instruction_seed stringlengths 2 10.7k | solution stringlengths 0 10.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
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Randomly arrange students \\(A\\), \\(B\\), \\(C\\), and \\(D\\) from left to right. What is the probability that \\(A\\) and \\(B\\) are adjacent and there is exactly one student between \\(A\\) and \\(C\\)?
A: \\( \dfrac {1}{2}\\)
B: \\( \dfrac {1}{4}\\)
C: \\( \dfrac {1}{6}\\)
D: \\( \dfrac {1}{8}\\) | Since we randomly arrange students \\(A\\), \\(B\\), \\(C\\), and \\(D\\) from left to right,
the total number of basic events is \\(n= A_{4}^{4}=4×3×2×1=16\\).
The basic events that include "A and B are adjacent and there is exactly one student between A and C" are:
\\(ABCD\\), \\(CBAD\\), \\(CDAB\\), \\(BADC\\)... | B | Combinatorics | MCQ | Yes | Yes | cn_k12 | false |
Given the function $f(x)= \begin{cases} x^2+x, & -2\leqslant x\leqslant c \\ \frac{1}{x}, & c < x\leqslant 3 \end{cases}$, if $c=0$, then the range of $f(x)$ is _______; if the range of $f(x)$ is $[-\frac{1}{4},2]$, then the range of the real number $c$ is _______. | When $c=0$, we have $f(x)=x^2+x=(x+\frac{1}{2})^2-\frac{1}{4}$. Since the function $f(x)$ is decreasing on the interval $[-2,-\frac{1}{2})$ and increasing on the interval $(-\frac{1}{2},0]$, we can deduce that $f(x)$ attains its maximum value of $2$ at $f(-2)$, and its minimum value at $-\frac{1}{4}$ located at $x = -\... | [\frac{1}{2},1] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given that the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $60^{\circ}$, $|\overrightarrow{b}|=4$, and $(\overrightarrow{a}+2\overrightarrow{b})\cdot(\overrightarrow{a}-3\overrightarrow{b})=-72$, find the magnitude of vector $\overrightarrow{a}$.
A: $2$
B: $4$
C: $6$
D: $12$ | Since the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $60^{\circ}$, and $|\overrightarrow{b}|=4$, we have:
$(\overrightarrow{a}+2\overrightarrow{b})\cdot(\overrightarrow{a}-3\overrightarrow{b})=|\overrightarrow{a}|^{2}-|\overrightarrow{a}||\overrightarrow{b}|\cos 60^{\circ}-6|\overrightarrow... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
In triangle $ABC$ we inscribe a square such that one of the sides of the square lies on the side $AC$, and the other two vertices lie on sides $AB$ and $BC$. Furthermore we know that $AC = 5$, $BC = 4$ and $AB = 3$. This square cuts out three smaller triangles from $\vartriangle ABC$. Express the sum of reciprocals of ... | null | null | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The sixty-seventh problem: As shown in the figure, $\triangle A B C$ is inscribed in $\odot O, I$ is the incenter of $\triangle A B C$, $J$ is the $A$-excenter of $\triangle A B C$, $A J$ intersects $\odot O$ at $K$, the perpendicular bisector of $A J$ intersects $B C$ at $D$, $A D$ intersects $\odot O$ at $E$, $M$ is ... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | notfound | Geometry | proof | Yes | Incomplete | olympiads | false |
If the length of a [diagonal](https://artofproblemsolving.com/wiki/index.php/Diagonal) of a [square](https://artofproblemsolving.com/wiki/index.php/Square) is $a + b$, then the area of the square is:
$\mathrm{(A) \ (a+b)^2 } \qquad \mathrm{(B) \ \frac{1}{2}(a+b)^2 } \qquad \mathrm{(C) \ a^2+b^2 } \qquad \mathrm{(D) \ \... | Let a side be $s$; then by the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $s^2 + s^2 = 2s^2 = (a+b)^2$. The area of a square is $s^2 = \frac{(a+b)^2}{2} \Rightarrow \mathrm{(B)}$.
Alternatively, using the area formula for a [kite](https://artofproblemsolving.com/wiki/inde... | \frac{1}{2}(+b)^2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Tara has been planning to buy a laptop which costs $1000. A computer shop accepts payment in installments of $65 per month provided that a 20% down payment is made. Tara wants to pay a certain additional amount for the down payment. Her balance after paying for 4 months is $520. How much additional amount did Tara pay ... | First, let's calculate the 20% down payment that Tara needs to make on the $1000 laptop.
20% of $1000 is:
0.20 * $1000 = $200
Now, let's calculate the total amount Tara would have paid after 4 months of installments without any additional down payment.
Each monthly installment is $65, so for 4 months, she would have... | \$20 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Given the sets $P=\{0,1,2\}$ and $Q=\{y|y=3^x\}$, then $P\cap Q=$ （ ）
A: $\{0,1\}$
B: $\{1,2\}$
C: $\{0,1,2\}$
D: $\emptyset$ | For $Q=\{y|y=3^x\}=\{y|y>0\}$,
then $P\cap Q=\{1,2\}$,
therefore, the correct choice is: $\boxed{\text{B}}$
**Analysis:** This problem can be solved by applying the basic operations of sets. | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Let $B := \mathbb{Q} \setminus \{0, 1, 2\}$ denote the set of all rationals other than 0, 1, and 2. A function $f : B \rightarrow \mathbb{R}$ has the property that for all $x \in B$,
\[
f\left( x\right) + f\left( 2 - \frac{1}{x}\right) = \log\lvert x\rvert,
\]
and additionally,
\[
f\left( x\right) + f\left( 1 + \frac{1... | Define $g: B \to B$ by $g(x) := 2 - \frac{1}{x}$ and $h: B \to B$ by $h(x) := 1 + \frac{1}{x}$. The key properties are:
- $g(g(g(x))) = x$
- $h(h(h(x))) = x$
From the first function property:
\[ f(x) + f(g(x)) = \log |x| \]
\[ f(g(x)) + f(g(g(x))) = \log |g(x)| \]
\[ f(g(g(x))) + f(x) = \log |g(g(x))| \]
From the sec... | \log 3 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Given $\alpha \in \left( \frac{\pi}{2}, \pi \right)$, the maximum value of the function $f(x) = (\sin \alpha)^{x^2 - 2x + 3}$ is $\frac{1}{4}$. Find $\alpha$. | Since $\alpha \in \left( \frac{\pi}{2}, \pi \right)$,
it follows that $\sin \alpha \in (0, 1)$,
thus $y = (\sin \alpha)^x$ is a decreasing function in its domain.
Because $x^2 - 2x + 3 \geq 2$,
we have $(\sin \alpha)^2 = \frac{1}{4}$,
therefore $\sin \alpha = \frac{1}{2}$,
hence $\alpha = \frac{5\pi}{6}$. ... | \frac{5\pi}{6} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
The length of a rectangular plot is 10 meters more than its breadth. The cost of fencing the plot at 26.50 per meter is a certain amount. The length of the plot is 55 meters. What is the total cost of fencing the plot? | Given that the length of the plot is 55 meters and it is 10 meters more than its breadth, we can find the breadth by subtracting 10 meters from the length.
Breadth = Length - 10 meters
Breadth = 55 meters - 10 meters
Breadth = 45 meters
Now that we have both the length and the breadth, we can calculate the perimeter ... | 5300 | Geometry | math-word-problem | Yes | Yes | orca_math | true |
A can finish a work in 15 days, B can do the same work in 20 days, and C can do the same work in 30 days. A worked for 10 days and left the job, then B worked for another 5 days and left the job. In how many days can C alone finish the remaining work? | First, let's find out how much work A, B, and C can do in one day.
A can finish the work in 15 days, so A's work rate is 1/15 of the work per day.
B can finish the work in 20 days, so B's work rate is 1/20 of the work per day.
C can finish the work in 30 days, so C's work rate is 1/30 of the work per day.
Now, let's ... | 2.5 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
The scale of a map is 1:3,000,000. The distance from City A to City B is measured to be 6 cm. A car travels from City A to City B at a speed of 30 km per hour. How many hours will it take to reach City B? | Given the scale and the distance on the map, we can find the actual distance using the formula "actual distance = map distance ÷ scale". Then, we can find the time it takes to travel this distance using the formula "time = distance ÷ speed". Let's solve this step by step.
This problem primarily tests the understanding... | 6,000\ \text{hours} | Algebra | math-word-problem | Yes | Incomplete | cn_k12 | false |
38. Find the limits of the following functions:
1) $f(x)=2 x-3-\frac{1}{x}$ as $x \rightarrow 1$:
2) $y=\frac{x^{3}-3 x^{2}+2 x-5}{x^{2}+2}$ as $x \rightarrow-1$;
3) $y=x \sin \frac{1}{x}$ as $x \rightarrow 0$. | Solution. Using the indicated theorems, we sequentially find:
1) $\lim _{x \rightarrow 1}\left(2 x-3-\frac{1}{x}\right)=\lim 2 \cdot \lim x-\lim 3-\frac{\lim 1}{\lim x}=$ $=2 \cdot 1-3-\frac{1}{1}=-2$
2) $\lim _{x \rightarrow-1} \frac{x^{3}-3 x^{2}+2 x-5}{x^{2}+2}=\frac{(\lim x)^{3}-3(\lim x)^{2}+2 \lim x-5}{(\lim x)^... | -2,-\frac{11}{3},0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Given a sequence $\{a_n\}$ with the sum of its first $n$ terms denoted as $S_n$, $a_1=1$, and the point $P(a_n,S_n)$ (where $n \geqslant 1$ and $n \in \mathbb{N}^*$) lies on the line $4x-3y-1=0$. The sequence $\left\{\frac{1}{{b_n}}\right\}$ is an arithmetic sequence with the first term $-1$ and common difference $-2$.... | $(1)$ Solution: Since the point $P(a_n,S_n)$ lies on the line $4x-3y-1=0$,
$\therefore 4a_n-3S_n-1=0$, which means $3S_n=4a_n-1$. Also, for $n \geqslant 2$, $3S_{n-1}=4a_{n-1}-1$. Subtracting these two equations gives $a_n=4a_{n-1}$,
$\therefore \frac{a_n}{a_{n-1}}=4$ (for $n \geqslant 2$), $\therefore \{a_n\}$ is a ... | T_n=- \frac{20}{9}+ \frac{6n+5}{9\times 4^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Five. (15 points) In a math competition for second-year junior high school students, a total of 99 middle schools registered to participate, with both male and female participants from each school. Prove: there exist 50 of these schools whose total number of male participants is not less than half of the total number o... | Five, number the participating middle schools as $1,2, \cdots, 99$, and let $x_{i}$ $(i=1,2, \cdots, 99)$ represent the number of male participants from the $i$-th school. Assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{99}$.
Divide the schools numbered $2 \sim 99$ into two groups.
The first group consist... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false |
A company produces some bottles of water every day. If a case can hold 10 bottles of water and the company requires 7,200 cases to hold its one day production, how many bottles of water does the company produce every day? | If one case holds 10 bottles of water and the company requires 7,200 cases to hold its one day production, then the total number of bottles produced in one day is:
7,200 cases * 10 bottles/case = 72,000 bottles
Therefore, the company produces $\boxed{72,000}$ bottles of water every day. | 72,000 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
(6) For any natural number $n$, let $r_{s}(n)$ denote the number of integer solutions $x_{1}, x_{2}, \cdots, x_{s}$ to the equation
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x^{2}=n
$$
and let $f_{s}(n)=\frac{r_{s}(n)}{2 s}$. It is known that when $s=$ $1,2,4,8$, the function $f_{s}(n)$ satisfies the multiplicative rule. That is,... | It can be proven that when $s \neq 1,2,4,8$, $f_{s}(2) f_{s}(3) \neq f_{s}(6)$. Note that if $x_{1}, x_{2}, \cdots, x_{s}$ are integer solutions to the equation
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{s}^{2}=2
$$
then $s-2$ of these solutions are 0, and the remaining two solutions are $\pm 1$. Therefore,
$$
r_{s}(2)=4 \mathr... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
A person puts 3 yellow and 3 white ping-pong balls of the same size into a box for people to draw. The rule is: if someone draws 3 balls of the same color, they will be given 5 yuan, but if they draw 3 balls of different colors, they have to pay 1 yuan. Assuming there are 100 people drawing balls per day, estimate from... | To solve this problem, we first need to calculate the probability of drawing 3 balls of the same color and the probability of drawing 3 balls of different colors.
There are a total of $\binom{6}{3} = 20$ ways to draw 3 balls out of 6. Among these, there are 2 ways to draw 3 balls of the same color (either all 3 are ye... | \text{D: 14400} | Combinatorics | MCQ | Yes | Yes | cn_k12 | false |
10. Vanya, Kostya, and Lesha are playing a game: there are 2008 matches on the table. In one move, Vanya and Lesha can take 1 or 2 matches, Kostya - 1, 2, or 3. Vanya moves first, Kostya second, and Lesha third. The player who takes the last match wins. Which two players can combine their efforts against the third to p... | Solution. We will show that any team with Kostya always wins. Note that any two players together can take any number of matches from 2 to 4 in one round (each moves once). If the third can only take 1 or 2, then the two who have teamed up can ensure taking 5 matches in one round. Then, if Vanya and Kostya team up, Vany... | proof | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
John believes that the amount of sleep he gets the night before a test and his score on that test are inversely related. On his first exam, he got 6 hours of sleep and scored 80 on the exam. To the nearest tenth, how many hours does John believe he must sleep the night before his second exam so that the average of his ... | First, to average an 85 on two exams, John must score a total of $85 \times 2 = 170$ across the two exams. Given he scored 80 on his first exam, he must score $170 - 80 = 90$ on his second exam.
Now, since the score and hours of sleep are inversely related, their product is a constant. Thus, we set up the relationship... | 5.3 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
The gross production value of a certain city has been continuously increasing for two consecutive years. The growth rate for the first year is $p$, and for the second year is $q$. The annual average growth rate of the city's gross production value over these two years is ______. | Let's denote the annual average growth rate of the city's gross production value over these two years as $x$. According to the problem, $(1+x)^2 = (1+p)(1+q)$,
Therefore, $x = \sqrt{(1+p)(1+q)} - 1$;
Hence, the answer is: $\boxed{\sqrt{(1+p)(1+q)} - 1}$.
To find the annual average growth rate $x$, we simply solve th... | \sqrt{(1+p)(1+q)} - 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
If the complex number $m^2(1+i) + (m+i)i^2$ is purely imaginary, then the value of the real number $m$ is
A: 0
B: 1
C: -1
D: 0 or 1 | Let's expand the given complex number:
$$
\begin{align*}
m^2(1+i) + (m+i)i^2 &= m^2 + m^2i + mi(-1) + i^2 \\
&= m^2 + m^2i - mi - 1 \\
&= (m^2 - m) + (m^2 - 1)i.
\end{align*}
$$
For this number to be purely imaginary, the real part must be equal to zero. Therefore, we must have:
$$
\begin{align*}
m^2 - m &= 0 \\
\en... | A: 0 | Algebra | MCQ | Yes | Yes | cn_k12 | false |
## Problem 3
$\mathrm{D}$ is a point inside the equilateral triangle $\mathrm{ABC}$. $\mathrm{E}$ is a point inside $\mathrm{DBC}$. Show that area $\mathrm{DBC} /($ perimeter $\mathrm{DBC})^{2}>$ area $\mathrm{EBC} /(\text { perimeter } \mathrm{EBC})^{2}$.
| ## Solution
Let us find an expression for $\mathrm{t}=($ area $\mathrm{DBC}) /(\text { perimeter } \mathrm{DBC})^{2}$. Let angle $\mathrm{B}=2 \mathrm{x}$, angle $\mathrm{C}=2 \mathrm{y}$ and angle $\mathrm{D}$ $=2 z$ and let the inradius be $r$. Then area $D B C=r / 2 x$ perimeter $D B C$. Also $B C=r \cot x+r \cot y... | proof | Geometry | proof | Yes | Yes | olympiads | false |
Let the sum of the first $n$ terms of the sequence $\{a_n\}$ be $S_n$, and $S_n + a_n = 1$. The sequence $\{b_n\}$ is an arithmetic sequence, and $b_1 + b_2 = b_3 = 3$.
$(1)$ Find $S_n$;
$(2)$ Find the sum of the first $n$ terms of the sequence $\{a_n b_n\}$, denoted as $T_n$. | Solution:
$(1)$ Since $S_n + a_n = 1$,
When $n=1$, $S_1 + a_1 = 1$, we get $S_1 = a_1 = \frac{1}{2}$,
When $n \geq 2$, substituting $a_n = S_n - S_{n-1}$,
into $S_n + a_n = 1$ gives $2S_n - S_{n-1} = 1$,
Thus, $2(S_n - 1) = S_{n-1} - 1$,
Since $S_1 - 1 = -\frac{1}{2} \neq 0$,
Thus, $\{S_n - 1\}$ is a geometric s... | 2 - (n+2)(\frac{1}{2})^n | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Example 11. What is the probability of event $A$ occurring in each trial, if the most probable number of occurrences of event $A$ in 120 trials is 32? | Solution. Given $n=120, k_{0}=32$. Inequalities (3.1.5) take the form
$$
120 p-(1-p) \leq 32 \leq 120 p+p
$$
or
$$
120 p-(1-p) \leq 32, \quad 120 p+p \geq 32
$$
Solving this system of inequalities, we find that
$$
\frac{32}{121} \leq p \leq \frac{33}{121}
$$ | \frac{32}{121}\leqp\leq\frac{33}{121} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Simplify and calculate
(1) $$2 \sqrt {3} \times \sqrt[3]{1.5} \times \sqrt[6]{12}$$;
(2) $$\lg25 + \frac {2}{3}\lg8 + \lg5 \cdot \lg20 + (\lg2)^{2}$$. | Solution:
(1) $$2 \sqrt {3} \times \sqrt[3]{1.5} \times \sqrt[6]{12}$$
= $$2 \cdot 3^{\frac {1}{2}} \cdot \left( \frac {3}{2} \right)^{\frac {1}{3}} \cdot (2^{2} \cdot 3)^{\frac {1}{6}}$$
= $$2^{1 - \frac {1}{3} + \frac {1}{3}} \cdot 3^{\frac {1}{2} + \frac {1}{3} + \frac {1}{6}}$$
= $2 \cdot 3 = 6$;
(2) $$\l... | 3 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Let
\[\bold{B} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 3 & 2 & 1 \end{pmatrix}.\]
There exist constants $p'$, $q'$, and $r'$ such that
\[\bold{B}^3 + p' \bold{B}^2 + q' \bold{B} + r' \bold{I} = \bold{0},\]
where $\bold{I}$ and $\bold{0}$ are the $3 \times 3$ identity matrix and zero matrix, respectively. Find the o... | First calculate $\bold{B}^2$:
\[\bold{B}^2 = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 3 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 14 & 11 & 10 \\ 11 & 9 & 10 \\ 10 & 11 & 14 \end{pmatrix}.\]
Then calculate $\bold{B}^3$:
\[\bold{B}^3 = \begin{pmatrix} 1 &... | (-5, -8, -6) | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Zach baked cookies for a bake sale over a period of six days. On Monday he baked 32 cookies. On Tuesday he ran out of sugar and could only bake half of the number of cookies he baked the day before. On Wednesday, he baked three times the number of cookies he did on Tuesday, but his brother ate 4 of those cookies. On Th... | Let's calculate the number of cookies Zach baked each day and then subtract the ones that were eaten or destroyed.
Monday: Zach baked 32 cookies.
Tuesday: He baked half of what he baked on Monday, so that's 32 / 2 = 16 cookies.
Wednesday: He baked three times the number of cookies he did on Tuesday, so that's 16 * 3... | 242 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
From a group of 10 students (including 6 females and 4 males), 3 students are randomly selected to take a test. The probability of each female student passing the test is $\frac{4}{5}$, and the probability of each male student passing the test is $\frac{3}{5}$. Find:
(I) The probability that among the 3 selected stud... | Solution:
(I) According to the problem, this is a classical probability model.
Since all possible events are selecting 3 students from the 10 students (including 6 females and 4 males) to take the test, there are $C_{10}^{3}$ outcomes.
The event that meets the condition is selecting 3 students with at least one m... | \frac{4}{125} | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false |
Fred had some baseball cards. Melanie bought 3 of Fred's baseball cards, and now Fred has 2 baseball cards left. How many baseball cards did Fred have initially? |
If Fred has 2 baseball cards left after Melanie bought 3, then Fred initially had 2 + 3 = $\boxed{5}$ baseball cards. | 5 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Example 5 Solve the inequality $|x|+|x-1|$ $>a x(a \in \mathbf{R})$ with respect to $x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Analysis: Consider the functions $y=|x|+|x-1|$ and $y=a x$. As shown in Figure 6, since the real number $a$ is a variable parameter, the line $y=a x$ rotates around the origin. Different positions of the line can lead to the following conclusions:
(i) When $-2 \leqslant a < \frac{1}{2-a}$. | not found | Inequalities | math-word-problem | Yes | Incomplete | cn_contest | false |
Prove that when $x \in [0, 1]$, $$\frac{\sqrt{2}}{2}x \leq \sin x \leq x.$$ | Let $F(x) = \sin x - \frac{\sqrt{2}}{2}x$, then $F'(x) = \cos x - \frac{\sqrt{2}}{2}$.
When $x \in (0, \frac{\pi}{4})$, $F'(x) > 0$, so $F(x)$ is increasing on $[0, \frac{\pi}{4}]$;
When $x \in (\frac{\pi}{4}, 1)$, $F'(x) 0$, thus when $x \in [0, 1]$, $F(x) \geq 0$, which means $\sin x \geq \frac{\sqrt{2}}{2}x$. ... | \frac{\sqrt{2}}{2}x \leq \sin x \leq x | Inequalities | proof | Yes | Yes | cn_k12 | false |
Example 2 (2004 College Entrance Examination Hunan Paper) Farmers' income consists of wage income and other income. In 2003, the per capita income of farmers in a certain area was 3150 yuan (of which wage income was 1800 yuan, and other income was 1350 yuan). It is expected that in the 5 years from 2004, farmers' wage ... | Choose B. Reason: According to the problem, the per capita income of farmers in the region in 2008 is $1800 \times(1+6 \%)^{5}+1350+5 \times 160$, and an approximate calculation is made for the expansion of $(1+6 \%)^{5}$.
$$
\begin{array}{l}
1800 \times(1+6 \%)^{5}+1350+5 \times 160>1800 \times\left(1+5 \times 0.06+10... | 4400 | Algebra | MCQ | Yes | Yes | olympiads | false |
Let $G{}$ be an arbitrary finite group, and let $t_n(G)$ be the number of functions of the form \[f:G^n\to G,\quad f(x_1,x_2,\ldots,x_n)=a_0x_1a_1\cdots x_na_n\quad(a_0,\ldots,a_n\in G).\]Determine the limit of $t_n(G)^{1/n}$ as $n{}$ tends to infinity. | 1. **Define the problem and the function count**: We are given a finite group \( G \) and need to determine the limit of \( t_n(G)^{1/n} \) as \( n \) tends to infinity, where \( t_n(G) \) is the number of functions of the form:
\[
f: G^n \to G, \quad f(x_1, x_2, \ldots, x_n) = a_0 x_1 a_1 \cdots x_n a_n \quad (a... | \frac{|G|}{|Z(G)|} | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Zander collected 100 stickers. He gave some of his stickers to his x friends, Andrew and Bill. Andrew received 1/5 of Zander's stickers, while Bill received 3/10 of the remaining stickers. How many stickers did Andrew give to his two friends?
If we know the answer to the above question is 44, what is the value of unkno... | To determine how many stickers Zander gave to his friends and find the value of \( x \), let's follow these steps:
1. **Calculate the number of stickers Andrew received:**
Andrew received \(\frac{1}{5}\) of Zander's 100 stickers.
\[
\text{Stickers Andrew received} = \frac{1}{5} \times 100 = 20
\]
2. **Cal... | 0 | Algebra | math-word-problem | Yes | Yes | metamath | true |
50 is 26 greater than a number. What is the value of the number multiplied by 9? | Let's call the unknown number "x". According to the statement, 50 is 26 greater than x. So we can write the equation:
50 = x + 26
Now, we want to find the value of x. To do this, we subtract 26 from both sides of the equation:
50 - 26 = x
24 = x
Now we know the value of x is 24. The question asks for the value of t... | 216 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
In the technological hotspot city X, data from 1992 reveals that 60 percent of families owned a personal computer. The city, being a thriving tech-hub, saw a surge in technology utilization. In 1993, there was a 50 percent increase in families owning computers from the previous year. The overall number of families in c... | Let's break down the information year by year and calculate the percentages step by step.
**1992:**
- 60% of families owned a personal computer.
**1993:**
- There was a 50% increase in the number of families owning computers. This means that the number of families owning computers is now 150% of what it was in 1992.
... | 10.8\% | Logic and Puzzles | math-word-problem | Yes | Yes | orca_math | true |
Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ satisfy $|\overrightarrow{a}|=2$, $|\overrightarrow{b}|=1$, and $|\overrightarrow{a}-\overrightarrow{b}|=2$.
$(1)$ Find the value of $\overrightarrow{a}\cdot\overrightarrow{b}$;
$(2)$ Find the value of $|\overrightarrow{a}+\overrightarrow{b}|$;
$(3)$ Find... | Solution:
$(1)$ From $|\overrightarrow{a}|=2$, $|\overrightarrow{b}|=1$, and $|\overrightarrow{a}-\overrightarrow{b}|=2$, we get
$|\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2-2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2=4+1-2\overrightarrow{a}\cdot\overrightarrow{b}=4$,
$\... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Let $P$ and $Q$ be two polynomials with real coeficients such that $P$ has degree greater than $1$ and \[P(Q(x)) = P(P(x)) + P(x).\]Show that $P(-x) = P(x) + x$. | null | null | Algebra | proof | Yes | Yes | aops_forum | false |
1. On the side $B C$ of triangle $A B C$, a point $D$ is taken in such a way that the perpendicular bisector of segment $A D$ passes through the center of the inscribed circle of triangle $A B C$. Prove that this perpendicular bisector passes through the vertex of triangle $A B C$. (L. Emelyanov) | Solution. Drop a perpendicular $J N$ from the center $J$ of the inscribed circle of the triangle to the side $C B$. Suppose the point $N$ lies on the segment $D C$ (the case where the point $N$ lies on the segment $D B$ is considered similarly.). Drop a perpendicular $J M$ from the center $J$ to the side $C A$. Since $... | proof | Geometry | proof | Yes | Yes | olympiads | false |
Given sets $A=\{-2, 0, 2\}$ and $B=\{x|x^2-x-2=0\}$, then $A \cap B = \_\_\_\_\_\_$. | Since $A=\{-2, 0, 2\}$ and $B=\{x|x^2-x-2=0\}=\{-1, 2\}$,
Therefore, $A \cap B = \{2\}$.
Hence, the answer is $\boxed{\{2\}}$.
This problem is solved by using the definition of intersection.
It tests the method of finding intersections, which is a basic question. When solving, pay attention to the proper use of... | \{2\} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Here are two functions:
$$
\begin{array}{ccc}
f(x) & = & 4x^3 - 3x^2 + 2x + 5 \\
g(x) & = & x^3 - (k+1)x^2 - 7x - 8
\end{array}
$$
If \( f(5) - g(5) = 24 \), what is the value of \( k \)? | We have:
\[
f(x) - g(x) = (4x^3 - 3x^2 + 2x + 5) - (x^3 - (k+1)x^2 - 7x - 8) = 3x^3 + (k+4)x^2 + 9x + 13.
\]
Substituting \( x = 5 \), we get:
\[
f(5) - g(5) = 3 \times 5^3 + (k+4) \times 5^2 + 9 \times 5 + 13 = 24.
\]
Calculate:
\[
375 + 25(k + 4) + 45 + 13 = 24.
\]
Simplify:
\[
433 + 25k = 24 \implies 25k = 24 - 433 ... | -16.36 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
What is the product of the least common multiple and the greatest common factor of $30$ and $75$? | 1. **Finding the Greatest Common Factor (GCD)**:
To find the GCD of $30$ and $75$, we list the factors of each:
- Factors of $30$: $1, 2, 3, 5, 6, 10, 15, 30$
- Factors of $75$: $1, 3, 5, 15, 25, 75$
The greatest common factor is the highest number that appears in both lists, which is $15$.
2. **Finding th... | 2250 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Let $x, y, p, n$, and $k$ be positive integers such that $x^n + y^n = p^k$. Prove that if $n > 1$ is odd, and $p$ is an odd prime, then $n$ is a power of $p$.
[i]A. Kovaldji, V. Senderov[/i] | 1. Assume for contradiction that \( n \) is not a power of \( p \). Then we can write \( n = p^{\alpha} q \) where \( \gcd(p, q) = 1 \).
2. Consider the equation \( x^n + y^n = p^k \). Substituting \( n = p^{\alpha} q \), we get:
\[
x^{p^{\alpha} q} + y^{p^{\alpha} q} = p^k
\]
3. We can factorize \( x^{p^{\a... | null | Number Theory | proof | Yes | Yes | aops_forum | false |
A track and field team has 56 male athletes and 42 female athletes. If a stratified sampling method is used to draw a sample of 28 from all athletes, then the number of female athletes in the sample is ( )
A: 9
B: 10
C: 11
D: 12 | The probability of each individual being selected is $$\frac{28}{56+42} = \frac{2}{7}$$, so the number of female athletes in the sample is $42 \times \frac{2}{7} = 12$.
First, calculate the probability of each individual being selected, then multiply this probability by the number of female athletes to find the answer... | D: 12 | Combinatorics | MCQ | Yes | Yes | cn_k12 | false |
Evaluate
$$ \lim_{x\to \infty} \left( \frac{a^x -1}{x(a-1)} \right)^{1\slash x},$$
where $a>0$ and $a\ne 1.$ | To evaluate the limit
\[ \lim_{x\to \infty} \left( \frac{a^x -1}{x(a-1)} \right)^{1/x}, \]
where \( a > 0 \) and \( a \ne 1 \), we will consider two cases: \( a > 1 \) and \( 0 < a < 1 \).
### Case 1: \( a > 1 \)
1. Let \( L = \lim_{x\to \infty} \left( \frac{a^x -1}{x(a-1)} \right)^{1/x} \). Taking the natural logar... | 1 | Calculus | other | Yes | Yes | aops_forum | false |
Of 96 oranges, half were ripe. If 1/4 of the ripe oranges were eaten and 1/8 of the unripe oranges were eaten, how many oranges were left uneaten in total? | First, let's find out how many oranges were ripe and how many were unripe.
Half of the 96 oranges were ripe, so:
Ripe oranges = 96 / 2 = 48 oranges
The other half were unripe, so:
Unripe oranges = 96 / 2 = 48 oranges
Now, let's find out how many ripe oranges were eaten:
1/4 of the ripe oranges were eaten, so:
Eaten ... | 78 | Logic and Puzzles | math-word-problem | Yes | Yes | orca_math | true |
Jamshid can paint a fence in 50 percent less time than Taimour can when each works alone. When they work together, they can paint the fence in some hours. It would take Taimour 15 hours to paint the fence alone. How long does it take for Jamshid and Taimour to paint the fence together? | If Jamshid can paint the fence in 50 percent less time than Taimour, that means Jamshid takes half the time Taimour takes when working alone.
Since Taimour takes 15 hours to paint the fence alone, Jamshid would take 50% of 15 hours, which is 7.5 hours.
Now, let's calculate the rate at which they work together.
Taimo... | 5 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A regular octagon is inscribed in a circle with a radius of 3 units. Calculate the area of the regular octagon using exact values. | Since the regular octagon is inscribed in a circle with radius \( r = 3 \) units, the side length \( s \) of each side of the octagon can be determined using the relationship \( s = 2r \sin(\pi/8) \), based on the octagon being divided into 8 isosceles triangles at the center.
1. Calculation of the Side Length \( s \)... | 18\sqrt{2} | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
A rectangular parallelepiped has dimensions $a,b,c$ that satisfy the relation $3a+4b+10c=500$, and the length of the main diagonal $20\sqrt5$. Find the volume and the total area of the surface of the parallelepiped. | null | null | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
44. (INA 4)
(a) Let $g(x)=x^{5}+x^{4}+x^{3}+x^{2}+x+1$. What is the remainder when the polynomial $g\left(x^{12}\right)$ is divided by the polynomial $g(x)$ ?
(b) If $k$ is a positive integer and $f$ is a function such that for every positive number $x, f\left(x^{2}+1\right)^{\sqrt{x}}=k$, find the value of $f\left(\fr... | None | notfound | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
Let $ABC$ be an isosceles triangle with base $AC$. Points $D$ and $E$ are chosen on the sides $AC$ and $BC$, respectively, such that $CD = DE$. Let $H, J,$ and $K$ be the midpoints of $DE, AE,$ and $BD$, respectively. The circumcircle of triangle $DHK$ intersects $AD$ at point $F$, whereas the circumcircle of triangle ... | 1. **Identify the midpoints:**
- \( H \) is the midpoint of \( DE \), so \( DH = HE \).
- \( J \) is the midpoint of \( AE \), so \( AJ = JE \).
- \( K \) is the midpoint of \( BD \), so \( BK = DK \).
2. **Parallel lines and midpoints:**
- The line through \( K \) parallel to \( AC \) intersects \( AB \) ... | null | Geometry | proof | Yes | Yes | aops_forum | false |
How many square inches are in one square meter? (Note: 1 inch equals 0.0254 meters) | First, we know from the unit conversion that $1 \text{ inch} = 0.0254 \text{ meters}$. Therefore, to find the number of square inches in one square meter, we need to convert both dimensions of a square from meters to inches, requiring the use of the relationship squared.
We therefore square the conversion factor:
$$ (... | 1550 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
Let $a$ and $b$ be positive real numbers with $a \geq b$. Let $\rho$ be the maximum possible value of $\frac{a}{b}$ for which the system of equations
$$
a^2 + y^2 = b^2 + x^2 = (a + x)^2 + (b - y)^2
$$
has a solution in $(x, y)$ satisfying $0 \leq x < a$ and $0 \leq y < b$. Find $\rho^2$. | Expanding the given equations, we get:
\[ b^2 + x^2 = a^2 + 2ax + x^2 + b^2 - 2by + y^2. \]
Thus,
\[ a^2 + y^2 = -2ax + 2by. \]
Since \( 2by > 2y^2 \geq y^2 \), it follows that \( 2by - y^2 \geq 0 \). Therefore:
\[ a^2 - 2ax \geq 0 \Rightarrow a \geq 2x. \]
Next, we find:
\[ a^2 \leq a^2 + y^2 = b^2 + x^2 \leq b^2 + \f... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
If $x^{2} y z^{3}=7^{4}$ and $x y^{2}=7^{5}$, then $x y z$ equals
(A) 7
(B) $7^{2}$
(C) $7^{3}$
(D) $7^{8}$
(E) $7^{9}$ | Solution
By multiplying the given equations together we obtain
$$
\begin{aligned}
\left(x^{2} y z^{3}\right)\left(x y^{2}\right) & =\left(7^{4}\right)\left(7^{5}\right) \\
x^{3} y^{3} z^{3} & =7^{9}
\end{aligned}
$$
Taking the cube root of each side gives $x y z=7^{3}$.
ANSWER: (C)
# | 7^3 | Algebra | MCQ | Yes | Yes | olympiads | false |
Haley had 27 music files and some video files on her flash drive. She deleted 11 of the files, and now there are 58 files left on her flash drive. How many video files did she have initially? | Let's denote the number of video files Haley initially had as V.
Before deleting any files, the total number of files on her flash drive was the sum of the music files and the video files, which is 27 (music files) + V (video files).
After deleting 11 files, the total number of files left on her flash drive is 58.
S... | 42 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
15$\cdot$18 Mapping $f:\{a, b, c, d\} \rightarrow\{1,2,3\}$. If $10<f(a) \cdot f(b)$. $f(c) \cdot f(d)<20$, the number of such mappings is
(A) 23.
(B) 24.
(C) 25.
(D) 26.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991) | [Solution] . If $f(a)=f(b)=f(c)=f(d)=2$, it is clearly a mapping that fits the problem's requirements.
If $f(a), f(b), f(c), f(d)$ are not all 2, then at least one of them must be 1, and at most one of them can be 1 (note that $2^{3}=820$).
We discuss the following two scenarios:
(1) Among $f(a), f(b), f(c), f(d)$, th... | 25 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
The coefficients of the polynomial
\[ b_{8} x^8 + b_7 x^7 + b_6 x^6 + \dots + b_2 x^2 + b_1 x + b_0 = 0 \]
are all integers, and its roots $s_1, s_2, \dots, s_8$ are all integers. Furthermore, the roots of the polynomial
\[ b_0 x^8 + b_1 x^7 + b_2 x^6 + \dots + b_6 x^2 + b_7 x + b_8 = 0 \]
are also $s_1, s_2, \dots, s_... | Let $s$ be an integer root of the first polynomial $p(x) = b_8 x^8 + b_7 x^7 + \dots + b_1 x + b_0 = 0$, so
\[ b_8 s^8 + b_7 s^7 + \dots + b_1 s + b_0 = 0. \]
Since $b_0 \neq 0$, $s$ cannot be 0. By dividing both sides by $s^8$, we get
\[ b_8 + b_7 \frac{1}{s} + \dots + b_1 \frac{1}{s^7} + b_0 \frac{1}{s^8} = 0. \]
Thu... | 9 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
When a red die and a blue die are thrown at the same time, let event $M$ be "the number of points facing up on the red die is a multiple of 3", and $N$ be "the sum of the points on both dice is greater than 8". Calculate $P(N|M)$.
A: $\frac{5}{12}$
B: $\frac{7}{12}$
C: $\frac{1}{2}$
D: $\frac{1}{3}$ | This problem tests our understanding of classical probability computation, application, and conditional probability.
Using classical probability computation, we get $P(MN)=\frac{5}{36}$, $P(M)=\frac{12}{36}$. Then, using the conditional probability formula, we can calculate the conclusion.
Let's consider the simultan... | \frac{5}{12} | Combinatorics | MCQ | Yes | Yes | cn_k12 | false |
In how many ways can 8 rooks be placed on a chessboard so that they do not attack each other?
# | In each vertical, there is one rook. Their position is determined by the permutation of the horizontals.
## Answer
8! ways. | 8! | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
Example 6.1.5. Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers. Prove that
$$x_{1}+\sqrt{x_{1} x_{2}}+\ldots+\sqrt[n]{x_{1} x_{2} \ldots x_{n}} \leq e\left(x_{1}+x_{2}+\ldots+x_{n}\right)$$ | Solution. Suppose that $a_{1}, a_{2}, \ldots, a_{n}$ are positive real numbers. According to AM-GM inequality, we have
$$\begin{array}{c}
\sqrt[k]{\left(a_{1} x_{1}\right) \cdot\left(a_{2} x_{2}\right) \cdots\left(a_{k} x_{k}\right)} \leq \frac{a_{1} x_{1}+a_{2} x_{2}+\ldots+a_{k} x_{k}}{k} \\
\Rightarrow \sqrt[k]{x_{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false |
The number of common points between the hyperbola $\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$ and the line $y=-\dfrac{2}{3}x+m(m\in\mathbb{R})$ is ( ).
A: $0$
B: $1$
C: $0$ or $1$
D: $0$ or $1$ or $2$ | From the equation of the hyperbola $\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$, we find $a=3$, $b=2$.
Therefore, the equations of the asymptotes of the hyperbola are $y=±\dfrac{2}{3}x$.
When $m=0$, the line $y=-\dfrac{2}{3}x$ has no common points with the hyperbola.
When $m\neq 0$, the line $y=-\dfrac{2}{3}x+m$ is paralle... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
34 Prove: Every positive integer can be expressed as the difference of two positive integers, and these two positive integers have the same number of prime factors.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 34. Let $n$ be a positive integer. If $n$ is even, then the expression $n=(2 n)-n$ satisfies the requirement. If $n$ is odd, let $p$ be the smallest odd prime that does not divide $n$. Then the expression $n=p n-(p-1) n$ shows that the number of prime factors of $p n$ is one more than the number of prime factors of $n$... | null | Number Theory | proof | Yes | Yes | number_theory | false |
In the arithmetic sequence $\{a_n\}$, if $a_1 < 0$ and $S_9 = S_{12}$, then when $n$ equals \_\_\_\_\_\_\_\_, $S_n$ reaches its minimum value. | Given that $S_9 = S_{12}$ in an arithmetic sequence, it implies that the sum of the terms from the 10th to the 12th is zero. This means that the 11th term, which is the middle term of these three, must be zero. Since the sequence is decreasing (as $a_1 < 0$ and it's an arithmetic sequence), the sum of the sequence reac... | 10 \text{ or } 11 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A bag has 5 red marbles, 6 white marbles, 7 blue marbles, and 3 green marbles. Two marbles are drawn from the bag (without replacement). What is the probability that they are both the same color? | The total number of marbles in the bag is \(5 + 6 + 7 + 3 = 21\) marbles. We need to calculate the probability of drawing two marbles of each color, and sum these probabilities to find the overall probability of drawing two marbles of the same color.
1. **Probability of both red**:
\[
\frac{5}{21} \times \frac{4... | \frac{7}{30} | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
Let $S_{n}$ be the sum of the first $n$ terms of the sequence $\{a_{n}\}$, and when $n \geqslant 2$, the point $(a_{n-1}, 2a_{n})$ is on the line $y = 2x + 1$. Also, the first term of the sequence $\{a_{n}\}$, $a_{1}$, is the minimum value of the quadratic function $y = x^{2} - 2x + 3$. Find the value of $S_{9}$. | Since the point $(a_{n-1}, 2a_{n})$ lies on the line $y = 2x + 1$, we have:
$$2a_{n} = 2a_{n-1} + 1$$
This simplifies to:
$$a_{n} - a_{n-1} = \frac{1}{2}, \quad (n \geqslant 2)$$
The minimum value of the quadratic function $y = x^{2} - 2x + 3$ can be found by completing the square:
$$y = x^{2} - 2x + 3 = (x - 1)^{... | 36 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
There are 6 balls of each of the four colors: red, blue, yellow, and green. Each set of 6 balls of the same color is numbered from 1 to 6. If 3 balls with different numbers are randomly selected, and these 3 balls have different colors and their numbers are not consecutive, the number of ways to do this is ______. | To solve this problem, we first consider the ways in which the numbers can be non-consecutive. The possible combinations are: $135$, $136$, $146$, and $246$, totaling $4$ methods.
For these 3 balls to have different colors, there are $C_{4}^{3}A_{3}^{3} = 4 \times 3 \times 2 \times 1 = 24$ ways.
Therefore, the number... | 96 | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false |
Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a man in the slower train in 10 seconds. What is the length of the faster train? | To find the length of the faster train, we need to calculate the relative speed at which the faster train crosses the man in the slower train. Since both trains are moving in the same direction, we subtract the speed of the slower train from the speed of the faster train to get the relative speed.
Relative speed = Spe... | 100 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Find the number of moles of AgOH formed on combining 3 moles of AgNO3 and some moles of NaOH. The number of moles of AgOH formed is 3. How many moles of NaOH were combined? | To determine the number of moles of NaOH combined, we need to look at the balanced chemical equation for the reaction between AgNO3 (silver nitrate) and NaOH (sodium hydroxide) to form AgOH (silver hydroxide) and NaNO3 (sodium nitrate).
The balanced chemical equation is:
AgNO3 + NaOH → AgOH + NaNO3
From the balanced... | 3 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Josh wants to build a square sandbox that is 3 ft long, 3 ft wide for his son. He can buy sand in 3 sq ft bags for $4.00 a bag. How much will it cost him to fill up the sandbox? | First, we need to calculate the volume of the sandbox to determine how much sand is needed to fill it up. Since the sandbox is a square, we can use the formula for the volume of a cube, which is:
Volume = side length × side width × side height
Given that the sandbox is 3 ft long, 3 ft wide, and we'll assume a standar... | \$12.00 | Geometry | math-word-problem | Yes | Yes | orca_math | true |
Given a monotonically increasing function $f(x)$ defined on $\mathbb{R}$, satisfying $f(-x) + f(x) = 0$, and $x_1, x_2, x_3 \in \mathbb{R}$, with $x_1 + x_2 > 0$, $x_2 + x_3 > 0$, and $x_3 + x_1 > 0$, then the value of $f(x_1) + f(x_2) + f(x_3)$ is ( )
A: Definitely greater than 0
B: Definitely less than 0
C: Equal... | The correct answer is $\boxed{\text{A}}$.
Explanation is omitted. | \text{A} | Algebra | MCQ | Yes | Problem not solved | cn_k12 | false |
Zoe was unboxing some of her old winter clothes. She found 8 boxes of clothing and inside each box there were some scarves and 6 mittens. Zoe had a total of 80 pieces of winter clothing. How many scarves were in each box? | Let's denote the number of scarves in each box as S.
Since Zoe found 8 boxes and each box contains 6 mittens, the total number of mittens is 8 boxes * 6 mittens/box = 48 mittens.
Zoe has a total of 80 pieces of winter clothing, which includes both scarves and mittens. If we subtract the total number of mittens from t... | 4 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Given the complex number $z=x+yi$ $(x, y\in R)$, and $\frac{x}{1-i}=1+yi$, find $|z|=$ ( ).
A: $5$
B: $\sqrt{5}$
C: $3$
D: $\sqrt{3}$ | We have the complex number $z=x+yi$ $(x, y\in R)$, and $\frac{x}{1-i}=1+yi$.
First, let's simplify $\frac{x}{1-i}$ by multiplying both the numerator and denominator by the conjugate of the denominator:
$\frac{x}{1-i} \times \frac{1+i}{1+i} = \frac{x(1+i)}{(1-i)(1+i)} = \frac{x+xi}{1^2+1^2} = \frac{x+xi}{2}$
Now, let... | \sqrt{5} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
3. In the test, there are 4 sections, each containing the same number of questions. Andrey answered 20 questions correctly. The percentage of his correct answers was more than 60 but less than 70. How many questions were in the test?
OTBET: 32. | Solution. According to the condition $\frac{60}{100}<\frac{20}{x}<\frac{70}{100}$, from this $28 \frac{4}{7}=\frac{200}{7}<x<\frac{100}{3}=33 \frac{1}{3}$, that is $29 \leq x \leq 33$. From the first condition of the problem, it follows that the number of questions must be divisible by 4. | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Which of the following statements is correct ( )
A: Every line segment has exactly one golden ratio point
B: The golden ratio point divides a line segment into two line segments, with the longer one being approximately 0.618 times the length of the original segment
C: If point C divides line segment AB at the golden ra... | Solution:
A, Every line segment has two golden ratio points, so this option is incorrect;
B, The golden ratio point divides a line segment into two line segments, with the longer one being approximately 0.618 times the length of the original segment, which is correct;
C, If point C divides line segment AB at the golden... | \text{B} | Geometry | MCQ | Yes | Yes | cn_k12 | false |
16. What is the maximum number of numbers that can be chosen from the set $1,2,3, \ldots, 1985$ such that the difference between any two of the chosen numbers is not a prime number? | 85.16. Answer. 497 numbers. | 497 | Number Theory | math-word-problem | Yes | Incomplete | olympiads | false |
Lucy went to the grocery store. She bought 4 packs of cookie, 22 packs of cake and 16 packs of chocolate.. How many packs of groceries did she buy in all? | Lucy bought a total of:
4 packs of cookies + 22 packs of cake + 16 packs of chocolate
= 4 + 22 + 16
= $\boxed{42}$ packs of groceries in all. | 42 | Logic and Puzzles | math-word-problem | Yes | Yes | orca_math | true |
When Sally asked Johnny how old he was, he replied that in 2 years he would be twice as old as he was 3 years ago. How old is Johnny? | Let's call Johnny's current age "x" years.
According to Johnny, in 2 years, he will be x + 2 years old.
He also says that this age (x + 2) will be twice as old as he was 3 years ago.
Three years ago, Johnny was x - 3 years old.
So, we can set up the equation:
x + 2 = 2(x - 3)
Now, let's solve for x:
x + 2 = 2x -... | 8 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Given a random variable $X \sim B(2, p)$. If $P(X \geq 1) = \frac{3}{4}$, then find the value of $p$. | $X$ follows a binomial distribution $B(2, p)$. By definition of the binomial distribution, the probability that $X$ is greater than or equal to 1 is given by:
\[ P(X \geq 1) = 1 - P(X = 0) \]
We know that $P(X \geq 1) = \frac{3}{4}$. Hence:
\[ 1 - P(X = 0) = \frac{3}{4} \]
\[ P(X = 0) = 1 - \frac{3}{4} \]
\[ P(X = 0... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A quadrilateral in the plane has vertices at $(0, 0)$, $(0, 2)$, $(3, 2)$, and $(5, 0)$. How many square units is the area of the quadrilateral? | 1. **Divide the quadrilateral into two triangles**: Consider the line connecting $(0,0)$ to $(3,2)$.
2. **Calculate the area of triangle formed by vertices $(0, 0)$, $(0, 2)$, $(3, 2)$**:
- Base = Distance between $(0, 2)$ and $(3, 2)$ = $3$ units.
- Height = Distance from $(0, 0)$ to line through $(0, 2)$ and ... | 8 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
The graph of the equation $y = |x^2| - 4$ is translated three units to the right and four units down. What are the coordinates of the minimum point of the new graph? | Start with the original function $y = |x^2| - 4$. Since $x^2$ is nonnegative, and taking the absolute value of a nonnegative number leaves it unchanged, we have $y = x^2 - 4$. The minimum value of $x^2$ is 0, which occurs when $x = 0$.
Hence, the minimum point on the graph of $y = x^2 - 4$ is at $(0, -4)$.
Now, apply... | (3, -8) | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
In $\triangle ABC$, $a$, $b$, $c$ are the sides opposite to angles $A$, $B$, $C$ respectively, and $A=60^{\circ}$. If $S_{\triangle ABC}= \frac {15 \sqrt {3}}{4}$, and $5\sin B=3\sin C$, then the perimeter of $\triangle ABC$ is equal to $\_\_\_\_\_\_$. | Given that the area of the triangle is $S_{\triangle ABC}= \frac{1}{2}bc\sin A= \frac{1}{2}bc\times \frac{ \sqrt {3}}{2}= \frac {15}{4} \sqrt {3}$, we can find that $bc=15$.
Also, we have $5\sin B=3\sin C$, which by the sine law gives us $5b=3c$.
Solving the system of equations:
$\begin{cases} bc=15 \\ 5b=3c \end{cas... | 8+ \sqrt {19} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
A quadrilateral $ABCD$ has a right angle at $\angle ABC$ and satisfies $AB = 12$, $BC = 9$, $CD = 20$, and $DA = 25$. Determine $BD^2$.
.
| 1. Given the quadrilateral \(ABCD\) with a right angle at \(\angle ABC\), we know:
\[
AB = 12, \quad BC = 9, \quad CD = 20, \quad DA = 25
\]
We need to determine \(BD^2\).
2. Let \(\angle BCA = \theta\). Since \(\triangle ABC\) is a right triangle at \(B\), we can use trigonometric identities to find \(\si... | 769 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let the sequence $\{a_n\}$ have a sum of the first $n$ terms denoted by $S_n$. If the point $(n, S_n)$ (where $n$ is a positive integer) lies on the graph of the inverse function of $y = \log_2(x + 1)$, then $a_n =$ ______. | Given the problem, let's find the relationship between $S_n$ and $n$. Since the point $(n, S_n)$ lies on the graph of the inverse function of $y = \log_2(x + 1)$, this means that $(S_n, n)$ lies on the graph of $y = \log_2(x + 1)$. Therefore, we can write:
\[ n = \log_2(S_n + 1) \Rightarrow S_n = 2^n - 1 \]
Next, let'... | a_n = 2^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
In the arithmetic sequence $\{a_n\}$, $a_2 = -5$ and $d = 3$. Find $a_1$. | Since in the arithmetic sequence $\{a_n\}$, $a_2 = -5$ and $d = 3$,
we have $a_1 + d = a_2$. Substituting the given values, we get $a_1 + 3 = -5$,
solving this equation, we find $a_1 = -8$.
Therefore, the answer is $\boxed{-8}$.
**Analysis:** The solution is obtained by using the general formula of an arithmet... | -8 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Julia has a parrot, a rabbit, a turtle, and a guinea pig. She buys food for all of these animals at different costs per pound: $4 for the parrot food, $3 for the rabbit food, $2 for the turtle food, and $1 for the guinea pig food. Each week, Julia's parrot needs 3.75 pounds of food, her rabbit needs 4 pounds of food, h... | To calculate the total amount of money Julia spent on food for her animals, we need to multiply the cost per pound of each type of food by the amount of food needed per week and then by the number of weeks she has each animal.
For the parrot:
Cost per week = 3.75 pounds * $4/pound = $15
Total cost for 3 weeks = $15/we... | \$151 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
In triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, with $A=60^{\circ}$, and the area of $\triangle ABC$ is $\frac{{3\sqrt{3}}}{2}$. If $b+c=3\sqrt{3}$, then $a=\_\_\_\_\_\_$. | Given that in triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, with $A=60^{\circ}$, and the area of $\triangle ABC$ is $\frac{3\sqrt{3}}{2}$. We also know that $b+c=3\sqrt{3}$. We are asked to find the value of $a$.
Step 1: Use the formula for the area of a t... | 3 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
In the quadrilateral $OABC$ in space, $\overrightarrow{OA}=\overrightarrow{a}, \overrightarrow{OB}=\overrightarrow{b}, \overrightarrow{OC}=\overrightarrow{c},$ and point $M$ is on $OA$ such that $OA=2MA$. Point $N$ is the midpoint of $BC$. Express $\overrightarrow{MN}$ in terms of $\overrightarrow{a}, \overrightarrow{b... | This problem requires the understanding of vector addition and subtraction, the concept of a base, and the application of the fundamental theorem of space vectors. Utilizing the triangle rule for vector addition and subtraction, we can write $\overrightarrow{MN}=\overrightarrow{MO}+\overrightarrow{OC}+\overrightarrow{C... | -\frac{1}{2}\overrightarrow{a} + \frac{1}{2}\overrightarrow{b} + \frac{1}{2}\overrightarrow{c} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
Given that $a$ and $b$ are positive real numbers, the condition for "$a > b > 1$" is "\_\_\_\_\_\_\_\_\_\_\_". | The inequality $\log_{2}a > \log_{2}b > 0$ can be equivalently transformed into $a > b > 1$. Therefore, "$a > b > 1$" is the necessary and sufficient condition for "$\log_{2}a > \log_{2}b > 0$".
Key Point: Necessary Condition and Sufficient Condition
The final answer to fill in the blank is $\boxed{a > b > 1}$. | a > b > 1 | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false |
## Task 5 - 261235
Two people, A and B, play the following game with $n$ lamps arranged in a straight line $(n>3)$:
At the start of the game, all lamps are turned off. An integer $k$ with $13$ and every $k$ with $1<k<n-1$ can enforce a win through a suitable approach (strategy). | We denote the lamps in sequence as $1, \ldots, n$ and a configuration of the on/off states of the lamps as symmetric if the two lamps in each of the lamp pairs $(1, n); (2, n-1); \ldots$; $\left(\left\lceil\frac{n}{2}\right\rceil, \left\lceil\frac{n+1}{2}\right\rceil\right)$ have the same on/off state (if $n$ is odd, t... | proof | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
In the game of Frood, dropping $n$ froods gives a score of the sum of the first $n$ positive even integers. For example, dropping five froods scores $2 + 4 + 6 + 8 + 10 = 30$ points. Eating $n$ froods earns $8n$ points. For example, eating five froods earns $8(5) = 40$ points. What is the least number of froods for whi... | For dropping $n$ froods, the score formula is the sum of the first $n$ even positive numbers. The $n$-th even number is $2n$, so the sequence is $2, 4, ..., 2n$. The sum of an arithmetic series is given by $\text{Number of terms} \times (\text{First term} + \text{Last term}) / 2$. Here, it simplifies to:
$$
n \times (2... | 8 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Given circle $C_1: x^2+y^2-6x+5=0$, and parabola $C_2: y^2=x$. A line $l$ passing through point $M(m,0)$ intersects circle $C_1$ at points $A$ and $B$, and intersects parabola $C_2$ at points $C$ and $D$.
(1) If $m=0$, as line $l$ rotates around point $M$, find the equation of the trajectory of the midpoint $R$ of segm... | (1) Let $R(x,y)$, circle $C_{1}: (x-3)^{2}+y^{2}=4$, with the center of circle $C_1$ at $(3,0)$,...(1 point)
$\overrightarrow {OR}=(x,y)$, $\overrightarrow {C_{1}R}=(x-3,y)$...(2 points)
By the property of the circle, we know $\overrightarrow {OR} \cdot \overrightarrow {C_{1}R}=0$...(3 points)
This gives $x(x-3)+... | x=2 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
6. Xiao Fang calculates the least common multiple of two two-digit numbers $\overline{A B}$ and $\overline{C D}$, while Xiao Yuan calculates the least common multiple of two two-digit numbers $\overline{B A}$ and $\overline{D C}$ (the process is as follows). It is known that Xiao Fang's answer is 1.75 times that of Xia... | 【Analysis】That is, $D \times E \times B \times D=1.75 \times D \times B \times B \times D \Rightarrow E=1.75 \times B$, so $\mathrm{B}$ can be 4 or 8. If $\mathrm{B}$ is 8, then $\mathrm{E}$ is 14, $\overline{B A}=B \times B \times D \Rightarrow \overline{8 A}=64 \times D$, which is obviously impossible.
Therefore, $\m... | 252 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
If $f\left(x\right)=x^{3}+x-1$, ${f'}(x_{0})=4$, then the value of $x_{0}$ is ( )
A: $1$
B: $-1$
C: $3\sqrt{3}$
D: $-3\sqrt{3}$ | Given that $f\left(x\right)=x^{3}+x-1$ and ${f'}(x_{0})=4$, we need to find the value of $x_{0}$. Let's proceed step by step:
1. First, we determine the derivative of $f(x)$, which involves differentiating each term of the function separately:
$$f\left(x\right)=x^{3}+x-1$$
Therefore, using the power rule for different... | AB | Calculus | MCQ | Yes | Yes | cn_k12 | false |
If infinitely many values of \( y \) satisfy the equation \( 3(5 + 2cy) = 18y + 15 \), then what is the value of \( c \)? | Start by simplifying both sides of the equation:
\[ 3(5 + 2cy) = 18y + 15. \]
Distribute \( 3 \) on the left side:
\[ 15 + 6cy = 18y + 15. \]
Subtract \( 15 \) from both sides to simplify:
\[ 6cy = 18y. \]
Dividing both sides by \( 6 \) gives:
\[ cy = 3y. \]
Assuming \( y \neq 0 \) (since infinitely many values of ... | 3 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Given $0 -1$；②$\log_{2}a+\log_{2}b > -2$
(B) ②$\log_{2}a+\log_{2}b > -2$；③$\log_{2}(b-a) 1$
(D) ①$\log_{2}a > -1$；④$\log_{2}\left( \frac{b}{a}+ \frac{a}{b}\right) > 1$ | Since $0 2\sqrt{ab}$, which implies $ab 2\sqrt{\frac{b}{a}\cdot \frac{a}{b}}=2$,
thus $\log_{2}\left( \frac{b}{a}+ \frac{a}{b}\right) > \log_{2}2=1$, so option ④ is correct.
Therefore, the correct answer is $\boxed{C}$.
This problem is solved using properties of logarithms, the mean value theorem, properties of... | C | Algebra | MCQ | Incomplete | Yes | cn_k12 | false |
Let $A B C D$ be a parallelogram. Let $E$ be a point such that $A E=B D$ and $(A E)$ is parallel to $(B D)$. Let $F$ be a point such that $E F=A C$ and $(E F)$ is parallel to $(A C)$. Express the area of the quadrilateral $A E F C$ in terms of the area of $A B C D$. | As $A E / / D B$ and $A E = D B$, $A E B D$ is a parallelogram. Therefore, $E B = A D = B C$ and $E B / / A D / / B C$, in other words, $B$ is the midpoint of segment $E C$. Furthermore, since $E F = A C$ and $E F / / A C$, $A E F C$ is a parallelogram, and thus $B$ is the point of intersection of its diagonals. Theref... | 2\timesAreaofABCD | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A certain percentage of the world's forested area is represented by Finland. The world has 8.076 billion hectares of forested land. If Finland represents approximately 0.66% of the world's forested area, how many hectares of forested land does Finland have? | To find out how many hectares of forested land Finland has, we can calculate 0.66% of the world's total forested area.
0.66% of 8.076 billion hectares = 0.0066 * 8,076,000,000 hectares
Let's do the math:
= 53,301,600 hectares
So, Finland has approximately $\boxed{53,301,600}$ hectares of forested land. | 53,301,600 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Objects $A$ and $B$ simultaneously move in the coordinate plane following sequences of steps, each of length one. Object $A$ starts at $(0,0)$ and can only move right or up, both equally likely. Object $B$ starts at $(10,10)$, with each step either left or down, both equally likely. What is the probability that they me... | For $A$ and $B$ to meet, the number of steps taken must lead them to the same point. Given B's starting point is $(10,10)$, it means they have to travel a total of 20 steps split equally between them to potentially meet (10 steps each).
The possible meeting points can be derived by the possible overlaps of their movem... | 0.352 | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
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