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Line $b$ is parallel to line $y=3x-2$ and passes through the point $(5, -1)$. What are the y-intercept and x-intercept of line $b$? | 1. **Determine the Slope**: Since line $b$ is parallel to the line $y=3x-2$, it has the same slope, $m = 3$.
2. **Use Point-Slope Form**: Apply the point-slope form of the equation of a line using the point $(5, -1)$:
\[
y + 1 = 3(x - 5) \implies y = 3x - 15 - 1 \implies y = 3x - 16
\]
3. **Find Y-Intercept**:... | \frac{16}{3} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
The condition "$$α= \frac {π}{6}+2kπ(k∈Z)$$" is a \_\_\_\_\_\_ condition for "$$cos2α= \frac {1}{2}$$". | When "$$α= \frac {π}{6}+2kπ(k∈Z)$$", the condition "$$cos2α= \frac {1}{2}$$" holds.
However, when "$$cos2α= \frac {1}{2}$$", the condition "$$α= \frac {π}{6}+2kπ(k∈Z)$$" does not necessarily hold, for example, when $$α=- \frac {π}{6}$$.
Therefore, "$$α= \frac {π}{6}+2kπ(k∈Z)$$" is a sufficient but not necessary con... | \text{sufficient but not necessary} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A shooting athlete is undergoing shooting training. The landing points of the first three shots on the target, A, B, and C, are exactly the three vertices of an equilateral triangle with a side length of 3cm.
(I) The scores (ring numbers) of the athlete's first three shots are all within the interval [7.5, 8.5). After... | (I) Let the scores of the first three shots be $x_{1}$, $x_{2}$, and $x_{3}$, and the scores of the subsequent three shots be $y_{1}$, $y_{2}$, and $y_{3}$.
From these six shooting scores, two are randomly selected. The basic events are: $\{x_{1},x_{2}\}$, $\{x_{1},x_{3}\}$, $\{x_{2},x_{3}\}$, $\{y_{1},y_{2}\}$, $\{y_... | \frac{2 \sqrt{3}π}{27} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
[ Lemoine Point ]
Lines $A M$ and $A N$ are symmetric with respect to the bisector of angle $A$ of triangle $A B C$ (points $M$ and $N$ lie on line $B C$). Prove that $B M \cdot B N /(C M \cdot C N)=c^{2} / b^{2}$. In particular, if $A S-$ is the symmedian, then $B S / C S=c^{2} / b^{2}$. | By the Law of Sines, $A B / B M = \sin A M B / \sin B A M$ and $A B / B N = \sin A N B / \sin B A N$. Therefore,
$$
\frac{A B^{2}}{B M \cdot B N} = \frac{\sin A M B \sin A N B}{\sin B A M \sin B A N} = \frac{\sin A M C \sin A N C}{\sin C A N \sin C A M} = \frac{A C^{2}}{C M \cdot C N}
$$ | proof | Geometry | proof | Yes | Yes | olympiads | false |
Determine the coefficient of the $x^{2}$ term in the expansion of $\left(1+x+\frac{1}{x^{2015}}\right)^{10}$. (Express the answer as a numerical value.) | 1. Observe that only the first part of the expansion contains the $x^{2}$ term: $\left(1+x+\frac{1}{x^{2015}}\right)^{10}=C_{10}^{0}(1+x)^{10}\cdot\left(\frac{1}{x^{2015}}\right)^{0}+C_{10}^{1}(1+x)^{9}\cdot\left(\frac{1}{x^{2015}}\right)^{1}+\ldots$
2. Focus on the first part of the expansion, which is $(1+x)^{10}$. W... | 45 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
1. In this problem, we consider positive integers consisting of 5 digits, where the first and the last digit are not zero. We call such a number a palindrome product if it meets the following two conditions:
- the number is a palindrome (i.e., reads the same from left to right as from right to left);
- the number is a... | 1. Note that the two numbers mentioned in the second condition must not end in 0; otherwise, the product would also end in 0. Therefore, the two numbers must have the same number of digits. However, the product of two numbers with four or more digits is too large (it would have at least seven digits), and the product o... | 10201,12321,14641,20502,23632,26962,40804,44944 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In an election with only two candidates, a candidate secured 70% of the valid votes and won by a certain majority. The total number of valid votes was 435. What was the majority of votes by which the candidate won? | Let's denote the winning candidate's votes as \( W \) and the losing candidate's votes as \( L \).
Since the winning candidate secured 70% of the valid votes, we can calculate their number of votes as follows:
\[ W = 70\% \times 435 \]
\[ W = 0.70 \times 435 \]
\[ W = 304.5 \]
Since we can't have a fraction of a vote... | 173 | Algebra | math-word-problem | Yes | Incomplete | orca_math | true |
The equation of the line that passes through the point $P(1,2)$ and is at the maximum distance from the origin is ( )
A: $x+2y-5=0$
B: $2x+y-4=0$
C: $x+3y-7=0$
D: $3x+y-5=0$ | **Analysis**
This question examines the solution of the equation of a line. To determine when the distance is maximum in order to find the equation of the line is a basic problem.
**Solution**
According to the problem, the distance is maximum when it is perpendicular to line $OA$.
Since the slope of line $OA$ is $... | A | Geometry | MCQ | Yes | Yes | cn_k12 | false |
Zhang Chao, a student at a certain school, has a student ID number of 200608251, where 2006 represents the year of enrollment, 08 represents the class he is in, 25 represents his number in the class, and 1 represents male (2 represents female). If Huang Yan, who enrolled in the school this year, is assigned to class 12... | To find the sum of the digits in Huang Yan's student ID number, we first need to construct her ID based on the given information. Since she enrolled this year, the first four digits will represent the current year. Assuming the current year is the same as Zhang Chao's enrollment year for the sake of example, her ID sta... | 22 | Number Theory | math-word-problem | Yes | Yes | cn_k12 | false |
3. Let $p>3$ be a given prime number. For a set $S \subseteq \mathbb{Z}$ and $a \in \mathbb{Z}$, define
$$
S_{a}=\left\{x \in\{0,1, \ldots, p-1\} \mid(\exists s \in S) x \equiv_{p} a \cdot s\right\}
$$
(a) How many sets $S \subseteq\{1,2, \ldots, p-1\}$ are there for which the sequence $S_{1}, S_{2}, \ldots, S_{p-1}$... | 3. Let $g$ be a primitive root modulo $p$. The sequence $S_{1}, S_{2}, \ldots, S_{p-1}$ is a permutation of $S_{1}, S_{g}, \ldots, S_{g^{p-2}}$. Since the sequence $S_{1}, S_{g}, S_{g^{2}}, \ldots$ is periodic with period $p-1$, its minimal period $d$ divides $p-1$ and the sets $S_{1}, S_{g}, \ldots, S_{g^{d-1}}$ are m... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a circle with a radius of 2cm, the central angle of a sector with an area of 4cm2 is ( ) rad.
A: 1
B: 2
C: 3
D: 4 | Given a sector with a radius of 2cm and an area of 4cm2, let the arc length be $l$,
then $$\frac {1}{2}l \times 2 = 4$$, thus $l=4$,
Therefore, the central angle of the sector is: $$\frac {4}{2} = 2 \text{ rad}.$$
Hence, the correct option is: $\boxed{B}$.
By using the formula for the area of a sector to find the a... | B | Geometry | MCQ | Yes | Yes | cn_k12 | false |
Find the values of $C$ and $D$ such that
\[\frac{6x-3}{x^2-8x+15} = \frac{C}{x-3} + \frac{D}{x-5}.\] | First, factorize the denominator:
\[\frac{6x-3}{x^2-8x+15} = \frac{6x-3}{(x-3)(x-5)}.\]
We propose that the equivalence holds:
\[\frac{6x-3}{(x-3)(x-5)} = \frac{C}{x-3} + \frac{D}{x-5}.\]
Multiplying through by $(x-3)(x-5)$ gives:
\[6x - 3 = C(x - 5) + D(x - 3).\]
To find $C$, substitute $x = 3$:
\[6(3) - 3 = C(3 - 5),... | \left(-\frac{15}{2}, \frac{27}{2}\right) | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
7 Let $m_{1}, m_{2}, \cdots, m_{r}$ (which can be the same) and $n_{1}, n_{2}, \cdots, n_{s}$ (which can be the same) be two sets of integers, satisfying: for any integer $d$ greater than 1, the number of elements in the array $m_{1}, m_{2}, \cdots, m_{r}$ that are divisible by $d$ (counting multiplicities, for example... | Let $p$ be a prime, and $m_{1}, \cdots, m_{r}$ have exactly $a_{i}$ numbers divisible by $p^{i}$, and $n_{1}, \cdots, n_{s}$ have exactly $b_{i}$ numbers divisible by $p^{i}$. From the given, we have $a_{i} \geqslant b_{i} \quad(i=1$, $2, \cdots)$. The number of elements in $m_{1}, \cdots, m_{r}$ that are divisible by ... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
Let the length of the two diagonals of a rhombus be 4 centimeters (cm). Find the area of this rhombus. | The area of a rhombus can be calculated using the formula:
Area = (d1 * d2) / 2
where d1 and d2 are the lengths of the diagonals.
In this case, both diagonals are 4 cm long, so:
Area = (4 cm * 4 cm) / 2
Area = 16 cm² / 2
Area = 8 cm²
Therefore, the area of the rhombus is $\boxed{8}$ square centimeters. | 8 | Geometry | math-word-problem | Yes | Yes | orca_math | true |
Given that the moving circle $P$ passes through the point $M(-1,0)$ and is tangent to the circle $N: x^{2}+y^{2}-2x-15=0$.
$(1)$ Find the equation of the locus $\Omega$ of the center of circle $P$.
$(2)$ Let $A$ and $C$ be two moving points on the locus $\Omega$ in the first quadrant (not on the line $x=1$). Lines ... | ### Solution:
#### Part (1):
Given the equation of circle $N$ is $x^{2}+y^{2}-2x-15=0$, we can rewrite it as:
\[
(x-1)^{2}+y^{2}=16
\]
This represents a circle centered at $(1,0)$ with a radius of $4$.
Since the point $M(-1,0)$ satisfies the inequality:
\[
(-1-1)^{2}+0^{2}=4 |MN| = 2
\]
This implies that the locus ... | |EN|=|FN| | Geometry | proof | Yes | Yes | cn_k12 | false |
William is a jockey. He can ride his horse for 6 hours a day. Over 6 days, he only used the maximum riding time twice. On two days he rode his horse for a certain number of hours a day and half the maximum time for the next two days. William rode his horse for 21 hours during those 6 days. How many hours did William ri... | Let's denote the number of hours William rode his horse on those two days as x hours each day.
According to the problem, William used the maximum riding time of 6 hours twice, so that accounts for 2 * 6 = 12 hours.
On the next two days, he rode for x hours each day, so that's 2 * x hours.
On the last two days, he r... | 1.5 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
5. In the office, each computer was connected by wires to exactly 5 other computers. After some computers were infected by a virus, all wires from the infected computers were disconnected (a total of 26 wires were disconnected). Now, each of the uninfected computers is connected by wires to only 3 others. How many comp... | # 5. Answer: 8.
Let $\mathrm{m}$ be the number of infected computers, and $\mathrm{n}$ be the number of uninfected computers. Then, before the infection, there were $5(\mathrm{~m}+\mathrm{n}) / 2$ cables, and after the disconnection, there were $3 \mathrm{n} / 2$ cables (from which it follows that $\mathrm{n}$ is even... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Given the sequence $\{a_n\}$ where $a_1 = 2$ and $a_{n+1} = 4a_n - 3n + 1$ for $n \in \mathbb{N}$:
(1) Prove that the sequence $\{a_n - n\}$ is a geometric sequence;
(2) Find the explicit formula for the $n$-th term of the sequence $\{a_n\}$ and the sum of its first $n$ terms $S_n$. | (1) Proof:
Since $a_1=2$ and $a_{n+1} = 4a_n - 3n + 1$, we can calculate the following term:
$$
a_{n+1} - (n+1) = 4a_n - 3n + 1 - n - 1 = 4(a_n - n).
$$
This implies:
$$
\frac{a_{n+1} - (n+1)}{a_n - n} = 4.
$$
Therefore, the sequence $\{a_n - n\}$ is a geometric sequence with a common ratio of 4.
\[\boxed{\text{The se... | S_n = \frac{n(n+1)}{2} + \frac{4^n - 1}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
In the diagram, $D$ is on side $A C$ of $\triangle A B C$ so that $B D$ is perpendicular to $A C$. Also, $\angle B A C=60^{\circ}$ and $\angle B C A=45^{\circ}$. If the area of $\triangle A B C$ is $72+72 \sqrt{3}$, what is the length of $B D$ ?
=x^{n}+a_{1} x^{n-1}+\cdots+a_{n}
$$
can be expressed as \( P(x)=\sum_{j=0}^{n}\left(\prod_{\substack{i \neq j \\ 0 \leqslant i \leqslant n}} \frac{x-x_{i}}{x_{j}-x_{i}}\right) P\left(x_{j}\right) \).
By contradiction. Suppose the conclusio... | proof | Algebra | proof | Yes | Yes | olympiads | false |
Given that the area of triangle $\triangle ABC$ is $10\sqrt{3}$, and the lengths of the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, with $a=7$, $\sin ^{2}A=\sin ^{2}B+\sin ^{2}C-\sin B\sin C$. Find the values of $A$, $b$, and $c$. | To solve this problem, we start by analyzing the given equation $\sin ^{2}A=\sin ^{2}B+\sin ^{2}C-\sin B\sin C$. This equation can be rearranged to express a relationship between the sides of the triangle using the Law of Cosines.
First, we note that the given equation can be rewritten as:
\[
\sin ^{2}B+\sin ^{2}C-\s... | A = \frac{\pi}{3}, b = 8, c = 5 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
a) Given a prime number $p$ and $2$ polynomials$$P(x)=a_nx^n+...+a_1x+a_0; Q(x)=b_mx^m+...+b_1x+b_0.$$
We know that the product $P(x)Q(x)$ is a polynomial whose coefficents are all divisible by $p.$
Prove that: at least $1$ in $2$ polynomials $P(x),Q(x)$ has all coefficents are all divisible by $p.$
b) Prove that the ... | null | null | Algebra | proof | Yes | Yes | aops_forum | false |
A regular $15$-gon has $L$ lines of symmetry, and the smallest positive angle for which it has rotational symmetry is $R$ degrees. What is $L+R$?
$\textbf{(A) }24\qquad\textbf{(B) }27\qquad\textbf{(C) }32\qquad\textbf{(D) }39\qquad\textbf{(E) }54$ | null | null | Geometry | MCQ | Yes | Yes | aops_forum | false |
Mark does a gig every other day for 2 weeks. For each gig, he plays a certain number of songs. 2 of the songs are 5 minutes long and the last song is twice that long. He played for 280 minutes. How many songs does he play at each gig? | Let's break down the information given:
- Mark does a gig every other day for 2 weeks. Since there are 7 days in a week, 2 weeks would have 14 days. If he does a gig every other day, that means he does a gig on 7 of those days.
- Each gig consists of a certain number of songs, with 2 songs that are 5 minutes long each... | 7 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Sean has 55 blocks. 29 are eaten by a hippopotamus. How many blocks will Sean have? | If Sean starts with 55 blocks and a hippopotamus eats 29 of them, then Sean will have:
55 - 29 = $\boxed{26}$ blocks remaining. | 26 | Other | math-word-problem | Yes | Yes | orca_math | true |
The average age of an adult class is 40 years. 2 new students with an average age of 32 years join the class. This causes the average age to decrease by a certain number of years. The original strength of the class was 2. By how many years did the average age decrease? | Let's denote the original total age of the class as \( T \) and the original number of students as \( N \). We are given that the average age of the class is 40 years, so we can write:
\[ T = 40N \]
We are also told that the original strength of the class was 2, so \( N = 2 \). Therefore, the total age of the class b... | 4 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
If $f(x)$ is an odd function and for $x 0$. | Since $f(x)$ is an odd function, it satisfies the property $f(-x) = -f(x)$ for all $x$ in its domain.
Given that for $x 0$, which means $-x 0$:
$$
f(x) = -f(-x) = -(-3\sin(x) + 4\cos(x) + 1) = 3\sin(x) - 4\cos(x) - 1.
$$
Hence, the expression for $f(x)$ when $x > 0$ is $\boxed{3\sin(x) - 4\cos(x) - 1}$. | 3\sin(x) - 4\cos(x) - 1 | Algebra | math-word-problem | Incomplete | Yes | cn_k12 | false |
The numbers $1,2,...,100$ are written in a board. We are allowed to choose any two numbers from the board $a,b$ to delete them and replace on the board the number $a+b-1$.
What are the possible numbers u can get after $99$ consecutive operations of these? | 1. Let's denote the initial set of numbers on the board as \( S = \{1, 2, 3, \ldots, 100\} \).
2. We are allowed to choose any two numbers \( a \) and \( b \) from the board, delete them, and replace them with \( a + b - 1 \).
3. We need to determine the final number on the board after 99 operations.
To solve this, we... | 4951 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
$4.9 \sin ^{2}\left(\frac{7 \pi}{8}-2 \alpha\right)-\sin ^{2}\left(\frac{9 \pi}{8}-2 \alpha\right)=\frac{\sin 4 \alpha}{\sqrt{2}}$. | ## 4.9 We have
$$
\begin{aligned}
& \sin ^{2}\left(\frac{7 \pi}{8}-2 \alpha\right)-\sin ^{2}\left(\frac{9 \pi}{8}-2 \alpha\right)= \\
& =\frac{1-\cos \left(\frac{7 \pi}{4}-4 \alpha\right)-1+\cos \left(\frac{9 \pi}{4}-4 \alpha\right)}{2}= \\
& =-\sin (2 \pi-4 \alpha) \sin \frac{\pi}{4}=\frac{\sin 4 \alpha}{\sqrt{2}}
\e... | \frac{\sin4\alpha}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
316. A Glazed Pastry. A pastry made in the shape of a triangular prism with a very small height is covered with a thin layer of glaze on all sides. How should this pastry be divided among 11 friends so that each gets an equal amount of both the pastry and the glaze? | 316. Divide the triangular perimeter into 11 equal parts, and then draw cuts along the lines connecting the center of the inscribed circle with the resulting division points. This problem was first posed by H. S. M. Coxeter for a square. This method also applies to the case of any polygon in which a circle can be inscr... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In $\triangle A B C$, if $A D$ is the angle bisector of $\angle A$, and the function $f(x)=$ $x^{2}-a x+\frac{a^{2} b c}{(b+c)^{2}}$, prove:
$$
f(\overline{B D})=f(\overline{C D})=0 \text {. }
$$ | 4. $B D + C D = a, B D \cdot C D = \frac{a^{2} b c}{(b+c)^{2}}$. | proof | Algebra | proof | Yes | Yes | olympiads | false |
Find all integers $a$ and $b$ satisfying
\begin{align*}
a^6 + 1 & \mid b^{11} - 2023b^3 + 40b, \qquad \text{and}\\
a^4 - 1 & \mid b^{10} - 2023b^2 - 41.
\end{align*} | 1. We start with the given conditions:
\[
a^6 + 1 \mid b^{11} - 2023b^3 + 40b
\]
\[
a^4 - 1 \mid b^{10} - 2023b^2 - 41
\]
2. Note that \(a^2 + 1 \mid a^6 + 1\) and \(a^2 + 1 \mid a^4 - 1\). This is because:
\[
a^6 + 1 = (a^2)^3 + 1 \quad \text{and} \quad a^4 - 1 = (a^2)^2 - 1
\]
By the fa... | (a, b) = (0, b) | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
A point $Q$ is chosen inside $\triangle DEF$ such that lines drawn through $Q$, parallel to the sides of $\triangle DEF$, divide it into three smaller triangles with areas $16$, $25$, and $36$ respectively. Find the area of $\triangle DEF$. | Since the lines through $Q$ are parallel to the sides of $\triangle DEF$, all four triangles ($\triangle DEF$ and the three smaller triangles) are similar to each other.
Let the side lengths corresponding to the smaller triangles with areas $16$, $25$, and $36$ be $4x$, $5x$, and $6x$ respectively. This is because the... | 77 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
Let's calculate the area of a regular star polygon if we know the distance from the line connecting two adjacent outer vertices to the farthest vertex from this line.
untranslated portion: (Note: The original text did not contain any additional formatting or line breaks beyond the initial sentence, so the translation... | The figure is shown below.
Given $AB = p, AC = AD = a, AO = m_1, AE = m_2$. Since $\angle AOC = \frac{360^\circ}{10} = 36^\circ$, it follows that $\angle ABC = \angle OCE = 18^\circ$, $\angle... | 2\sqrt{85-38\sqrt{5}}p^2\approx0.343p^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $S_{n}$ be the sum of the first $n$ terms of a geometric sequence ${a_{n}}$. Given that $S_{3}=2$ and $S_{6}=6$, find the value of $a_{4}+a_{5}+...+a_{12}=$ _____ . | By the properties of a geometric sequence, the following sequences are also geometric: $S_{3}$, $S_{6}-S_{3}$, $S_{9}-S_{6}$, $S_{12}-S_{9}$.
Thus, we have:
$(S_{6}-S_{3})^{2}=S_{3} \cdot(S_{9}-S_{6})$
$(S_{6}-S_{3})(S_{12}-S_{9})=(S_{9}-S_{6})^{2}$
Given that $S_{3}=2$ and $S_{6}=6$, we solve to find $S_{9}=14$ an... | 28 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
The polynomial \( Q(x) \) is a monic, quartic polynomial with real coefficients, having roots at \( e^{i\phi} \) and \( i e^{i\phi} \), where \( 0 < \phi < \frac{\pi}{6} \). When all four roots of \( Q(x) \) are plotted in the complex plane, they form a rectangle. If \( Q(0) \) is twice the area of this rectangle, find... | Given the roots \( e^{i\phi} \) and \( i e^{i\phi} \), the other two roots are their conjugates \( e^{-i\phi} \) and \( -i e^{-i\phi} \) considering \( Q(x) \) has real coefficients and uses complex conjugates.
These roots form a rectangle since:
- \( e^{i\phi} \) and \( e^{-i\phi} \) are on the unit circle at angles ... | 1 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Calculate the sum of the proper divisors of $540$. | To find the sum of the proper divisors of $540$, we first determine its prime factorization:
\[ 540 = 2^2 \cdot 3^3 \cdot 5^1. \]
The sum of all divisors $\sigma(n)$ of a number $n$ given by its prime factorization $n = p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}$ can be calculated using:
\[ \sigma(n) = (1 + p_1 + p_1^2 + \l... | 1140 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Given $\sin x + \sin y = \frac{15}{17}$ and $\cos x + \cos y = \frac{8}{17}$, find the value of $\tan x + \tan y$ while also knowing $\sin(x - y) = \frac{1}{5}$. | From the angle addition formulas and sum-to-product identities, we can write:
\[
\tan x + \tan y = \frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = \frac{\sin(x + y)}{\cos x \cos y}
\]
Squaring and adding the given trigonometric equations:
\[
(\sin x + \sin y)^2 + (\cos x + \cos y)^2 = \left(\frac{15}{17}\right)^2... | \frac{195\sqrt{6}}{328} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Being his favorite, Michael has 4 times more flying robots than Tom. For every 3 flying robots Tom has, He gives 1 to James, and James gives all the robots he receives to Michael. If Tom has 12 flying robots, how many flying robots does Michael have in total, including his original amount and those given by James? | First, let's find out how many flying robots Michael originally has. Since Michael has 4 times more flying robots than Tom, and Tom has 12 flying robots, we calculate Michael's original amount by multiplying Tom's amount by 4:
Michael's original amount = Tom's amount * 4
Michael's original amount = 12 * 4
Michael's or... | 52 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Let H be a complex separable Hilbert space and denote $B(H)$ the algebra of bounded linear operators on H. Find all *-subalgebras C of $B(H)$ for which for all $A \in B(H)$ and $T \in C$ there exists $S \in C$ that
$$TA-AT^{\ast} = TS-ST^{\ast}$$
note: *-algebra is also known as involutive algebra. | null | null | Other | math-word-problem | Yes | Yes | aops_forum | false |
A rectangular park measures $30$ meters by $60$ meters. A landscaper has $2400$ meters of fencing that can be used for internal fencing to partition the park into congruent, square garden plots. The entire park must be partitioned, and the sides of the squares must be parallel to the edges of the park. What is the larg... | Let each square plot have a side length of $s$ meters. Since the plots are square and the entire area must be completely covered without overlaps or gaps:
- The number of squares along the width ($30$ meters) is $\frac{30}{s}$
- The number of squares along the length ($60$ meters) is $\frac{60}{s}$
The number of verti... | 2 | Geometry | math-word-problem | Yes | Problem not solved | synthetic_math | true |
Two points $A$ and $B$ move with uniform speed on the plane along lines $a$ and $b$, respectively. The points $B_{1}$ and $B_{2}$ on line $b$ corresponding to the positions $A_{1}$ and $A_{2}$ of point $A$ are given. Construct the corresponding points $A$ and $B$ for which the distance $AB$ is minimized. | I. solution: We assume that $a$ and $b$ are not parallel lines. Translate the line $a$ parallel to itself to the position $a^{\prime}$, such that two corresponding points, for example, $A_{1}$ and $B_{1}$, coincide (Fig. 1).
 with no common edge. Does the statement hold for six points? | null | null | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Putnam 1997 Problem A4 G is a group, not necessarily abelian. We write the operation as juxtapostion and the identity as 1. There is a function φ : G → G such that if abc = def = 1, then φ(a)φ(b)φ(c) = φ(d)φ(e)φ(f). Prove that there exists an element k ∈ G such that kφ(x) is a homomorphism. | Easy. x -1 1 x = 1 = x -1 x 1, so φ(x -1 )φ(1)φ(x) = φ(x -1 )φ(x)φ(1). Hence φ(1)φ(x) = φ(x)φ(1), so φ(1) commutes with all φ(x), and hence k = φ(1) -1 also commutes with all φ(x). x y (y -1 x -1 ) = 1 = (x y)(y -1 x -1 ) 1, so φ(x)φ(y)φ(y -1 x -1 ) = φ(xy)φ(y -1 x -1 )φ(1). But φ(1) commutes, so φ(xy)φ(y - 1 x -1 )φ(1... | proof | Algebra | proof | Yes | Yes | olympiads | false |
The average age of 40 fifth-graders is 10 years. The average age of 60 of their parents is 35 years. Additionally, there are 5 teachers with an average age of 45 years. What is the average age of all these fifth-graders, parents, and teachers combined? | First, calculate the total age of the fifth-graders:
\[ 40 \times 10 = 400 \, \text{years} \]
Next, calculate the total age of their parents:
\[ 60 \times 35 = 2100 \, \text{years} \]
Then, calculate the total age of the teachers:
\[ 5 \times 45 = 225 \, \text{years} \]
Now, sum all the ages:
\[ 400 + 2100 + 225 = 2... | 26 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Translate the given propositions:
(1) The negation of the statement "If $x > 2$, then $x > 3$";
(2) The negation of "For all $a \in (0, +\infty)$, the function $y = a^x$ is monotonically increasing in its domain";
(3) "π is a period of the function $y = \sin x$" or "2π is a period of the function $y = \sin 2x$";
(4) "$... | For (1), the negation of the statement "If $x > 2$, then $x > 3$" is "If $x \leq 2$, then $x \leq 3$," which is a true statement.
For (2), if $a = 1$, then $y = 1$ is a constant function, so the statement "For all $a \in (0, +\infty)$, the function $y = a^x$ is monotonically increasing in its domain" is false. Therefo... | (1), (2), (3) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Luca has a quadratic of the form $x^2 + bx - 36$, where $b$ is a specific negative number. He wants to rewrite this quadratic in the form $(x + n)^2 - 20$. Discover the value of $b$. | 1. Expand the target quadratic form:
\[
(x + n)^2 - 20 = x^2 + 2nx + n^2 - 20
\]
where $n^2 - 20$ must correspond to the constant term in Luca's original equation, and $2n$ is equal to $b$.
2. Equate the constants from both forms and solve for $n$:
\[
n^2 - 20 = -36 \implies n^2 = 16 \implies n = 4 \... | b = -8 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
The ages of Patrick and Michael are in the ratio of 3 : 5, and that of Michael and Monica are in the ratio of 3 : 4. If the sum of their ages is 88, what is the difference between the ages of Patrick and Monica? | Let's denote the ages of Patrick, Michael, and Monica as P, M, and Mo, respectively.
According to the problem, the ages of Patrick and Michael are in the ratio of 3:5, so we can write:
P/M = 3/5
P = (3/5)M (1)
The ages of Michael and Monica are in the ratio of 3:4, so we can write:
M/Mo = 3/4
Mo = (4/3)M (2)
The s... | 22 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
The focal length of the hyperbola $\dfrac{x^{2}}{10} - \dfrac{y^{2}}{2} = 1$ is _______. | Consider the standard form of a hyperbola's equation ${\dfrac {x^{2}}{a^{2}} - \dfrac {y^{2}}{b^{2}} = 1}$.
Given the hyperbola $\dfrac{x^{2}}{10} - \dfrac{y^{2}}{2} = 1$, we identify $a^{2} = 10$ and $b^{2} = 2$. Thus, we have $a = \sqrt{10}$ and $b = \sqrt{2}$.
For a hyperbola, the focal distance $c$ is found usi... | 4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
In the classic work "Sunzi Suanjing" from the Northern and Southern Dynasties period of China, there is a deep study of congruence arithmetic. Let $a$, $b$, and $m (m > 0)$ be integers. If $a$ and $b$ have the same remainder when divided by $m$, then $a$ and $b$ are said to be congruent modulo $m$, denoted as $a \equiv... | This problem mainly tests the application of the binomial theorem and the general term formula of binomial expansion, which is a basic problem. According to the problem, $a = (10 - 1)^{10}$. Expanding it according to the binomial theorem, we can find its remainder when divided by $10$. Combined with $a \equiv b \pmod{1... | 2011 | Number Theory | MCQ | Yes | Yes | cn_k12 | false |
Divide a circle evenly into 11 parts, and express 8 of those parts as a fraction. | To divide a circle evenly into 11 parts is equivalent to dividing the unit "1" evenly into 11 parts, where each part is
$$
\frac{1}{11}
$$
Taking 8 of these parts, we express it as a fraction
$$
\frac{8}{11}
$$
Therefore, the answer is:
$$
\boxed{\frac{8}{11}}
$$ | \frac{8}{11} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
If $r$ is a number such that $r^2-6r+5=0$, find $(r-3)^2$ | null | null | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Among the rational numbers $1$, $0$, $\frac{1}{2}$, $-2$, the negative number is ( ).
A: $1$
B: $0$
C: $-2$
D: $\frac{1}{2}$ | To identify the negative number among the given rational numbers $1$, $0$, $\frac{1}{2}$, $-2$, we compare each number to zero:
- $1 > 0$ (positive)
- $0 = 0$ (neither positive nor negative)
- $\frac{1}{2} > 0$ (positive)
- $-2 < 0$ (negative)
Therefore, the only negative number among the given options is $-2$.
Hen... | C | Number Theory | MCQ | Yes | Yes | cn_k12 | false |
Ms. Mary sold three properties, X, Y, and Z.
She sold property X for $40,000, which was 15% more than she paid for it, and incurred a $2,000 renovation cost. In addition, she had to pay a 5% commission on the sale of property X.
She sold property Y for $55,000, which was 30% less than she paid for it, and incurred a ... | Let's calculate the net gain or loss for each property individually and then sum them up to find the total net gain or loss.
**Property X:**
- Sale price: $40,000
- Renovation cost: $2,000
- Commission: 5% of $40,000 = $2,000
- Purchase price: Let's denote it as P. She sold it for 15% more, so $40,000 = P + 0.15P = 1.... | \$23,003.64 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
XXIX OM - I - Problem 8
Let $ (a_n) $ be a non-decreasing sequence of natural numbers such that the sequence $ \left(\frac{n}{a_n}\right) $ is unbounded. Prove that among the terms of the sequence $ \left(\frac{n}{a_n}\right) $ there are infinitely many integers. | The first term of the sequence $ \binom{n}{a_n} $ is not greater than $ 1 $, since $ a_1 \geq 1 $. The sequence is unbounded, and therefore for every natural number $ m $ there exists a natural number $ r $ such that
Indeed, it suffices to choose $ r $ so that $ \frac{r+1}{a_{r+1}} $ is the earliest term of the sequen... | proof | Number Theory | proof | Yes | Incomplete | olympiads | false |
Last year, Dallas was 3 times the age of his sister Darcy. This year, Darcy is twice as old as Dexter, who is 8 years old. Their brother Derek is 4 years older than Dallas. How old is Dallas now? | Let's start by figuring out Darcy's age this year. Since Darcy is twice as old as Dexter, who is 8 years old, Darcy must be:
2 * 8 = 16 years old this year.
Now, let's find out how old Dallas was last year when he was three times the age of Darcy. Since Darcy is 16 years old this year, she must have been:
16 - 1 = 1... | 46 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
1.006. $\frac{\left(13.75+9 \frac{1}{6}\right) \cdot 1.2}{\left(10.3-8 \frac{1}{2}\right) \cdot \frac{5}{9}}+\frac{\left(6.8-3 \frac{3}{5}\right) \cdot 5 \frac{5}{6}}{\left(3 \frac{2}{3}-3 \frac{1}{6}\right) \cdot 56}-27 \frac{1}{6}$. | ## Solution.
$$
\begin{aligned}
& \frac{\left(13.75+9 \frac{1}{6}\right) \cdot 1.2}{\left(10.3-8 \frac{1}{2}\right) \cdot \frac{5}{9}}+\frac{\left(6.8-3 \frac{3}{5}\right) \cdot 5 \frac{5}{6}}{\left(3 \frac{2}{3}-3 \frac{1}{6}\right) \cdot 56}-27 \frac{1}{6}=\frac{\left(\frac{55}{4}+\frac{55}{6}\right) \cdot \frac{6}{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If the function $f(x)$ satisfies $f(x-1)=x^2+1$, then $f(-1)=$ （ ）
A: 1
B: 2
C: 3
D: 4 | Given that the function $f(x)$ satisfies $f(x-1)=x^2+1$,
Let $x=0$,
Then $f(-1)=1$,
Therefore, the correct answer is: $\boxed{A}$
Based on the function $f(x)$ satisfying $f(x-1)=x^2+1$, by setting $x=0$, we can find the value of $f(-1)$.
This question tests the application of abstract functions and function e... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Robin's hair was 16 inches long. He cut off 11 inches. Then it again grew by 12 inches.. How long is his hair now? | Robin's hair was initially 16 inches long. After cutting off 11 inches, his hair would be:
16 inches - 11 inches = 5 inches long.
Then, his hair grew by 12 inches, so the new length would be:
5 inches + 12 inches = 17 inches.
Robin's hair is now $\boxed{17}$ inches long. | 17 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Let $f(x) = \sqrt{4x^2 - 4x^4}$. Let $A$ be the number of real numbers $x$ that satisfy
$$f(f(f(\dots f(x)\dots ))) = x,$$ where the function $f$ is applied to $x$ 2020 times. Compute $A \pmod {1000}$.
[i]Proposed by Timothy Qian[/i] | null | null | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
To ensure the secure transmission of information, there is a secret key cryptosystem, whose encryption and decryption principles are as follows:
Plaintext $$\xrightarrow{\text{Encryption Key}}$$ Ciphertext $$\xrightarrow{\text{Send}}$$ Ciphertext $$\xrightarrow{\text{Decryption Key}}$$ Plaintext
Now the encryption key ... | Let us denote the encryption key by the function $y=\log_a (x+2)$.
Given that when the plaintext $x=6$, the ciphertext $y=3$, we can find the base $a$ by solving for $a$ in this logarithmic equation:
$$ 3 = \log_a (6+2) $$
$$ 3 = \log_a 8 $$
Since $8$ can be represented as $2^3$, we can deduce that $a=2$.
Now, let's ... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
3. Cut a $6 \times 6$ square (see the figure) along the grid lines into 4 equal parts so that each part contains all the digits. Parts are considered equal if they can be perfectly superimposed on each other by rotation only.
| 2 | 9 | 9 | 8 | 7 | 3 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | 2 | 4 | 4 | 3 | 6... | Solution. See the figure
| 2 | 9 | 9 | 8 | 7 | 3 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | 2 | 4 | 4 | 3 | 6 |
| 7 | 5 | 1 | 1 | 5 | 6 |
| 8 | 5 | 1 | 1 | 5 | 9 |
| 8 | 3 | 4 | 4 | 2 | 8 |
| 3 | 9 | 6 | 7 | 7 | 2 |
Comment. A correct example is provided - 7 points. A correct figure is sufficient justificati... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Given $f(n)=\sin \frac{n\pi }{3}$, then the value of $f(1)+f(2)+f(3)+\cdots +f(2014)$ is equal to
A: $\sqrt{3}$
B: $\frac{\sqrt{3}}{2}$
C: $0$
D: $-\frac{\sqrt{3}}{2}$ | **Analysis**
This problem tests the simplification and evaluation of trigonometric functions, and the solution can be found by identifying the period.
**Solution**
We have: $f(1)+f(2)+f(3)+...+f(2014)=\sin \frac{\pi}{3}+\sin \frac{2\pi}{3}+\sin \frac{3\pi}{3}+\sin \frac{4\pi}{3}+\sin \frac{5\pi}{3}+\sin \frac{6\pi}{... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false |
On Wednesday, Joan spent 4 half-dollars playing pinball. The next day, she spent 14 half-dollars on pinball. What was the total amount Joan spent playing pinball? | On Wednesday, Joan spent 4 half-dollars, which is 4 * $0.50 = $2.00.
On the next day, she spent 14 half-dollars, which is 14 * $0.50 = $7.00.
The total amount Joan spent playing pinball is $2.00 + $7.00 = $\boxed{\$9.00}$ . | \$9.00 | Logic and Puzzles | math-word-problem | Yes | Yes | orca_math | true |
Given circles $⊙O_{1}$: $(x-1)^{2}+y^{2}=4$, and $⊙O_{2}$: $x^{2}+(y- \sqrt {3})^{2}=9$.
(1) Find the equation of the line containing the common chord.
(2) Calculate the length of the common chord. | (1) To find the equation of the line containing the common chord, subtract the equation of circle $⊙O_{1}$ from that of circle $⊙O_{2}$:
$$(x^{2}+(y- \sqrt {3})^{2}) - ((x-1)^{2}+y^{2}) = 9-4$$
Expand and simplify the expression:
$$(x^{2}+y^{2}-2 \sqrt {3}y+3)-(x^{2}-2x+1+y^{2})=5$$
$$2 \sqrt {3}y - 2x + 2 = 5$$
$... | \frac{3 \sqrt{7}}{2} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
Consider two sets, $A$ and $B$, defined as follows: $A = \{(x, y) \mid y = a|x|\}$ and $B = \{(x, y) \mid y = x + a\}$. If the intersection $A \cap B$ contains exactly two elements only, find the range of the real number $a$. | First, we examine different cases for the value of $a$.
Case 1: When $a=0$, we have $A = \{(0, 0)\}$ as the only element, since for all $x$, $a|x| = 0$. Hence, the intersection $A \cap B$ would at most contain one element. This conflicts with the condition that the intersection must have exactly two elements. So $a \n... | (-\infty, -1) \cup (1, +\infty) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
What is the largest integer with distinct digits such that no two of its digits sum to a perfect square? | null | null | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
7. (30 points) On the side $AC$ of triangle $ABC$, where $\angle ACB=45^{\circ}$, a point $K$ is marked such that $AK=2KC$. On the segment $BK$, there is a point $S$ such that $AS \perp BK$ and $\angle AKS=60^{\circ}$. Prove that $AS=BS$.
, find $|z|=$ \_\_\_\_\_\_. | Since $z=3+4i$,
we have $|z|= \sqrt{3^{2}+4^{2}}= \sqrt{25}=5$.
Therefore, the answer is $\boxed{5}$.
This solution directly uses the formula for the magnitude of a complex number.
This question tests the operation of multiplication and division in the algebraic form of complex numbers and the method of finding... | 5 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
3. There are $4 n$ pebbles, with weights $1,2, \cdots, 4 n$. Each pebble is painted one of $n$ colors, such that there are exactly four pebbles of each color. Prove: It is possible to divide these pebbles into two piles, simultaneously satisfying the following two conditions:
(1) The two piles of pebbles have the same ... | 3. Let the $n$ colors be $c_{1}, c_{2}, \cdots, c_{n}$, and the set of small stones be $\left\{p_{1}, p_{2}, \cdots, p_{4 n}\right\}$, where $p_{i}$ represents a small stone with weight $i$. We construct a graph $G=(V, E)$, where the vertex set $V=\left\{c_{1}, c_{2}, \cdots, c_{n}\right\}$, and the edge set is defined... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
Determine the number of vertical asymptotes for the function \( y = \frac{x+2}{x^2 - 4} \). | First, factor the denominator:
\[ y = \frac{x+2}{x^2-4} = \frac{x+2}{(x-2)(x+2)} \]
The expression simplifies to:
\[ y = \frac{1}{x-2} \quad \text{for} \quad x \neq -2 \]
Now, identify the values of \( x \) that make the denominator zero:
1. \( x - 2 = 0 \) gives \( x = 2 \)
2. \( x + 2 = 0 \) gives \( x = -2 \)
How... | 1 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
The first three terms of an arithmetic sequence $\{a_n\}$ are $a-1$, $a+1$, and $2a+3$, respectively. The $n$-th term $a_n$ of this sequence is ( )
A: $2n-5$
B: $2n-3$
C: $2n-1$
D: $2n+1$ | Since the first three terms of the arithmetic sequence $\{a_n\}$ are $a-1$, $a+1$, and $2a+3$, we have $2(a+1) = (a-1) + (2a+3)$. Solving this equation, we get $a=0$.
Therefore, the first three terms of the arithmetic sequence $\{a_n\}$ are $-1$, $1$, and $3$,
which means the first term is $-1$ and the common diffe... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Let $p$ be a prime number and $q$ a prime divisor of the number $1+p+\ldots+p^{p-1}$. Show that $q \equiv 1(\bmod p)$. | First, let's recall the following factorization:
$$
a^{n}-1=(a-1)\left(a^{n-1}+a^{n-2}+\cdots+a^{2}+a+1\right)
$$
Thus,
$$
\begin{gathered}
q \mid 1+p+\cdots+p^{p-1} \\
\Rightarrow q \mid(p-1)\left(1+p+\cdots+p^{p-1}\right) \\
\Longrightarrow q \mid p^{p}-1
\end{gathered}
$$
We therefore have $p^{p} \equiv 1(\bmod ... | q\equiv1(\bmodp) | Number Theory | proof | Yes | Yes | olympiads | false |
Given that $i$ is the imaginary unit, find the value of $i^{2018}=\_\_\_\_\_\_\_\_.$ | **Analysis**
The value can be obtained by considering the periodicity of the powers of $i$.
**Enhanced Step-by-Step Solution**
Step 1: Recognize the pattern of the powers of $i$.
We know that $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$. This shows that the powers of $i$ cycle through the following values: $i, -1, -i, 1$.... | -1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Yanni has 5 paintings. 3 of the paintings are 5 feet by 5 feet. 1 painting is 10 feet by 8 feet. The final painting is 5 feet tall and 9 feet wide. What is the total area occupied by all the paintings combined? | To find the total area occupied by all the paintings combined, we need to calculate the area of each painting and then sum them up.
For the three paintings that are 5 feet by 5 feet:
Area of each painting = 5 feet * 5 feet = 25 square feet
Since there are three such paintings, the total area for these three paintings ... | 200 | Geometry | math-word-problem | Yes | Yes | orca_math | true |
5. If the two roots of the equation $x^{2}-2 x+\frac{\sqrt{3}}{2}=0$ are $\alpha, \beta$, and they are also the roots of the equation $x^{4}+p x^{2}+q=0$, then $p=$ | $5 . \sqrt{3}-4$ | \sqrt{3}-4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
nina has exactly enough money to purchase 5 widgets . if the cost of each widget were reduced by $ 1.25 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | Let's denote the original cost of each widget as \( C \) dollars. Nina has enough money to purchase 5 widgets, so the total amount of money she has is \( 5C \).
If the cost of each widget is reduced by $1.25, the new cost of each widget would be \( C - 1.25 \) dollars. With the reduced price, Nina would have enough mo... | \$16.66\overline{6} | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A distributor sells a product through an online store, which takes a commission of 20% of the price set by the distributor. The distributor obtains the product from a producer at the price of $19 per item. The price that the buyer observes online, if the distributor wants to maintain a certain profit on the cost of the... | To find the profit percentage the distributor wants to maintain, we first need to calculate the actual profit per item in dollars, and then express that as a percentage of the cost price.
Let's denote:
- \( P \) as the price the buyer observes online.
- \( C \) as the commission rate (20% or 0.2).
- \( S \) as the pri... | 87.5\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
The line $l$ passes through the point $P(1,1)$ and is tangent to the curve $C$: $y=x^{3}$. If the line $l$ does not pass through the fourth quadrant, what is the equation of the line $l$? | Let the point of tangency be $(x\_0, y\_0)$. From the problem, we have $y=3x^{2}$ and $y\_0=x\_0^{3}$.
The slope of the tangent line is $k=3x\_0^{2}$.
So, the equation of the tangent line is: $y-x\_0^{3}=3x\_0^{2}(x-x\_0)$, $(①)$
Given that the tangent line passes through the point $(1,1)$, we have $1-x\_0^{3}=3x\_0... | 3x-4y+1=0 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
Three, Buzhi:
$$
\left\{\begin{array}{c}
1988(x-y)+1989(y-z)+1990(z-x) \\
=0, \\
1988^{2}(x-y)+1989^{2}(y-z) \\
+1990^{2}(z-x)=1989,
\end{array}\right.
$$
Find \( z-y \). | Three, $z-y=1989$.
翻译结果：
Three, $z-y=1989$. | z-y=1989 | Algebra | math-word-problem | Yes | Incomplete | cn_contest | false |
Some boys went down the slide. Thirteen more boys went down the slide. In total, 35 boys went down the slide. How many boys went down the slide initially? | Let's call the number of boys who initially went down the slide "B". According to the information given:
B + 13 = 35
To find out how many boys went down the slide initially, we need to subtract 13 from 35:
35 - 13 = 22
So, initially, $\boxed{22}$ boys went down the slide. | 22 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Find the distance between the vertices of the hyperbola
\[\frac{x^2}{144} - \frac{y^2}{64} = 1.\] | From the equation of the hyperbola,
\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]
we identify that \( a^2 = 144 \).
Calculating \( a \):
\[ a = \sqrt{144} = 12. \]
Therefore, the distance between the vertices, which is \( 2a \), is:
\[ 2a = 2 \times 12 = 24. \]
Thus, the distance between the vertices is \( \boxed{24} ... | 24 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Find the angle between 0° and 360° whose terminal side is the same as that of -100°. | To find an angle between 0° and 360° that shares the same terminal side as -100°, we can express the general solution for any angle $\alpha$ with the same terminal side as -100° using the formula:
$$ \alpha = k \times 360° - 100°, \quad k \in \mathbb{Z} $$
Now we need to choose an integer $k$ such that the angle $\alp... | 260° | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
James writes a 3-page letter to 2 different friends twice a week. How many pages does he write a year? | James writes a 3-page letter to each of 2 friends, which totals 3 pages x 2 friends = 6 pages per writing session.
Since he writes twice a week, he writes 6 pages x 2 times = 12 pages per week.
There are 52 weeks in a year, so over the course of a year, he writes 12 pages x 52 weeks = $\boxed{624}$ pages. | 624 | Other | math-word-problem | Yes | Yes | orca_math | true |
A regular hexagon and an equilateral triangle have equal perimeter.
If the area of the triangle is $4\sqrt3$ square units, the area of the hexagon is
(A): $5\sqrt3$, (B): $6\sqrt3$, (C): $7\sqrt3$, (D): $8\sqrt3$, (E): None of the above. | null | null | Geometry | MCQ | Yes | Yes | aops_forum | false |
Mark, who lives in the United States, buys a Magic card for $100 USD. He is subject to a 5% sales tax on the initial purchase. The card then triples in value, but Mark wants to sell it to a friend in Europe. On the day he purchased the card, the exchange rate was 1 USD = 0.9 EUR. On the day he sells the card, the excha... | Let's break down the problem step by step:
1. Mark buys the Magic card for $100 USD and pays a 5% sales tax.
Sales tax = 100 * 0.05 = $5
Total initial cost = 100 + 5 = $105
2. The card triples in value.
New value of the card = 100 * 3 = $300
3. Mark sells the card to a friend in Europe when the exchange rat... | \$158.10 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Example 22 A circle is tangent to the line $2 x-y+5=0$ at point $M(-2,1)$, and passes through point $N(3,2)$. Find the equation of this circle. | Solve: Represent point $M$ in the form of a point circle: $(x+2)^{2}+(y-1)^{2}=0$. Let the equation of the required circle be $(x+2)^{2}+(y-1)^{2}+\lambda(2 x-y+5)=0$.
Substitute the coordinates of point $N$ into the above equation, and solve to get $\lambda=-\frac{26}{9}$.
Thus, the equation of the required circle is ... | 9x^{2}+9y^{2}-16x+8y-85=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 4.1.7 Let $a, b, c>0$, prove: $\sqrt{\frac{a}{4 a+4 b+c}}+\sqrt{\frac{b}{4 b+4 c+a}}+\sqrt{\frac{c}{4 c+4 a+b}} \leq 1$ | Proof: Note that the function $f(x)=\sqrt{x}$ is a convex function, so by Jensen's inequality, we have
$$\begin{array}{l}
\sum_{c y c} \sqrt{\frac{a}{4 a+4 b+c}}=\sum_{c y c} \frac{(4 a+4 b+c)}{9(a+b+c)} \cdot \sqrt{\frac{81 a(a+b+c)^{2}}{(4 a+4 b+c)(4 a+4 c+b)^{2}}} \\
\leq \sqrt{\sum_{c y c} \frac{(4 a+4 b+c)}{9(a+b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false |
It is now 3:00:00 PM, as read on a 12-hour digital clock. In 317 hours, 58 minutes, and 33 seconds, the time will be $X:Y:Z$. What is the value of $X + Y + Z$? | 1. **Identify the number of complete 12-hour cycles in 317 hours**:
- Each day has two 12-hour cycles (AM and PM).
- First, calculate how many full 12-hour cycles are covered by 317 hours:
\[
\text{Full cycles} = \left\lfloor \frac{317}{12} \right\rfloor = 26 \text{ cycles}
\]
- Hours after thes... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | synthetic_math | true |
The analytical expression of the function of the distance $y$ (unit: $m$) that the plane travels after landing with respect to the time $t$ (unit: $s$) is $y=60t-\frac{6}{5}t^{2}$. The total distance the plane travels from landing to stopping is ______ $m$. | To solve for the total distance the plane travels from landing to stopping, we start with the given function of distance with respect to time:
$$y = 60t - \frac{6}{5}t^2.$$
We can rewrite this equation by completing the square to make it easier to identify the maximum value of $y$:
\begin{align*}
y &= 60t - \frac{6}... | 750 \, \text{m} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Barbara went shopping in a supermarket. She bought 5 packs of tuna for $2 each and 4 bottles of water for $1.5 each. She also received a 10% discount on the total amount of her purchases. After the discount, she paid $56 for her shopping. How much did Barbara spend on different goods than the mentioned ones before the ... | First, let's calculate the total cost of the tuna and water before the discount.
The cost of the tuna is:
5 packs * $2/pack = $10
The cost of the water is:
4 bottles * $1.5/bottle = $6
The total cost of the tuna and water before the discount is:
$10 (tuna) + $6 (water) = $16
Now, let's calculate the total amount Ba... | \$46.22 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
All values of the quadratic trinomial $a x^{2}+b x+c$ on the interval $[0,1]$ do not exceed 1 in absolute value.
What is the maximum value that the quantity $|a|+|b|+|c|$ can have in this cas... | Let $f(x)=a x^{2}+b x+c$. By the condition $|f(0)| \leq 1,\left|f\left(\frac{1}{2}\right)\right| \leq 1,|f(1)| \leq 1$, that is $|c| \leq 1,|a+2 b+4 c| \leq 4, \mid a+b+$ $c \mid \leq 1$. Therefore,
$$
|a|=|2(a+b+c)-(a+2 b+4 c)+2 c| \leq 2|a+b+c|+|a+2 b+4 c|+2|c| \leq 8
$$
$$
|b|=|(a+2 b+4 c)-3 c-(a+b+c)| \leq|a+2 b+... | 17 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
and $b$ be natural numbers such that
$$
\frac{3 a^{2}+b}{3 a b+a}
$$
is an integer. Determine all the values that the above expression can take.
## Solution: | $3 a b+a=a(3 b+1)$ divides $3 a^{2}+b$, so in particular $a$ is a divisor of $3 a^{2}+b$. Now $a$ also divides $3 a^{2}+b-(3 a) a=b$. We can therefore write $b=a s$. Substituting into the expression yields $3 a^{2} s+a \mid 3 a^{2}+a s$ and after truncating with $a$ we get $3 a s+1 \mid 3 a+s$. Since $a$ and $s$ are na... | 1 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
Chad has 100 apples and each apple has different sizes and different price ranges. Each small apple cost x, medium apple cost $2, and big apples cost $3. If Donny bought 6 small and medium apples and also 8 big apples, how much will Donny have to pay for all of it?
If we know the answer to the above question is 45, wha... | To determine how much Donny will have to pay for all the apples, we need to calculate the total cost of the small and medium apples and the big apples separately, and then sum these amounts.
First, let's denote:
- \( x \) as the cost of one small apple.
- \( 2 \) as the cost of one medium apple.
- \( 3 \) as the cost ... | 3.5 | Algebra | math-word-problem | Yes | Yes | metamath | true |
Through vertices $A$ and $C$ of triangle $ABC$, a circle $K$ is drawn, the center of which lies on the circumcircle of triangle $ABC$. Circle $K$ intersects side $AB$ at point $M$. Find the a... | Let $Q$ be the center of the circle $K, AB=7, AM=2, MN$ and $CK$ be the altitudes of triangle $AMC$. Then $MN=4, BN=3$,
$\angle AMC = \frac{1}{2} \angle AQC = \frac{1}{2}(180^{\circ} - \angle B) = 90^{\circ} - \frac{1}{2} \angle B$.
Since $QA = QC$, $BQ$ is the bisector of angle $B$. Therefore, $\angle QBM = \frac{1}... | 45 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
21. If $a, b, c$ are positive real numbers, prove:
$$3(a+\sqrt{a b} \sqrt[3]{a b c}) \leqslant\left(8+\frac{2 \sqrt{a b}}{a+b}\right) \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}$$
(Generalization of Kiran - Kellaya Inequality) | 21. By Hölder's inequality, we have
$$\begin{array}{l}
a+\sqrt{a b}+\sqrt[3]{a b c}=a^{\frac{1}{3}} a^{\frac{1}{3}} a^{\frac{1}{3}}+a^{\frac{1}{3}}(\sqrt{a b})^{\frac{1}{3}} b^{\frac{1}{3}}+a^{\frac{1}{3}} b^{\frac{1}{3}} c^{\frac{1}{3}} \leqslant \\
(a+a+a)^{\frac{1}{3}}(a+\sqrt{a b}+b)^{\frac{1}{3}}(a+b+c)^{\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false |
The trigonometric form of the complex number $z=\sqrt{2}-\sqrt{2}i$ is:
A: $2(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
B: $2(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$
C: $2(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$
D: $2(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})$ | To convert the complex number $z=\sqrt{2}-\sqrt{2}i$ into its trigonometric form, we follow these steps:
1. Express $z$ in terms of its real and imaginary parts:
$$z = \sqrt{2} - \sqrt{2}i$$
2. Factor out the modulus of $z$:
$$z = \sqrt{2}(1 - i) = \sqrt{2}\left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right)$$... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
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