vertical line through the points where the tangent function is undefined (asymptotes).
For each period of $\pi$, the tangent function has a vertically-oriented symmet... | cn_k12 | 939d6f6b-56d3-512b-ab85-8b5dbf4483b5 | open-r1/OpenR1-Math-220k | [
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Given the sets $A=\{1,2,m\}$ and $B=\{2,4\}$, if the union of A and B is $A \cup B = \{1,2,3,4\}$, then the real number $m=$ ______. | The definition of the union of two sets $A \cup B$ includes all the elements that are in $A$, in $B$, or in both. According to the problem, $A \cup B = \{1,2,3,4\}$. Since set $A=\{1,2,m\}$ and set $B=\{2,4\}$, to satisfy the union, $m$ must be an element that is not already in set $B$ and has to be in the resulting se... | cn_k12 | 79235a61-8e49-5c05-99f8-c182f0fea90e | open-r1/OpenR1-Math-220k | [
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In an opaque box, there are a total of 60 balls of two different colors, black and white. Apart from the color, all the balls are the same. Wang Ying conducted a ball-picking experiment. After stirring the balls in the box, she randomly picked one ball, noted its color, put it back into the box, and repeated this proce... | ### Solution:
#### Part 1:
Given that the frequency of picking a white ball stabilized at 0.25 after many trials, we can estimate the probability of picking a white ball to be close to this frequency. Therefore, the probability of picking a white ball is estimated to be:
$$\boxed{0.25}$$
#### Part 2:
Let's calculate... | cn_k12 | d6206f97-aa18-530c-b771-1319484ffe7a | open-r1/OpenR1-Math-220k | [
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To commemorate Earth Day, the senior class of Fuxing High School visited the Earth Natural Museum. After the visit, a group of 7 students from a class took a group photo. If student A and student B stand together, and student C stands on the edge, the probability that student C is not adjacent to student A or B is ____... | To solve this problem, we'll break it down into two main parts as described in the solution. Let's follow each step closely to understand how the final answer is derived.
**Part 1: Calculate the total number of ways where "student A and student B stand together, and student C stands on the edge"**
1. **Positioning st... | cn_k12 | 8a55183f-bab9-59c3-afbe-62165e8c3ef4 | open-r1/OpenR1-Math-220k | [
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## Task B-4.4.
A sequence of real numbers $\left(a_{n}\right)$ is defined recursively by
$$
a_{1}=1, \quad a_{2}=\frac{3}{5}, \quad a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}} \quad \text { for } n \geqslant 3
$$
Determine $a_{2020}$. | ## First Solution.
The terms of the sequence $\left(a_{n}\right)$ are $1, \frac{3}{5}, \frac{3}{7}, \frac{3}{9}, \ldots$, and the sequence of their reciprocals is $1, \frac{5}{3}, \frac{7}{3}, \frac{9}{3}, \ldots$ We will show that the sequence $\left(\frac{1}{a_{n}}\right)$ is an arithmetic sequence. The general term... | olympiads | c432838a-c045-541b-bf59-732fabaac634 | open-r1/OpenR1-Math-220k | [
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Given the function $f(x)=x^{3}+bx^{2}+cx+d$ whose graph passes through the point $P(0,2)$, and the equation of the tangent line at point $M(-1,f(-1))$ is $6x-y+7=0$.
$(1)$ Find the value of $f(-1)$
$(2)$ Find the expression for the function $y=f(x)$; | Solution:
$(1)$ Since the equation of the tangent line at $M(-1,f(-1))$ is $6x-y+7=0$,
it follows that $-6-f(-1)+7=0$,
$(2)$ Since the graph of $f(x)$ passes through $P(0,2)$, we know $d=2$,
thus $f(x)=x^{3}+bx^{2}+cx+2$, and then $f'(x)=3x^{2}+2bx+c$,
Given $f(-1)=1$, $f'(-1)=6$,
$\therefore \begin{cases} ... | cn_k12 | 8ce97311-8763-527d-9599-d69a41e83f2d | open-r1/OpenR1-Math-220k | [
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The reciprocal of $-7$ is
A: $7$
B: $\frac{1}{7}$
C: $-7$
D: $-\frac{1}{7}$ | To find the reciprocal of a number, we divide 1 by that number. Thus, for $-7$, the reciprocal is calculated as follows:
\[ \text{Reciprocal of } -7 = \frac{1}{-7} = -\frac{1}{7} \]
Therefore, the correct answer is $D: -\frac{1}{7}$.
So, we encapsulate the final answer as: $\boxed{D}$. | cn_k12 | 650305ad-5aa9-5c2e-9a9b-58d6ce2980ef | open-r1/OpenR1-Math-220k | [
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12. (5 points) Teacher Wang drives from home to location $A$. On the way there, the first $\frac{1}{2}$ of the journey is traveled at a speed of 50 kilometers per hour, and the remaining distance is traveled at a speed increased by $20 \%$; On the return trip, the first $\frac{1}{3}$ of the journey is traveled at a spe... | 12. (5 points) Teacher Wang drives from home to location $A$. On the way there, the first $\frac{1}{2}$ of the journey is traveled at 50 kilometers/hour, and the remaining distance is traveled at a speed increased by $20 \%$; On the return trip, the first $\frac{1}{3}$ of the journey is traveled at 50 kilometers/hour, ... | olympiads | 595cc72b-a662-5576-a6a3-a5c9ecee38e6 | open-r1/OpenR1-Math-220k | [
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## Task B-3.6.
Inside the square $A B C D$ there is a point $M$ such that $|M A|=7 \mathrm{~cm}, |M B|=13 \mathrm{~cm}$ and $|M C|=17 \mathrm{~cm}$. Calculate the area of the square $A B C D$. | ## Solution.
Let $a$ be the length of the side of the square. Set up the coordinate system as shown in the figure:
For the accurate image with all points labeled with coordinates:
(2 point... | olympiads | 7724a74d-1382-5c58-83b7-d4f6c95460ac | open-r1/OpenR1-Math-220k | [
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Given that the terminal side of angle $\alpha$ passes through the point $(3a, 4a)$ ($a < 0$), then $\sin\alpha=$ ______, $\tan(\pi-2\alpha)=$ ______. | From the given conditions, we have $x=3a$, $y=4a$, thus $r=|5a|=-5a$
Therefore, $\sin\alpha= \frac{y}{r} = -\frac{4}{5}$, $\tan\alpha= \frac{y}{x} = \frac{4}{3}$
Therefore, $\tan(\pi-2\alpha)=-\tan(2\alpha)=-\frac{2\tan\alpha}{1-\tan^2\alpha}=-\frac{\frac{8}{3}}{1-\frac{16}{9}}=\frac{24}{7}$
Hence, the answers ar... | cn_k12 | c649c672-1632-5e77-96f8-43781b14e893 | open-r1/OpenR1-Math-220k | [
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Given two points A (-2, 0) and B (3, 1), if a moving point P satisfies $|PA| = 2|PB|$, then the area of the shape enclosed by the trajectory of point P is \_\_\_\_\_\_. | Let the coordinates of point P be $(x, y)$,
Since A(-2, 0), B(3, 1), and the moving point P satisfies $|PA| = 2|PB|$,
Therefore, $\sqrt{(x+2)^{2}+y^{2}} = 2\sqrt{(x-3)^{2}+(y-1)^{2}}$, squaring both sides gives $(x+2)^{2}+y^{2} = 4[(x-3)^{2}+(y-1)^{2}]$,
Simplifying yields $(x- \frac{14}{3})^{2}+(y- \frac{4}{3})^{2}... | cn_k12 | 126c9ad9-82a2-5725-aee2-1df268c05631 | open-r1/OpenR1-Math-220k | [
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Convert $25_{(7)}$ to binary. | To convert $25_{(7)}$ to binary, we first convert it to decimal:
$25_{(7)} = 2 \times 7^{1} + 5 \times 7^{0} = 19$.
Next, we convert 19 to binary:
- $19 \div 2 = 9$ remainder $1$
- $9 \div 2 = 4$ remainder $1$
- $4 \div 2 = 2$ remainder $0$
- $2 \div 2 = 1$ remainder $0$
- $1 \div 2 = 0$ remainder $1$
Therefore, $2... | cn_k12 | 69ce8a70-9dab-5251-9bc2-4cc662cbb4aa | open-r1/OpenR1-Math-220k | [
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1. The number of proper subsets of the set $\left\{n \left\lvert\,-\frac{1}{2}<\log _{\frac{1}{n}} 2<-\frac{1}{3}\right., n \in N\right\}$ is ().
(A) 7
(B) 8
(C) 31
() 32 | $$
\begin{array}{l}
-\sqrt{1}(\mathrm{~A}) \\
-\frac{1}{2}<\log _{\frac{1}{n}} 2<-\frac{1}{3} \\
\Leftrightarrow \log _{\frac{1}{n}}\left(\frac{1}{n}\right)^{-\frac{1}{2}}<\log _{\frac{1}{n}} 2<\log _{\frac{1}{n}} \cdot\left(\frac{1}{n}\right)^{-\frac{1}{3}}, \\
\log _{\frac{1}{n}} \sqrt{n}<\log _{\frac{1}{2}} 2<\log _... | cn_contest | 1f7cefa9-d99e-5fdd-9a54-5c45319b79c7 | open-r1/OpenR1-Math-220k | [
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13. A student participates in military training and must shoot 10 times. In the 6th, 7th, 8th, and 9th shots, he scored 9.0, 8.4, 8.1, and 9.3 points, respectively. The average score of his first 9 shots is higher than the average score of his first 5 shots. If he wants the average score of 10 shots to exceed 8.8 point... | Three, 13. From the given information, the average score of the first 5 shots is less than $\frac{9.0+8.4+8.1+9.3}{4}=8.7$, the total score of the first 9 shots is at most $8.7 \times 9-0.1=78.2$. Therefore, the 10th shot must score at least $8.8 \times 10+0.1-78.2=9.9$ (rings). | cn_contest | 2f1c0719-e1f8-5ab1-a18c-f5e20b8050ab | open-r1/OpenR1-Math-220k | [
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On the Island of Misfortune, only knights, who always tell the truth, and liars, who always lie, live. In the Parliament of the island, there are 101 deputies. To reduce the budget, it was decided to reduce the Parliament by one deputy. However, each deputy stated that if they were removed from the Parliament, the majo... | Write two inequalities corresponding to the statements of the deputy liar and the deputy knight.
## Solution
Let $P$ be the number of knights in the Duma, and $L$ be the number of liars ($P+L=101$). According to the statement of the deputy knight, $P-1<50$. Therefore, $P<51$ (from this, in particular, it follows that... | olympiads | 6b372010-2d5d-59c8-8938-b6e2ff7ac80e | open-r1/OpenR1-Math-220k | [
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7. If a non-zero complex number $x$ satisfies $x+\frac{1}{x}=1$, then $x^{2014}+\frac{1}{x^{2014}}=$ $\qquad$ . | II.7. -1.
From $x+\frac{1}{x}=1 \Rightarrow x=\mathrm{e}^{ \pm \frac{x}{3} \mathrm{i}}$.
Therefore, $x^{2014}=\mathrm{e}^{\mathrm{E}^{\frac{2014 \pi}{3}} \mathrm{i}}=\mathrm{e}^{ \pm \frac{4 \pi}{3} \mathrm{i}}$.
Thus, $x^{2014}+\frac{1}{x^{2014}}=2 \cos \frac{4 \pi}{3}=-1$. | olympiads | 2325152c-45cd-531c-ab9d-2ed7070f27f5 | open-r1/OpenR1-Math-220k | [
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SUBIECTUL II Let the sequence $\left(I_{n}\right)_{n \geq 1}$ with the general term:
$$
I_{n}=\int_{0}^{\frac{\pi}{4}} e^{n x}\left(\operatorname{tg}^{n-1} x+\operatorname{tg}^{n} x+\operatorname{tg}^{n+1} x\right) d x, n \geq 1
$$
## www.edums.ro
Calculate: $\lim _{n \rightarrow \infty} n\left(\sqrt[n^{2}]{n I_{n}}... | ## Solution:
We have
$$
\begin{aligned}
& I_{n}=\int_{0}^{\frac{\pi}{4}}\left(e^{x} \tan x\right)^{n-1} e^{x}\left(1+\tan x+\tan^{2} x\right) d x=\ldots \ldots \ldots \\
& =\int_{0}^{\frac{\pi}{4}}\left(e^{x} \tan x\right)^{n-1}\left(e^{x} \tan x\right)^{\prime} x=\left.\frac{1}{n}\left(e^{x} \tan x\right)^{n}\right|... | olympiads | 75d462d6-de45-5068-b97a-a348dd53c577 | open-r1/OpenR1-Math-220k | [
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Given the set $M=\{x|1\leqslant x\leqslant 10$,$x\in N^{*}\}$, for its non-empty subset $A$, each element $k$ in $A$ is multiplied by $\left(-1\right)^{k}$ and then summed. For example, if $A=\{1,4,7\}$, the sum can be calculated as $\left(-1\right)^{-1}\times 1+\left(-1\right)^{4}\times 4+\left(-1\right)^{7}\times 7=-... | Given the set $M=\{x|1\leqslant x\leqslant 10, x\in N^{*}\}$, we can explicitly write it as $M=\{1,2,3,4,5,6,7,8,9,10\}$. We are interested in finding the total sum of sums for all non-empty subsets of $M$, where each element $k$ in a subset $A$ is multiplied by $\left(-1\right)^{k}$ before summing.
First, let's consi... | cn_k12 | c4d5c301-42ba-5b00-b281-fa202f9128a6 | open-r1/OpenR1-Math-220k | [
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6. Let $0<a<1$, if $x_{1}=a, x_{2}=a^{x_{1}}, x_{3}=a^{x_{2}}, \cdots, x_{n}=a^{x_{n-1}}, \cdots$, then the sequence $\left\{x_{n}\right\}$
(A) is increasing;
(B) is decreasing;
(C) odd terms are increasing, even terms are decreasing;
(D) even terms are increasing, odd terms are decreasing.
Answer ( ) | 6. (C)
First, consider the first three terms of the sequence,
$$
\begin{array}{l}
x_{1}=a, \quad x_{2}=a^{a}, \quad x_{3}=a^{a} \\
\because \quad 0<a<1, \\
\therefore \quad a<a^{a}<1. \\
a<a^{a^{a}}<a^{a} . \\
\text { i.e., } x_{1}<x_{3}<x_{2} \text {. } \\
\end{array}
$$ | olympiads | 110462bf-2160-51d0-980c-f5324076d5c5 | open-r1/OpenR1-Math-220k | [
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5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point ( $25 ; 60$ ). Find the number of such squares. | Answer: 650.
Solution. Draw through the given point $(25 ; 60)$ vertical and horizontal lines $(x=25$ and $y=60)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "left" vertex of the square can be located in 25 ways: $(0... | olympiads | 8ff2ecd8-4a67-558a-b379-586552599c22 | open-r1/OpenR1-Math-220k | [
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In space, four statements are given below (where $m, n$ are two distinct lines, and $\alpha, \beta$ are two distinct planes):
1. If $m \perp \alpha$ and $n \parallel \alpha$, then $m \perp n$.
2. If $m \parallel n$ and $n \parallel \alpha$, then $m \parallel \alpha$.
3. If $m \parallel n$, $n \perp \beta$, and $m \par... | 1. According to the properties of line-perpendicular-to-plane and line-parallel-to-plane, we know that if $m \perp \alpha$ and $n \parallel \alpha$, then the direction of $m$ is perpendicular to $\alpha$, while $n$ is parallel to $\alpha$ and hence cannot be perpendicular to $m$. Therefore, statement ① is incorrect.
2... | cn_k12 | d5ae2b40-d30a-5ee5-8cf9-8b761575ecc6 | open-r1/OpenR1-Math-220k | [
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Find all odd prime $p$ such that $1+k(p-1)$ is prime for all integer $k$ where $1 \le k \le \dfrac{p-1}{2}$. | 1. **Restate the problem**: We need to find all odd primes \( p \) such that \( 1 + k(p-1) \) is prime for all integers \( k \) where \( 1 \le k \le \frac{p-1}{2} \).
2. **Consider the case \( p \ge 13 \)**:
- Let \( q \) be an odd prime such that \( q \le \frac{p-1}{2} \).
- We need \( 1 + k(p-1) \) to be prime... | aops_forum | c9cd2c77-e75a-595e-bdbc-4222c94fdcb0 | open-r1/OpenR1-Math-220k | [
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In triangle $ABC$ with $AB < AC$, let $H$ be the orthocenter and $O$ be the circumcenter. Given that the midpoint of $OH$ lies on $BC$, $BC = 1$, and the perimeter of $ABC$ is 6, find the area of $ABC$. | 1. **Reflection Argument**: We start by claiming that $H$ and $A$ are reflections of each other over $BC$. To prove this, consider the reflections of $A$ and $O$ over $BC$, denoted as $A'$ and $O'$, respectively. Clearly, $AA' \perp BC$, the midpoint of $A'O$ lies on $BC$, and $A'$ lies on the circle through $B$ and $C... | aops_forum | 0b1e5dd3-cddd-5564-9625-93c14a5593c3 | open-r1/OpenR1-Math-220k | [
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Given $a>1$, $0<b<1$, and $c<0$, arrange $\log_a b$, $0$, $1$, $a^c$ in ascending order. (Use “<” to connect) | Since $a>1$ and $01$ and $c<0$, we have $0<a^c<1$.
Therefore, $\log_a b<0<a^c<1$.
Hence, the correct order is: $\boxed{\log_a b<0<a^c<1}$. | cn_k12 | 04808631-ef1d-51a9-bf4d-2b1edb54ff41 | open-r1/OpenR1-Math-220k | [
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Problem 4. Students from a school were supposed to go on a trip. $\frac{2}{9}$ more students registered than the planned number. Before departure, $\frac{3}{11}$ of the registered students canceled due to illness, so 5 fewer students went on the trip than the planned number. How many students went on the trip? | Solution. Let's denote the planned number of students by $x$. Then after registration, there were $x+\frac{2}{9} x=\frac{11}{9} x$. Since before departure, $\frac{3}{11}$ of the registered students canceled, i.e., $\frac{3}{11} \cdot \frac{11}{9} x=\frac{3}{9} x=\frac{1}{3} x$ of the planned number of students canceled... | olympiads | ab758671-c2de-5840-96a3-70f6249154cb | open-r1/OpenR1-Math-220k | [
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7. (10 points) Xiao Ming is preparing to make dumplings. He added 5 kilograms of water to 1.5 kilograms of flour and found the dough too runny. His grandmother told him that the dough for making dumplings should be mixed with 3 parts flour and 2 parts water. So, Xiao Ming added the same amount of flour three times, and... | 7. (10 points) Xiao Ming is preparing to make dumplings. He added 5 kilograms of water to 1.5 kilograms of flour and found the dough too runny. His grandmother told him that the dough for dumplings should be mixed in a ratio of 3 parts flour to 2 parts water. So, Xiao Ming added the same amount of flour three times, an... | olympiads | 1c8ba091-0c17-5ed3-a741-951e78f3aef7 | open-r1/OpenR1-Math-220k | [
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$17 \cdot 7$ In $\triangle A B C$, $\angle A=100^{\circ}, \angle B=50^{\circ}, \angle C=30^{\circ}$, and $A H$ is an altitude, $B M$ is a median, then $\angle M H C$ equals
(A) $15^{\circ}$.
(B) $22.5^{\circ}$.
(C) $30^{\circ}$.
(D) $40^{\circ}$.
(E) $45^{\circ}$.
(40th American High School Mathematics Examination, 198... | [Solution] In the right triangle $\triangle A H C$, $M$ is the midpoint of $A C$, so $H M$ is the median to the hypotenuse, thus $M H = M C$.
$\therefore$
$$
\angle M H C = \angle C = 30^{\circ} \text{.}
$$
Therefore, the answer is $(C)$. | olympiads | 67b6ed38-8172-579d-849e-a04c582a0c10 | open-r1/OpenR1-Math-220k | [
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What is the ratio of the sides of a triangle in which the sum of the lengths of two altitudes, taken in pairs, is in the ratio 5:7:8? | Let the heights in the first two sums be denoted by $m_{a}$ and $m_{b}$, and $m_{b}$ and $m_{c}$, respectively. Then the third is equal to $m_{a}+m_{c}$, and according to the problem,
\[
\begin{aligned}
m_{a}+m_{b} & =5 x \\
m_{b}+m_{c} & =7 x \\
m_{a}+\quad m_{c} & =8 x
\end{aligned}
\]
where $x$ is an unknown posit... | olympiads | a263c3ab-2ad6-5830-a134-279bf945fdf3 | open-r1/OpenR1-Math-220k | [
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4. As shown in Figure 1, given that $E$ is a point on side $BC$ of rectangle $ABCD$, and folding along $AE$ makes vertex $B$ coincide with point $F$ on side $CD$. If $AD=16, BE=10$, then the length of $AE$ is $\qquad$ | 4. $10 \sqrt{5}$.
From the problem, we have
$$
E F=B E=10, E C=A D-B E=6 \text {. }
$$
By the Pythagorean theorem, $C F=\sqrt{E F^{2}-E C^{2}}=8$.
Clearly, Rt $\triangle A F D \backsim \mathrm{Rt} \triangle F E C$.
Thus, $\frac{D F}{D A}=\frac{E C}{C F}=\frac{3}{4}$. Therefore, $D F=12$.
At this point, $A B=C D=C F+D... | cn_contest | f3b85243-56a7-5213-a932-a85d39314ce5 | open-r1/OpenR1-Math-220k | [
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] |
Given that $x= \frac {1}{3-2 \sqrt {2}}$ and $y= \frac {1}{3+2 \sqrt {2}}$, find: (1) the value of $x^2y-xy^2$; (2) the value of $x^2-xy+y^2$. | Since $x= \frac {1}{3-2 \sqrt {2}} = \frac {3+2 \sqrt {2}}{(3-2 \sqrt {2})(3+2 \sqrt {2})} = 3+2 \sqrt {2}$ and $y= \frac {1}{3+2 \sqrt {2}} = \frac {3-2 \sqrt {2}}{(3+2 \sqrt {2})(3-2 \sqrt {2})} = 3-2 \sqrt {2}$,
We have $xy= \frac {1}{3-2 \sqrt {2}} \cdot \frac {1}{3+2 \sqrt {2}} = 1$ and $x+y=3+2 \sqrt {2}+3-2 \sq... | cn_k12 | 5bedfdac-fb13-57e4-b527-6a907f63f98e | open-r1/OpenR1-Math-220k | [
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] |
Condition of the problem
Find the differential $d y$.
$$
y=\ln \left|x^{2}-1\right|-\frac{1}{x^{2}-1}
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\ln \left|x^{2}-1\right|-\frac{1}{x^{2}-1}\right)^{\prime} d x=\left(\frac{1}{x^{2}-1} \cdot 2 x+\frac{1}{\left(x^{2}-1\right)^{2}} \cdot 2 x\right) d x= \\
& =\frac{2 x}{x^{2}-1} \cdot\left(1+\frac{1}{x^{2}-1}\right) d x=\frac{2 x}{x^{2}-1} \cdot \frac{x... | olympiads | 52e4bc53-1a69-5c15-aed8-11c4e9f787e0 | open-r1/OpenR1-Math-220k | [
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] |
5. Let $f(x)=\frac{x^{2}}{2 x-1}$, and denote
$$
f_{1}(x)=f(x), f_{n+1}(x)=f\left(f_{n}(x)\right) \text {. }
$$
Then $f_{2019}(x)=$ $\qquad$ . | 5. $\frac{x^{2^{2019}}}{x^{2^{2019}}-(x-1)^{2^{20119}}}$.
$$
\begin{array}{l}
\text { Given } f(x)=\frac{x^{2}}{2 x-1}=\frac{x^{2}}{x^{2}-(x-1)^{2}} \\
=\frac{1}{1-\left(1-\frac{1}{x}\right)^{2}}(x \neq 0), \\
\text { then } \frac{1}{f(x)}=1-\left(1-\frac{1}{x}\right)^{2} \\
\Rightarrow 1-\frac{1}{f(x)}=\left(1-\frac{1... | olympiads | 5cb2e090-4598-5669-8727-67f4384017c4 | open-r1/OpenR1-Math-220k | [
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] |
80. Given the function $y=2 x+3, x \in[-1.5 ; 1]$. Find the function that is the inverse of the given one. | Solution. Solving the given equation for $x$, we have $2 x=$ $=y-3$, from which $x=0.5 y-1.5$. Transitioning to conventional notation, i.e., replacing $x$ with $y$ and $y$ with $x$ in the last equation, we obtain the inverse function: $y=0.5 x-1.5, x \in[0,5]$.
In Fig. 79, the graphs of the given function and its inve... | olympiads | e4183910-4108-5617-aac9-b20d0606f657 | open-r1/OpenR1-Math-220k | [
""
] |
7. In the Cartesian coordinate system, it is known that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $P(3,2)$ and $Q$, and the product of the radii of the two circles is $\frac{13}{2}$. If both circles are tangent to the line $l: y=k x$ and the $x$-axis, then the equation of line $l$ is | 7. $y=2 \sqrt{2} x$.
By the problem, let the equations of the two circles be
$$
\left(x-a_{i}\right)^{2}+\left(y-r_{i}\right)^{2}=r_{i}^{2}(i=1,2) \text {. }
$$
Then $r_{1} r_{2}=\frac{13}{2}, k>0$.
Since both circles pass through the point $P(3,2)$, we have
$$
\begin{array}{l}
\left(3-a_{i}\right)^{2}+\left(2-r_{i}\... | cn_contest | 392a451a-66e8-59bc-858a-58ecd6bc9a14 | open-r1/OpenR1-Math-220k | [
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] |
28. There are three pairs of real numbers $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$ that satisfy both $x^{3}-3 x y^{2}=2005$ and $y^{3}-3 x^{2} y=2004$. Compute $\left(1-\frac{x_{1}}{y_{1}}\right)\left(1-\frac{x_{2}}{y_{2}}\right)\left(1-\frac{x_{3}}{y_{3}}\right)$. | Solution:
$$
1 / 1002
$$
By the given, $2004\left(x^{3}-3 x y^{2}\right)-2005\left(y^{3}-3 x^{2} y\right)=0$. Dividing both sides by $y^{3}$ and setting $t=\frac{x}{y}$ yields $2004\left(t^{3}-3 t\right)-2005\left(1-3 t^{2}\right)=0$. A quick check shows that this cubic has three real roots. Since the three roots are ... | olympiads | 60eb8ca3-577a-5fa9-b33c-d23f293ad1b3 | open-r1/OpenR1-Math-220k | [
""
] |
3. Given real numbers $x, y$ satisfy
$$
\left(x+\sqrt{x^{2}+2}\right)\left(y+\sqrt{y^{2}+4}\right)=16 .
$$
Then the value of $x \sqrt{y^{2}+4}+y \sqrt{x^{2}+2}$ is ( ).
(A) $\frac{27}{4}$.
(B) $\frac{29}{4}$
(C) $\frac{31}{4}$
(D) $\frac{33}{4}$ | 3. C.
From equation (1), we easily obtain
$$
\left(x-\sqrt{x^{2}+2}\right)\left(y-\sqrt{y^{2}+4}\right)=\frac{1}{2} \text {. }
$$
Expanding equations (1) and (2) respectively and then subtracting them, we get
$$
x \sqrt{y^{2}+4}+y \sqrt{x^{2}+2}=\frac{31}{4} .
$$ | cn_contest | ce7b773d-1231-553f-9f6f-b31fde644694 | open-r1/OpenR1-Math-220k | [
""
] |
15. Let the product of all positive divisors of a natural number $n$ be $2^{4} \cdot 3^{12}$. Then the value of $n$ is $\qquad$. | $15.54$ | cn_contest | 35989c17-fdb5-5aec-a571-1ba6a903912d | open-r1/OpenR1-Math-220k | [
""
] |
[ Arithmetic. Mental calculation, etc.]
$[\quad$ Invariants $]$
The rabbits are sawing the log again, but now both ends of the log are secured. Ten middle pieces fell, while the two end pieces remained secured. How many cuts did the rabbits make? | How many logs did the hares receive
## Solution
The hares received 12 logs - 10 fallen and 2 secured. Therefore, there were 11 cuts.
## Answer
The problem | olympiads | 63376adf-5499-5266-9406-18b19edda30e | open-r1/OpenR1-Math-220k | [
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] |
Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n$ but do not divide $n$? | 1. **Determine the prime factorization of \( n \):**
\[
n = 2^{31} \cdot 3^{19}
\]
2. **Calculate \( n^2 \):**
\[
n^2 = (2^{31} \cdot 3^{19})^2 = 2^{62} \cdot 3^{38}
\]
3. **Find the total number of divisors of \( n^2 \):**
The number of divisors of a number \( a^m \cdot b^n \) is given by \((m+1... | aops_forum | aa148a4a-2380-56a9-ac2b-79b09b8a28b2 | open-r1/OpenR1-Math-220k | [
""
] |
In space, three planes can divide the space into at most ______ parts. | Three planes in space can at most divide the space into $\boxed{8}$ parts. | cn_k12 | 42344d31-c483-5666-be33-ecb1d3c7ba9a | open-r1/OpenR1-Math-220k | [
""
] |
In the rules of oblique projection, which of the following statements is correct? ( )
A: The intuitive diagram of a horizontally placed angle is not necessarily an angle.
B: Equal angles remain equal in the intuitive diagram.
C: Equal line segments remain equal in the intuitive diagram.
D: If two line segments are par... | In the intuitive diagram, an angle can still be different from its original value but remains an angle, so option A is incorrect. By considering the intuitive diagram of a square, options B and C can be ruled out. Therefore, the correct answer is $\boxed{\text{D}}$. | cn_k12 | 68ca8320-b876-58e1-a7fa-d563625a65f6 | open-r1/OpenR1-Math-220k | [
""
] |
4. At a rectangular table, eight chairs are placed, four on one side and four opposite them on the other side. In how many ways can eight friends be seated at this table so that Ana and Bane sit opposite each other, and Vesna and Goran sit next to each other? (It is known that all friends have distinct names.)
| 4. We can choose the country for Vesna and Goran in 2 ways, their position as a pair in 3 ways, and their mutual arrangement in 2 ways, i.e., we can arrange them in 12 ways. Now let's arrange Ana and Banet. On the side where Vesna and Goran are, there are two free seats, so we can choose the position of Ana and Banet i... | olympiads | e571a177-ebb0-57c0-b3c3-0da20636bbd7 | open-r1/OpenR1-Math-220k | [
""
] |
Let $a_n=6^{n}+8^{n}$. Determine the remainder upon dividing $a_ {83}$ by $49$. | Solution 1
Firstly, we try to find a relationship between the numbers we're provided with and $49$. We notice that $49=7^2$, and both $6$ and $8$ are greater or less than $7$ by $1$.
Thus, expressing the numbers in terms of $7$, we get $a_{83} = (7-1)^{83}+(7+1)^{83}$.
Applying the [Binomial Theorem](https://artofprob... | amc_aime | 9084ad45-86bf-59cf-9010-d2fff3225d40 | open-r1/OpenR1-Math-220k | [
""
] |
20. A research institute transfers a high-tech technology to a company for production. Both parties agree: the research institute will withdraw 4 million yuan from the company's profit at the end of each year and invest it in other investments, thus earning a profit of 10% annually.
（1）Can the research institute obtain... | 20. (1) Let the annual income at the end of each year be $A$ million yuan, with an annual profit of $10\%$, then:
the income from the first year's $A$ million yuan to the $n$-th year is $Q_{1}=A(1+0.1)^{n-1}$;
the income from the second year's $A$ million yuan to the $n$-th year is $Q_{2}=A(1+0.1)^{n-2}$;
the income fr... | olympiads | 9c08e965-28b6-5fff-b6ed-9206449a637b | open-r1/OpenR1-Math-220k | [
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] |
2. $f(n)$ is defined on the set of positive integers, and: (1) for any positive integer $n, f[f(n)]=4 n+9$; (2) for any non-negative integer $k, f\left(2^{k}\right)=2^{k+1}+3$. Determine $f(1789)$.
(1989 Australian Olympiad Problem) | $\begin{array}{l}\text { 2. Since } 1789=9+4 \cdot 9+4^{2} \cdot 9+4^{3} \cdot 9+4^{4} \cdot 2^{2} \text {, and } f(4 n+9)=f[f(f(n))]=4 f(n)+9 \text {, } \\ \text { therefore } f(1789)=9+4 \cdot f\left(9+4 \cdot 9+4^{2} \cdot 9+4^{3} \cdot 2^{2}\right)=9+4\left[9+4 \cdot f\left(9+4 \cdot 9+4^{2} \cdot 2^{2}\right)\righ... | olympiads | 647e9118-cc49-5b50-8aa2-743d3f75949f | open-r1/OpenR1-Math-220k | [
""
] |
7. (8 points) In the equation $\frac{2 \frac{3}{5}-(8.5-\square) \div 3.5}{1 \div\left(3.05+4 \frac{9}{20}\right)}=2$, $\square$ represents a number. Therefore, $\square=$ | 【Solution】Solve: $\frac{2 \frac{3}{5}-(8.5-\square) \div 3.5}{1 \div\left(3.05+4 \frac{9}{20}\right)}=2$,
$$
\begin{array}{l}
\frac{13}{5}-\left(\frac{17}{2}-\square\right) \div \frac{7}{2}=2 \times\left[1 \div\left(\frac{61}{20}+\frac{89}{20}\right)\right], \\
\frac{13}{5}-\left(\frac{17}{2}-\square\right) \div \frac{... | olympiads | f4d28329-679d-5c41-af78-14a340a94f0c | open-r1/OpenR1-Math-220k | [
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] |
6. (3 points) On a $2 \times 4$ chessboard, there are knights, who always tell the truth, and liars, who always lie. Each of them said: "Among my neighbors, there are exactly three liars." How many liars are on the board?
Neighbors are considered to be people on cells that share a common side. | Answer: 6.
Solution: On the corner cells, liars are obviously standing: they simply do not have three neighbors.
A situation where all people on the board are liars is impossible, as it would mean that four of these liars are telling the truth.
Therefore, there is at least one knight. Then all his neighbors are liar... | olympiads | 9041276a-8238-5929-96fd-21a756c1d826 | open-r1/OpenR1-Math-220k | [
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] |
4th Junior Balkan 2001 Problem 1 Find all positive integers a, b, c such that a 3 + b 3 + c 3 = 2001. | 10,10,1 Solution Recollect that the cubes are 1 3 = 1, 2 3 = 8, 3 3 = 27, 4 3 = 64, 5 3 = 125, 6 3 = 216, 7 3 = 343, 8 3 = 512, 9 3 = 729, 10 3 = 1000, 11 3 = 1331, 12 3 = 1728, 13 3 = 2197. 8 3 +8 3 +8 3 = 1536 2001, so the largest of a, b, c must be 9, 10, 11 or 12. Assume a ≥ b ≥ c. If a = 9, then b 3 + c 3 = 1272,... | olympiads | cf2a1e6b-8e29-578d-acad-0a0c6bc5fb32 | open-r1/OpenR1-Math-220k | [
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] |
During a baby's "first grab" ceremony, there are 4 educational items, 3 living items, and 4 entertainment items placed in front of him. If he only grabs one of the items, there are $\_\_\_\_\_\_\_\_$ possible outcomes. | The different outcomes of grabbing an item can be divided into three categories. According to the principle of counting by classification, the total number of outcomes is $4 + 3 + 4 = 11$.
Therefore, the number of possible outcomes is $\boxed{11}$. | cn_k12 | 047f0b62-1c25-5d28-b13e-2367de06bfbb | open-r1/OpenR1-Math-220k | [
""
] |
Example 5 As shown in Figure $15-5$, in the prism $A B C-A_{1} B_{1} C_{1}$, $A B \perp$ side face $B B_{1} C_{1} C, E$ is a point on edge $C C_{1}$ different from $C$ and $C_{\text {: }}$. $E A \perp E B_{1}$. Given $A B=\sqrt{2}, B B_{1}=2 \cdot B C=1 \cdot \angle B C C_{1}=\frac{\pi}{3}$. Find the tangent value of t... | Solution 1: Since $A B \perp$ plane $B B_{1} C_{1} C$, hence $A B \perp B E$.
Also, $E B_{1} \perp E A$, and the projection of $E A$ in plane $B C C_{1} B_{1}$ is $E B$.
By the inverse of the theorem of two perpendiculars, we know $E B_{1} \perp B E$.
In parallelogram $B C C_{1} B_{1}$, let $E B=x$. Then $E B_{1}... | olympiads | fbb3cb25-cde4-51ff-8e6f-2f9fd765a89e | open-r1/OpenR1-Math-220k | [
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] |
If the inequality $(m^{2}-m)2^{x}-\left( \frac{1}{2} \right)^{x} < 1$ holds for all $x\in(-\infty,-1]$ , then the range of the real number $m$ is ______. | **Analysis**
This problem examines the issue of the constant validity of inequalities, the properties of exponential functions, linear and quadratic functions, the method of substitution, the idea of reduction and transformation, and the extremum of functions.
By using the idea of reduction and transformation, the pr... | cn_k12 | 263f4cb9-2a10-5564-8965-f574528e2fd2 | open-r1/OpenR1-Math-220k | [
""
] |
Calculate the following expressions:
1. $(1)(2 \frac{3}{5})^{0}+2^{-2}\cdot(2 \frac{1}{4})\;^{- \frac{1}{2}}+(\frac{25}{36})^{0.5}+\sqrt{(-2)^{2}}$;
2. $\frac{1}{2}\log_{\frac{32}{49}} -\frac{4}{3}\log_{\sqrt{8}}+\log_{\sqrt{245}}$. | 1. First, let's simplify each term in the expression:
- $(1)(2 \frac{3}{5})^{0} = 1$ (any non-zero number to the power of 0 is 1)
- $2^{-2}\cdot(2 \frac{1}{4})\;^{- \frac{1}{2}} = \frac{1}{4} \cdot \frac{1}{\sqrt{2 \frac{1}{4}}} = \frac{1}{4} \cdot \frac{1}{\sqrt{\frac{9}{4}}} = \frac{1}{4} \cdot \frac{2}{3} = \f... | cn_k12 | 7c37fc52-dc3f-5315-bcc9-a2db8b34de59 | open-r1/OpenR1-Math-220k | [
""
] |
Example 2 Find the integer solutions of the equation $\frac{1}{x^{2}}+\frac{1}{x y}+\frac{1}{y^{2}}=1$.
untranslated text remains the same as requested. | Solution: The original equation can be transformed into $y^{2}+x y+x^{2}=x^{2} y^{2}$. Multiplying both sides of the equation by 4, and then adding $4 x y+1$, we get $4(x+y)^{2}+1=(2 x y+1)^{2}$.
Let $p=2(x+y), q=(2 x y+1)$, then the original equation can be transformed into $p^{2}+1=q^{2}$, which is $(q+p)(q-p)=1$. T... | olympiads | e8d9de1f-8f82-5ace-a3a4-424aaabbdf5a | open-r1/OpenR1-Math-220k | [
""
] |
Find all functions $f$ from $\mathbb{N}$ to $\mathbb{N}$ satisfying the following condition:
$$
\forall n \in \mathbb{N}, f(f(n))<f(n+1)
$$ | Let $P_{n}$ be the proposition for all natural numbers $n$: " $\forall m \in 0, 1, \ldots, n, \forall k \in \mathbb{N}_{\geq m+1}, f(m)n+1$. Then, the inequality in the statement ensures that $f(f(\alpha-1))n$. Thus, $\alpha=n+1$, which ensures that $P_{n+1}$ is satisfied.
Thus, we have proved by induction that $P_{n}... | olympiads | c451d5bf-3073-5d4b-a394-abfa98faee04 | open-r1/OpenR1-Math-220k | [
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] |
(1) Given that $P$ is a point on the plane of $\triangle A B C$, satisfying $\overrightarrow{P A}+\overrightarrow{P B}+2 \overrightarrow{P C} = 3 \overrightarrow{A B}$, then the ratio of the area of $\triangle A B P$ to the area of $\triangle A B C$ is $\qquad$. | 1 $1: 2$
The text has been translated while preserving the original formatting and line breaks. | olympiads | ffe62add-a436-5584-b402-0899df3d70af | open-r1/OpenR1-Math-220k | [
""
] |
7. If $a, b, c, d$ are positive numbers, and $a+b+c+d=\sqrt{7960}$, then the minimum value of $a^{2}+b^{2}+c^{2}+d^{2}$ is | 7. By the convexity of $y=x^{2}$, we know that $\frac{1}{4}\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \geqslant\left(\frac{a+b+c+d}{4}\right)^{2}$. That is, $a^{2}+b^{2}+c^{2}+d^{2} \geqslant \frac{1}{4}(a+b+c+d)^{2}=\frac{1}{4} \cdot 7960=1990$, when $a=b=c=d=\frac{1}{4} \sqrt{7960}=\frac{1}{2} \sqrt{1990}$, the minimum val... | olympiads | fc6ac18c-7f34-573b-97e7-c64f0019a813 | open-r1/OpenR1-Math-220k | [
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] |
For the proposition "If $\angle 1 + \angle 2 = 90^{\circ}$, then $\angle 1 \neq \angle 2$", an example that proves it to be false is ( ).
A: $\angle 1 = 50^{\circ}$, $\angle 2 = 40^{\circ}$
B: $\angle 1 = 50^{\circ}$, $\angle 2 = 50^{\circ}$
C: $\angle 1 = 40^{\circ}$, $\angle 2 = 40^{\circ}$
D: $\angle 1 = \angle ... | To evaluate the proposition "If $\angle 1 + \angle 2 = 90^{\circ}$, then $\angle 1 \neq \angle 2$", we need to find an example where $\angle 1 + \angle 2 = 90^{\circ}$ but $\angle 1 = \angle 2$.
Let's examine the options:
A: $\angle 1 = 50^{\circ}$, $\angle 2 = 40^{\circ}$ implies $\angle 1 \neq \angle 2$ and their ... | cn_k12 | bd5f0ae7-37f1-5a49-a4db-77ff18abdd1c | open-r1/OpenR1-Math-220k | [
""
] |
7.1. Palm oil has increased in price by $10 \%$. Due to this, the cheese of one of the manufacturers has increased in price by $3 \%$. What is the percentage of palm oil in the cheese of this manufacturer? | Answer: $30 \%$.
Solution. Let's assume that the cheese cost 100 conditional rubles per kilogram. Then it increased by 3 rubles. Since this happened due to the increase in the price of palm oil, 3 rubles is $10 \%$ of the cost of palm oil in the cheese, meaning the cost of palm oil in a kilogram of cheese is 30 rubles... | olympiads | 85789c4f-7417-5b26-972d-8077c643e857 | open-r1/OpenR1-Math-220k | [
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] |
163 The number of solutions to the equation $\sin |x|=|\cos x|$ in the closed interval $[-10 \pi, 10 \pi]$ is $\qquad$ . | 163 20. Since $\sin |x|$ and $|\cos x|$ are both even functions, we only need to consider the number of solutions within $[0,10 \pi]$, at which point the equation becomes $\sin x=|\cos x|$. When $0 \leqslant x \leqslant \frac{\pi}{2}$, the equation becomes $\sin x=\cos x$, with only one solution $x=\frac{\pi}{4}$; when... | olympiads | 7822e3e4-0613-5ebe-af64-3c49520fae18 | open-r1/OpenR1-Math-220k | [
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] |
In the plane Cartesian coordinate system $xOy$, the Manhattan distance between two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ is defined as $d(A,B) = |x_{1}-x_{2}| + |y_{1}-y_{2}|$. Given points $M(2,0)$ and $N(a,b)$, if $d(M,N) = 2$, then the sum of the minimum and maximum values of $a^{2}+b^{2}-4a$ is:
A: $0$
B... | Given the problem statement, we start by analyzing the Manhattan distance between points $M(2,0)$ and $N(a,b)$, which is given as $d(M,N) = 2$. This translates to the equation $|a-2| + |b| = 2$. We need to find the sum of the minimum and maximum values of the expression $a^{2} + b^{2} - 4a$ under this condition.
First... | cn_k12 | 47b07b03-cc1e-5975-881a-d21a355d2866 | open-r1/OpenR1-Math-220k | [
""
] |
Example 10. Find the direction of the greatest increase of the scalar field $u=x y+y z+x z$ at the point $M_{0}(1,1,1)$ and the magnitude of this greatest increase at this point. | Solution. The direction of the greatest change in the field is indicated by the vector grad $\chi(M)$. Let's find it:
$$
\operatorname{grad} u(M)=(y+z) \dagger+(x+c) j+(y+x) \mathbf{k}
$$
Thus, $\operatorname{grad} u(M)=2(\mathbf{I}+\mathbf{J}+\mathbf{k})$. This vector determines the direction of the greatest increas... | olympiads | e842c768-f6bd-5e6e-b15c-6b60796b6cf5 | open-r1/OpenR1-Math-220k | [
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] |
Observe the following equations, and then answer the following questions based on the pattern you discovered.
The first equation: $1-\frac{1}{3}=\frac{2}{1×3}$
The second equation: $\frac{1}{2}-\frac{1}{4}=\frac{2}{2×4}$
The third equation: $\frac{1}{3}-\frac{1}{5}=\frac{2}{3×5}$
The fourth equation: $\frac{1}{... | ### Step-by-Step Solution
#### Part 1: Writing Down the Seventh Equation
Given the pattern from the provided equations:
- The first equation: $1-\frac{1}{3}=\frac{2}{1×3}$
- The second equation: $\frac{1}{2}-\frac{1}{4}=\frac{2}{2×4}$
- The third equation: $\frac{1}{3}-\frac{1}{5}=\frac{2}{3×5}$
- The fourth equation... | cn_k12 | 516f3744-6f62-5b8a-9089-af1acc30b5ee | open-r1/OpenR1-Math-220k | [
""
] |
When a line is tangent to a circle, the line connecting the center of the circle to the tangent point is perpendicular to the tangent line. By analogy, when a plane is tangent to a sphere, the line connecting the center of the sphere to the tangent point is perpendicular to the tangent plane. This uses:
A: Inductive re... | **Analysis of the problem**: From the analogy of a line to a plane, and from a circle to a sphere, that is, from a plane to space, analogical reasoning is used. Therefore, the answer is C.
**Key point**: Analogical reasoning.
$\boxed{\text{C}}$ | cn_k12 | 5e6b7cdc-cbee-52a8-99fb-8a9a94dd8cd2 | open-r1/OpenR1-Math-220k | [
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] |
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