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$9 \cdot 13$ If $\left(\log _{2} 3\right)^{x}-\left(\log _{5} 3\right)^{x} \geqslant\left(\log _{2} 3\right)^{-y}-\left(\log _{5} 3\right)^{-y}$, then
(A) $x-y \geqslant 0$
(B) $x+y \geqslant 0$
(C) $x-y \leqslant 0$
(D) $x+y \leqslant 0$
(China High School Mathematics League, 1999) | [Solution] Let $f(t)=\left(\log _{2} 3\right)^{t}-\left(\log _{5} 3\right)^{t}$, then $f(t)$ is a strictly increasing function on the real number domain $R$.
The original inequality is $f(x) \geqslant f(-y)$, hence $x \geqslant -y$, which means $x+y \geqslant 0$. Therefore, the answer is $(B)$. | x+y\geqslant0 | Inequalities | MCQ | Yes | Yes | olympiads | false |
Given proposition $p$: The minimum value of the function $f(x)=x+\frac{4}{x}$ when $x≠0$ is $4$; proposition $q$: In $\triangle ABC$, where the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, then "$A \gt B$" is a necessary and sufficient condition for "$a \gt b$". Which of the following ... | To analyze the propositions and determine the correct answer, we proceed as follows:
**Proposition $p$ Analysis:**
The function given is $f(x) = x + \frac{4}{x}$, where $x \neq 0$. To find its minimum value, we can use the AM-GM inequality, which states that for any positive numbers $a$ and $b$, the arithmetic mean i... | \text{A} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Given that the right focus of the hyperbola $\frac{x^2}{a^2} - y^2 = 1 (a > 0)$ coincides with the focus of the parabola $y^2 = 8x$, what is the equation of the asymptotes of this hyperbola?
A: $y = \pm \sqrt{5}x$
B: $y = \pm \frac{\sqrt{5}}{5}x$
C: $y = \pm \sqrt{3}x$
D: $y = \pm \frac{\sqrt{3}}{3}x$ | This problem tests our understanding of the properties and applications of hyperbolas. To solve it, we'll use the properties of the parabola to find the coordinates of its focus, which will give us one focus of the hyperbola. From there, we can find the value of $a$, which will lead us to the equations of the asymptote... | y = \pm \frac{\sqrt{3}}{3}x | Geometry | MCQ | Yes | Yes | cn_k12 | false |
7. Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$, such that for any positive real numbers $x, y$, we have
$$
\begin{array}{l}
f(f(x)+y) \\
=f(f(x))+2 y f(x)-f(y)+2 y^{2}+1 .
\end{array}
$$ | 7. First, in equation (1), replace $y$ with $f(y)$:
$$
\begin{array}{l}
f(f(x)+f(y)) \\
= f(f(x)) + 2 f(y) f(x) - f(f(y)) + \\
2 f^{2}(y) + 1 \\
\Rightarrow f(f(x)+f(y)) - f(f(x)) - \\
2 f(y) f(x) - f(f(y)) \\
= -2 f(f(y)) + 2 f^{2}(y) + 1 .
\end{array}
$$
Notice that the left side of the equation is symmetric with r... | f(x)=x^{2}+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Given sets $A=\{0, 1\}, B=\{a^2, 2a\}$, where $a\in\mathbb{R}$, we define the set $\{x | x = x_1 + x_2, x_1 \in A, x_2 \in B\}$ as $A+B$. If the largest element in the set $A+B$ is $2a+1$, find the range of values for $a$. | We have the set $A = \{0, 1\}$ and the set $B = \{a^2, 2a\}$. According to the definition of $A+B$, we combine each element from set $A$ with each element from set $B$ by addition. So the set $A+B$ will include the following elements:
1. $0 + a^2 = a^2$
2. $0 + 2a = 2a$
3. $1 + a^2 = a^2 + 1$
4. $1 + 2a = 2a + 1$
The... | a \in (1 - \sqrt{2}, 1 + \sqrt{2}) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A trader bought a car at a 20% discount on its original price. He sold it at some increase on the price he bought it. He made a 24% profit on the original price. What was the percentage increase on the price he bought it when he sold the car? | Let's assume the original price of the car is \( P \).
The trader bought the car at a 20% discount, so he paid:
\( P - 0.20P = 0.80P \)
He then sold the car at a 24% profit on the original price, so he sold it for:
\( P + 0.24P = 1.24P \)
Now, we want to find the percentage increase on the price he bought it for. Th... | 55\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Squares $CAKL$ and $CBMN$ are constructed on the sides of acute-angled triangle $ABC$, outside of the triangle. Line $CN$ intersects line segment $AK$ at $X$, while line $CL$ intersects line segment $BM$ at $Y$. Point $P$, lying inside triangle $ABC$, is an intersection of the circumcircles of triangles $KXN$ and $LYM$... | 1. **Construct the squares and identify intersections:**
- Construct squares \(CAKL\) and \(CBMN\) on the sides of the acute-angled triangle \(ABC\), outside the triangle.
- Line \(CN\) intersects line segment \(AK\) at \(X\).
- Line \(CL\) intersects line segment \(BM\) at \(Y\).
2. **Identify point \(P\):**... | null | Geometry | proof | Yes | Yes | aops_forum | false |
For any $x \in (-\infty, -1]$, the inequality $(3m - 1)2^x < 1$ always holds. Find the range of the positive real number $m$. | Given that $x \in (-\infty, -1]$, we have $2^x \in (0, \frac{1}{2}]$.
The inequality $(3m - 1)2^x < 1$ always holds, which implies $(3m - 1) < \frac{1}{2^x}$.
Since $2^x \in (0, \frac{1}{2}]$, we have $\frac{1}{2^x} \in [2, +\infty)$.
Therefore, $3m - 1 < 2$, which simplifies to $m < 1$.
Hence, the range of the pos... | (0, 1) | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false |
David obtained 61 marks in English, 65 in Mathematics, 82 in Physics, some marks in Chemistry, and 85 in Biology. His average marks are 72. What are his marks in Chemistry? | To find David's marks in Chemistry, we first need to calculate the total marks he obtained in the other subjects and then use the average to find the missing Chemistry marks.
The total marks obtained in the other subjects are:
English: 61
Mathematics: 65
Physics: 82
Biology: 85
Total marks in these four subjects = 61... | 67 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
If the inequality $(a^2 - 1)x^2 + (a - 1)x - 1 < 0$ holds true for any real number $x$, then the range of values for the real number $a$ is _______. | [Analysis]
This problem involves a quadratic inequality that always holds true. We need to discuss the possible values of $a$ and solve the problem using the discriminant ($\Delta$).
[Solution]
1. When $a = 1$, the inequality $(a^2 - 1)x^2 + (a - 1)x - 1 < 0$ becomes $0 < 1$, which is true and satisfies the condition... | (-\frac{3}{5}, 1] | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false |
The negation of the proposition "$\exists x_{0} \gt 0$, ${x}_{0}^{2}-5x_{0}+6 \gt 0$" is ( ).
A: $\forall x\leqslant 0$, $x^{2}-5x+6\leqslant 0$
B: $\forall x \gt 0$, $x^{2}-5x+6\leqslant 0$
C: $\exists x_{0}\leqslant R$, ${x}_{0}^{2}-5x_{0}+6\leqslant 0$
D: $\exists x_{0} \gt 0$, ${x}_{0}^{2}-5x_{0}+6\leqslant 0$ | To find the negation of the given proposition, we start with the original statement and then apply the rules of negation.
The original proposition is: $\exists x_{0} > 0$, such that ${x_{0}}^{2} - 5x_{0} + 6 > 0$.
The negation of an existential quantifier $\exists$ is a universal quantifier $\forall$, and vice versa... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false |
4. Let $O$ be the center of the inscribed circle of triangle $\triangle A B C$ and $D$ be the point on $A C$ with $O D \perp A C$. If $A B=10 A C=9 B C=11$, find $C D$.
(A) 4
(B) 4.5
(C) 5
(D) 5.5
(E) 6 | 4. Answer: (C)
$A C$ is tangent to the circle at $D$, by constructing $E$ and $F$ as shown, we have $C D=C F, A D=A E$ and $B E=B F$. Solving for the unknowns give $C D=5$. | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
If $2a^{m}b^{3}$ and $-3ab^{n}$ are like terms, then $m+n=\_\_\_\_\_\_$. | Given that $2a^{m}b^{3}$ and $-3ab^{n}$ are like terms, we can deduce the following:
For two algebraic expressions to be considered like terms, their variables and their corresponding exponents must match. This means that the powers of $a$ and $b$ in both terms must be the same.
From $2a^{m}b^{3}$, we see that the ex... | 4 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
## Problem 4
Let ABCD be a square that folds along the diagonal AC.
a) If $(D A C) \perp(A B C)$, prove that triangle DAB is equilateral.
b) If $m(\angle(D A C);(A B C))=60^{\circ}$ and $\mathrm{AB}=12 \mathrm{~cm}$, calculate the length of $[B D]$
Note
- Effective working time 3 hours.
- All problems are mandator... | ## Problem 4
B
a) $(D A C) \perp(A B C),(D A C) \cap(A B C)=[A C]$
Let $D E \perp A C \Rightarrow D E \perp(A B C)$.
$\triangle A D C$ is right-angled, $D E$ is the altitude $\Rightarrow D... | 6\sqrt{2}\, | Geometry | proof | Yes | Yes | olympiads | false |
Given the equation $||x-2|-1|=a$ has exactly three integer solutions, find the value of $a$. | To solve this, we consider two cases based on the absolute value expression:
1. If $|x-2|-1=a$,
- For $x \geq 2$, we have $x-2-1=a$, which simplifies to $x=a+3$, where $a \geq -1$.
- For $x -1$.
2. If $|x-2|-1=-a$,
- For $x \geq 2$, we have $x-2-1=-a$, which simplifies to $x=-a+3$, where $a \leq 1$.
- ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
In the arithmetic sequence $\{a_n\}$, it is known that $a_4+a_8=16$. Then, the sum of the first $11$ terms, $S_{11}$, is
A: $58$
B: $88$
C: $143$
D: $176$ | Since in the arithmetic sequence $\{a_n\}$, it is known that $a_4+a_8=16$,
Therefore, $a_1+a_{11}=a_4+a_8=16$,
Thus, $S_{11}= \dfrac{11(a_1+a_{11})}{2}=88$,
Hence, the correct option is $\boxed{B}$.
According to the definition and properties of an arithmetic sequence, we have $a_1+a_{11}=a_4+a_8=16$. Then, by calcu... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Layla is feeding her pet fish. She has 2 Goldfish which each get one teaspoon of fish food. Her 3 Swordtails each get 2 teaspoons of food. She has 8 Guppies which each eat some amount of food. She has to give her fish a total of 12 teaspoons of food. What is the ratio of the amount of food each Guppy gets to the amount... | First, let's calculate the total amount of food the Goldfish and Swordtails get:
Goldfish: 2 fish * 1 teaspoon each = 2 teaspoons
Swordtails: 3 fish * 2 teaspoons each = 6 teaspoons
Now, subtract the amount of food for the Goldfish and Swordtails from the total amount of food Layla has to give:
Total food - (Goldfis... | 1:2 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
In an $m\times n$ rectangular chessboard,there is a stone in the lower leftmost square. Two persons A,B move the stone alternately. In each step one can move the stone upward or rightward any number of squares. The one who moves it into the upper rightmost square wins. Find all $(m,n)$ such that the first person has a ... | 1. **Define the Problem in Terms of Coordinates:**
- Consider the stone's position on the chessboard as \((x, y)\), where \(x\) represents the number of steps upward needed to reach the top row, and \(y\) represents the number of steps rightward needed to reach the rightmost column.
- The initial position of the ... | m \neq n | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
a rectangular photograph is surrounded by a border that is 2 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 4 inches wide on each side , the total area would have been ( m + 94 ) square inches . what is the perimeter of the photograph , in inches ? | Let's denote the width of the photograph as w inches and the length as l inches. The border adds 2 inches to each side, so the total width of the photograph with the border is w + 2(2) = w + 4 inches, and the total length is l + 2(2) = l + 4 inches.
The area of the photograph with the 2-inch border is given by:
(w + 4... | 23 | Geometry | math-word-problem | Yes | Yes | orca_math | true |
A bag has 5 red marbles and 7 white marbles. Two marbles are drawn from the bag without replacement. What is the probability that both are red? | 1. Compute the probability that the first marble is red:
\[
P(\text{First is red}) = \frac{\text{number of red marbles}}{\text{total number of marbles}} = \frac{5}{12}
\]
2. After drawing one red marble, the bag now contains 4 red marbles and 7 white marbles, making a total of 11 marbles.
3. Compute the prob... | \frac{5}{33} | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
Four objects A, B, C, and D start moving from the same point in the same direction at the same time. Their distances $f_{i}(x)$ (i=1, 2, 3, 4) as functions of time $x$ ($x \geq 0$) are given by:
$$f_{1}(x)=2^{x}-1, \quad f_{2}(x)=x^{3}, \quad f_{3}(x)=x, \quad f_{4}(x)=\log_{2}(x+1)$$
The following conclusions are draw... | Solution: The distance functions $f_{i}(x)$ (i=1, 2, 3, 4) with respect to time $x$ ($x \geq 0$) are given by:
$$f_{1}(x)=2^{x}-1, \quad f_{2}(x)=x^{3}, \quad f_{3}(x)=x, \quad f_{4}(x)=\log_{2}(x+1)$$
Their corresponding function models are exponential, power, linear, and logarithmic models, respectively;
① When $x=2$... | ③④⑤ | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given that $sin(α+π)= \frac {3}{4}$, find the value of $cos(α+ \frac {π}{2})$. | If $sin(α+π)= \frac {3}{4}$, then $sin(α+π)= -sinα$.
According to the cofunction identity, $cos(α+ \frac {π}{2}) = -sinα$.
Since $-sinα = \frac{3}{4}$, we have $cos(α+ \frac {π}{2}) = \boxed{\frac{3}{4}}$. | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Find the smallest positive integer whose cube ends in $544$. | We require $n^3 \equiv 544 \pmod{1000}$, which requires $n^3 \equiv 0 \pmod 8$ because 544 is divisible by 8, and $n^3 \equiv 544 \pmod{125}$. The last digit of $n$ must be 4 to satisfy the condition that $n^3 \equiv 4 \pmod{5}$ (since $4^3 = 64 \equiv 4 \pmod{5}$). Setting $n = 5a + 4$, expanding and substituting, we ... | 69 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Lighters cost $1.75 each at the gas station, or a certain amount per pack of twelve on Amazon. Amanda would save $32 by buying 24 lighters online instead of at the gas station. How much does a pack of twelve lighters cost on Amazon? | Let's denote the cost of a pack of twelve lighters on Amazon as \( P \).
Amanda is considering buying 24 lighters. If she buys them at the gas station, it would cost her \( 24 \times $1.75 \). If she buys them online, she would buy two packs of twelve lighters, which would cost her \( 2 \times P \).
We are told that ... | \$5 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
While on a safari, Josie observed various animals grazing and playing together. She decided to count each animal and determine the total number. Her count included 80 antelopes, 34 more rabbits than antelopes, 42 fewer hyenas than the combined total of antelopes and rabbits, 50 more wild dogs than hyenas, and the numbe... | To determine the total number of animals that Josie counted, we will calculate the number of each type of animal step by step and then sum them up.
1. **Antelopes:**
Josie counted 80 antelopes.
\[
A = 80
\]
2. **Rabbits:**
Josie counted 34 more rabbits than antelopes.
\[
R = A + 34 = 80 + 34 = 11... | 521 | Algebra | math-word-problem | Yes | Yes | metamath | true |
The negation of the proposition "There exists an $x > 0$ such that $x^2 - x \leq 0$" is ( )
A: There exists an $x > 0$ such that $x^2 - x > 0$
B: There exists an $x \leq 0$ such that $x^2 - x > 0$
C: For all $x 0$
D: For all $x \leq 0$, $x^2 - x > 0$ | Given the proposition "There exists an $x > 0$ such that $x^2 - x \leq 0$":
To find the negation of this proposition according to the definition of negation in logic, the condition and the conclusion of the negated proposition must be the negation of the condition and conclusion of the original.
First, the existentia... | \text{D} | Inequalities | MCQ | Yes | Yes | cn_k12 | false |
After going down the falls, he then went to Rock Mist Mountains. If the mountain is 50 times farther from the city than Sky Falls and Sky Falls is 8 miles from the city, how far is the Rocky Mist from the city? | If Sky Falls is 8 miles from the city and Rock Mist Mountains are 50 times farther from the city than Sky Falls, then the distance to Rock Mist Mountains from the city is:
8 miles * 50 = 400 miles
So, Rock Mist Mountains are $\boxed{400}$ miles from the city. | 400 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A cab driver's income for 5 days was $400, $250, some amount, $400, and $500. His average income for these 5 days is $440. What was his income on the third day? | To find the cab driver's income on the third day, we first need to calculate his total income for the 5 days based on the average income.
If his average income for the 5 days is $440, then the total income for the 5 days is:
5 days * $440/day = $2200
We know the income for the first, second, fourth, and fifth days, s... | \$650 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
How many triangles with positive area have all their vertices at points \((i,j)\) in the coordinate plane, where \(i\) and \(j\) are integers between 1 and 4, inclusive? | First, we calculate the total number of points available, which is \(4 \times 4 = 16\). The total number of ways to choose three points from these 16 is given by \(\binom{16}{3} = \frac{16 \cdot 15 \cdot 14}{3 \cdot 2 \cdot 1} = 560\).
Next, we subtract the triangles that don't have a positive area, i.e., the points t... | 520 | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
For each real number $x$, let $\lfloor x \rfloor$ denote the largest integer not exceeding $x$.
A sequence $\{a_n \}_{n=1}^{\infty}$ is defined by $a_n = \frac{1}{4^{\lfloor -\log_4 n \rfloor}}, \forall n \geq 1.$ Let $b_n = \frac{1}{n^2} \left( \sum_{k=1}^n a_k - \frac{1}{a_1+a_2} \right), \forall n \geq 1.$
a) Find ... | ### Part (a)
1. **Define the sequence \( \{a_n\} \):**
\[
a_n = \frac{1}{4^{\lfloor -\log_4 n \rfloor}}, \quad \forall n \geq 1.
\]
For an integer \( l \) such that \( 4^l < n \leq 4^{l+1} \), we have:
\[
-(l+1) \leq \log_4 n < -l \implies \lfloor -\log_4 n \rfloor = -(l+1).
\]
Therefore,
\[... | P(x) = -\frac{1}{5}x^2 + x | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
The polynomial $g(x)=x^{2010}+14x^{2009}+1$ has distinct zeroes $s_1,\ldots,s_{2010}$. A polynomial $Q$ of degree $2010$ has the property that
\[Q\left(s_j+\dfrac{1}{s_j}\right)=0\] for $j=1,\ldots,2010$. Determine the value of $\frac{Q(1)}{Q(-1)}$. | We can write
\[g(x) = (x - s_1)(x - s_2) \dotsm (x - s_{2010})\] and
\[Q(z) = k \prod_{j=1}^{2010} \left(z - \left(s_j + \frac{1}{s_j}\right)\right)\] for some nonzero constant $k$.
We want to compute
\[\frac{Q(1)}{Q(-1)} = \frac{\prod_{j=1}^{2010} \left(1 - \left(s_j + \frac{1}{s_j}\right)\right)}{\prod_{j=1}^{2010} ... | \frac{49}{64} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
In the expansion of $(\sqrt{x} + \frac{1}{\sqrt[3]{x}})^n$, only the 13th term has the maximum binomial coefficient. The number of terms with an integer exponent of $x$ is $\boxed{\text{( )}}$.
A: 3 terms
B: 4 terms
C: 5 terms
D: 6 terms | According to the binomial expansion, the middle term has the maximum binomial coefficient.
$\therefore n = 24$
$(\sqrt{x} + \frac{1}{3x})^n = (\sqrt{x} + \frac{1}{3x})^{24}$. The general term of its expansion is:
$T_{r+1} = C_{24}^{r}x^{12 - \frac{5r}{6}}$
In order for the exponent of $x$ to be an integer, $r$ must... | \text{C} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Example 12 Let $x, y, z$ be real numbers, $k_{1}, k_{2}, k_{3} \in\left(0, \frac{1}{2}\right)$, and $k_{1}+k_{2}+k_{3}=1$, prove: $k_{1} k_{2} k_{3}(x+y+z)^{2} \geqslant x y k_{3}\left(1-2 k_{3}\right)+y z k_{1}\left(1-2 k_{1}\right)+z x k_{2}\left(1-2 k_{2}\right)$. (1999 China National Training Team Test Question) | Proof: First, we prove a lemma: In $\triangle ABC$, for any real numbers $x, y, z$, we have
$$x^{2}+y^{2}+z^{2} \geqslant 2 x y \cos C + 2 y z \cos A + 2 z x \cos B$$
In fact, since $\cos A = -\cos (B+C)$, we have
$$\begin{array}{l}
x^{2}+y^{2}+z^{2}-(2 x y \cos C + 2 y z \cos A + 2 z x \cos B) = \\
(x - y \cos C - z ... | proof | Inequalities | proof | Yes | Yes | inequalities | false |
The difference of two numbers is 1345. On dividing the larger number by the smaller, we get 6 as quotient and a certain remainder. The larger number is 1596. What is the remainder when the larger number is divided by the smaller number? | Let's denote the larger number as L and the smaller number as S.
We are given that the difference between the two numbers is 1345, so we can write:
L - S = 1345
We are also given that the larger number is 1596, so:
L = 1596
Now we can find the smaller number S by substituting the value of L into the first equation... | 90 | Number Theory | math-word-problem | Yes | Yes | orca_math | true |
Fill in the blanks in the following famous phrases.
(1) Govern it with virtue, regulate it with punishment, ; , regulate it with rites, and there will be a sense of shame and a sense of right. (《Governing with Virtue》)
(2) When Qiu heard of a country or... | This problem tests the ability to recall and write down common famous phrases. The key to solving this problem is to memorize the content of the work, understand the relationship between sentences, write down related content based on the preceding and following sentences, remember the correct character forms, and pay a... | \text{to cherish the young} | Other | math-word-problem | Not a problem | Yes | cn_k12 | false |
The range of values for $x$ that make the quadratic root $\frac{\sqrt{x+3}}{3}$ meaningful is ______. | To determine the range of values for $x$ that make the quadratic root $\frac{\sqrt{x+3}}{3}$ meaningful, we need to ensure that the expression under the square root, $x+3$, is non-negative. This is because the square root of a negative number is not defined in the set of real numbers. Therefore, we set up the inequalit... | x \geqslant -3 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Sally paid some amount for peaches, after a 3 dollar coupon, and $11.54 for cherries. In total, Sally spent $23.86. How much did Sally pay for the peaches after using the coupon? | To find out how much Sally paid for the peaches after using the coupon, we need to subtract the amount she spent on cherries from the total amount she spent.
Total amount spent: $23.86
Amount spent on cherries: $11.54
Amount spent on peaches after using the coupon = Total amount spent - Amount spent on cherries
= $23... | \$12.32 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Mrs. Hilt's favorite first grade classes are baking muffins. Mrs. Brier's class bakes 18 muffins, Mrs. MacAdams's class bakes 20 muffins, and Mrs. Flannery's class bakes some muffins. First grade bakes 55 muffins in all. How many muffins does Mrs. Flannery's class bake? | To find out how many muffins Mrs. Flannery's class bakes, we need to subtract the total number of muffins baked by Mrs. Brier's class and Mrs. MacAdams's class from the total number of muffins baked by all first grade classes.
Mrs. Brier's class bakes 18 muffins.
Mrs. MacAdams's class bakes 20 muffins.
Total muffins ... | 17 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
4. In the expression $S=\sqrt{x_{1}-x_{2}+x_{3}-x_{4}}$, $x_{1}, x_{2}, x_{3}, x_{4}$ are a permutation of $1,2,3,4$. The number of different permutations that make $S$ a real number is $\qquad$ | 4. 16 If $x_{1}, x_{3}$ take 1,4, there are $2 \times 2=4$ permutations; if $x_{1}, x_{3}$ take 2,3, there are $2 \times 2=4$ permutations; if $x_{1}, x_{3}$ take 2,4, there are $2 \times 2=4$ permutations; if $x_{1}, x_{3}$ take 3,4, there are $2 \times 2=4$ permutations. Therefore, the number of different permutation... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$(1)$ Calculate: $\sqrt{48}÷\sqrt{3}+\sqrt{\frac{1}{2}}\times \sqrt{12}-\sqrt{24}$.
$(2)$ Solve the system of equations: $\left\{\begin{array}{l}{3x-2y=3}\\{x+4y=1}\end{array}\right.$. | ### Step-by-Step Solution
#### Part 1: Calculate $\sqrt{48}÷\sqrt{3}+\sqrt{\frac{1}{2}}\times \sqrt{12}-\sqrt{24}$.
1. Simplify the expression using properties of square roots and arithmetic operations:
\[
\begin{align*}
\sqrt{48}÷\sqrt{3}+\sqrt{\frac{1}{2}}\times \sqrt{12}-\sqrt{24} &= \sqrt{\frac{48}{3}} +... | null | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A rope of which a calf is tied is increased from some length to 18 m. The calf can graze an additional 565.7142857142857 m² of grassy ground. What was the original length of the rope? | The area that a calf can graze is given by the formula for the area of a circle, A = πr², where r is the length of the rope. When the rope is increased to 18 m, the calf can graze an additional 565.7142857142857 m² of grassy ground. This means that the difference in area between the original circle (with the original l... | 12 | Geometry | math-word-problem | Yes | Yes | orca_math | true |
24. As shown in Figure 11, in $\triangle A B C$, $M$ is the midpoint of side $A B$, $N$ is a point on side $A C$, and $\frac{A N}{N C}=2, C M$ intersects $B N$ at point $K$. If the area of $\triangle B C K$ is 1, then the area of $\triangle A B C$ is ( ).
(A) 3
(B) $\frac{10}{3}$
(C) 4
(D) $\frac{13}{3}$ | 24. C | C | Geometry | MCQ | Yes | Incomplete | cn_contest | false |
Given vectors $\overrightarrow{a}, \overrightarrow{b}$ satisfy $|\overrightarrow{a}|=1$, $|\overrightarrow{a}+ \overrightarrow{b}|= \sqrt{7}$, $\overrightarrow{a}\cdot (\overrightarrow{b}- \overrightarrow{a})=-4$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is ( )
A: $\dfrac{5\pi}{6}$
B: $\dfr... | Since $\overrightarrow{a}(\overrightarrow{b}- \overrightarrow{a})= \overrightarrow{a}\cdot \overrightarrow{b}- \overrightarrow{a}^{2}=-4$, and $\overrightarrow{a}^{2}=|\overrightarrow{a}|^{2}=1$,
then $\overrightarrow{a}\cdot \overrightarrow{b}=-3$.
Since $|\overrightarrow{a}+ \overrightarrow{b}|= \sqrt{7}$, it mea... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Without using a calculator, find the largest prime factor of $17^4 + 2\times17^3 + 17^2 - 16^4$. | First, simplify the expression given. Notice that we can rearrange some terms:
\[
17^4 + 2 \times 17^3 + 17^2 - 16^4 = (17^2)^2 + 2 \times 17^3 + 17^2 - (16^2)^2
\]
We can factor a bit to simplify:
\[
= 17^2(17^2 + 2 \times 17 + 1) - 16^4
\]
Now, rewrite $16^4 = (16^2)^2$ and use the difference of squares:
\[
= 17^2 \t... | 17 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Let $S$ be the set of nonzero real numbers. Let $g : S \to \mathbb{R}$ be a function such that:
(i) $g(1) = 2,$
(ii) $g\left(\frac{1}{x+y}\right) = g\left(\frac{1}{x}\right) + g\left(\frac{1}{y}\right)$ for all $x, y \in S$ such that $x + y \in S,$ and
(iii) $(x+y) g(x+y) = x^2 y^2 g(x) g(y)$ for all $x, y \in S$ such ... | Setting $x = y = \frac{z}{2}$ in (ii), we obtain
\[g\left(\frac{1}{z}\right) = 2g\left(\frac{2}{z}\right).\quad (1)\]
Setting $x = y = \frac{1}{z}$ in (iii), we derive
\[\frac{2}{z} g\left(\frac{2}{z}\right) = \frac{1}{z^4} g\left(\frac{1}{z}\right)^2.\quad (2)\]
From (1) and (2),
\[2g\left(\frac{2}{z}\right) = \frac... | 1 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Rachel buys a stock that decreases by $30\%$ on the first day, and then on the second day, the stock increases by $50\%$ of its value at the end of the first day. What was the overall percent change in Rachel's stock over the two days? | Let the original value of the stock be $x$.
1. At the end of the first day, the stock has fallen by $30\%$:
\[
\text{Value at the end of the first day} = x - 0.30x = 0.70x
\]
2. On the second day, the stock increases by $50\%$ of its value at the end of the first day:
\[
\text{Value at the end of the se... | 5\% | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
A journalist wants to take a photo of 4 volunteers and 2 elderly people they are helping, arranged in a row. The requirement is that the 2 elderly people must be adjacent but not at the ends of the row. The number of different arrangements is ( )
A: 72
B: 144
C: 240
D: 480 | Based on the problem, this is a step-by-step problem that uses the slotting method.
First, arrange the 4 volunteers in a row.
Then, consider the 2 elderly people as a single entity and insert them into one of the three slots formed by the volunteers (excluding the two ends).
Next, arrange the 2 elderly people.
... | \text{B} | Combinatorics | MCQ | Yes | Yes | cn_k12 | false |
a number when divided by a certain divisor left remainder 255 , when twice the number was divided by the same divisor , the remainder was 112 . find the divisor ? | Let's denote the number as \( N \) and the divisor as \( D \). According to the problem, when \( N \) is divided by \( D \), the remainder is 255. This can be written as:
\[ N = kD + 255 \]
where \( k \) is some integer.
Now, when twice the number \( 2N \) is divided by \( D \), the remainder is 112. This can be wri... | 398 | Number Theory | math-word-problem | Yes | Yes | orca_math | true |
6. Evaluate the solution and the answer provided below. Indicate all errors and shortcomings.
Determine whether the function $f(x)=\cos x+\cos (x \sqrt{2})$ is periodic.
Solution: First, we will prove that for any rational number $a$, the function $g(x)=\cos x+\cos (a x)$ is periodic. Indeed, let $a=\frac{p}{q}$. Set... | 2.II.6. Comment. First of all, let us show that the answer given in the solution is incorrect, since the function $f(x)=\cos x+\cos (x \sqrt{2})$ is not periodic. Indeed, $f(0)=1+1=2$. However, at no other point can the value of $f(x)$ be equal to two, because if $x=2 \pi k$, where $k \in \mathbb{Z}$, then $x \sqrt{2} ... | proof | Algebra | proof | Incomplete | Incomplete | olympiads | false |
In a school, there are 3 student groups and 5 teachers. Each student group can make 195 recycled materials and each teacher can make 70. However, the fair organizers have imposed a rule that a maximum of 500 products can be displayed for sell from each participating unit. If, after quality checks, only 80% of the produ... | First, let's calculate the total number of recycled materials produced by the student groups and the teachers before the quality check.
The 3 student groups can make a total of:
3 groups * 195 materials/group = 585 materials
The 5 teachers can make a total of:
5 teachers * 70 materials/teacher = 350 materials
So, th... | 748 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Find the value of a constant 'b' such that (7 * x + b * y) / (x - 2 * y) = 25 when x / 2y = 3 / 2. | To find the value of the constant 'b', we need to solve the equation:
(7 * x + b * y) / (x - 2 * y) = 25
Given that x / 2y = 3 / 2, we can solve for x in terms of y:
x = (3 / 2) * 2y
x = 3y
Now, we can substitute x = 3y into the original equation:
(7 * (3y) + b * y) / ((3y) - 2 * y) = 25
(21y + by) / (3y - 2y) = 2... | 4 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
During the first two weeks of January, the total rainfall in Springdale was 25 inches. If the rainfall during the second week was 1.5 times the rainfall during the first week, what was the rainfall during the second week of January? | Let's denote the rainfall during the first week as x inches. According to the problem, the rainfall during the second week was 1.5 times the rainfall during the first week, so the rainfall during the second week would be 1.5x inches.
The total rainfall for the two weeks is the sum of the rainfall for the first week an... | 15 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Two 6-sided dice, one red and one green, are rolled. What is the probability that the red die shows an odd number and the green die shows a prime number? | In this problem, the total possible outcomes when two dice are rolled remains unchanged at $36$, since each die has $6$ faces.
For the red die to show an odd number (similar to the original problem), we consider three possible outcomes: it must show either 1, 3, or 5.
The scenario changes for the green die: now it sh... | \frac{1}{4} | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
Given \\(x < 1\\), the maximum value of the function \\(f(x) = x + \dfrac{1}{x-1}\\) is __________. | **Analysis**
This problem examines the use of basic inequalities to find the maximum or minimum value of a function, under certain conditions: the variable must be positive, the product must be constant, and equality must be achievable. We transform the function's expression to create a product of a constant value wit... | -1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A function $f: \mathbb{R}\to \mathbb{R}$ is [i]essentially increasing[/i] if $f(s)\leq f(t)$ holds whenever $s\leq t$ are real numbers such that $f(s)\neq 0$ and $f(t)\neq 0$.
Find the smallest integer $k$ such that for any 2022 real numbers $x_1,x_2,\ldots , x_{2022},$ there exist $k$ essentially increasing functions... | null | null | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Given an arithmetic sequence $\{a_{n}\}$ with the sum of the first $n$ terms denoted as $S_{n}$, where $S_{3}=15$ and $a_{4}+a_{5}=20$.
$(1)$ Find the general formula for $\{a_{n}\}$.
$(2)$ Let the sum of the first $n$ terms of the sequence $\{\frac{1}{{a}_{n}{a}_{n+1}}\}$ be denoted as $T_{n}$. Find $T_{n}$. | ### Solution:
#### Part (1): Finding the General Formula for $\{a_{n}\}$
Given the arithmetic sequence $\{a_{n}\}$ with a common difference $d$, we have two conditions:
1. The sum of the third term, $S_{3}=15$, which implies $3a_1 + 3d = 15$.
2. The sum of the fourth and fifth terms, $a_{4}+a_{5}=20$, which can be ex... | T_{n} = \frac{n}{6n+9} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Example 7 Divide $n$ stones into two piles at random, and record the product of the number of stones in the two piles; then divide one of the piles into two piles, and record the product of the number of stones in these two piles; then divide one of the three piles into two piles, and record the product of the number o... | When dividing into two piles, there is one product; when dividing into three piles, there are two products; $\cdots$; when dividing into $n$ piles, there are $n-1$ products. Now, we seek the sum of these $n-1$ products, denoted as $S(n)$.
When $n=2$, $S(2)=1 \times 1=1=\frac{2 \times 1}{2}$
When $n=3$, $S(3)=1 \times 2... | S(n)=\frac{n(n-1)}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 2 Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfy $f(0)=1$, and for any $x, y \in \mathbf{R}$, there is $f(x y+1)=f(x) f(y)-f(y)-x+2$, find $f(x)$
(2004 National High School League Question) | Given the condition $f(x y+1)=f(x) f(y)-f(y)-x+2$, if we swap $x$ and $y$, we get:
$$
f(x y+1)=f(x) f(y)-f(x)-y+2
$$
Thus, $f(x) f(y)-f(y)-x+2=f(x) f(y)-f(x)-y+2$
which simplifies to $f(x)+y=f(y)+x$. Setting $y=0$, we get: $f(x)=x+1$ | f(x)=x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
䛈. Such a point $C$ is $\qquad$.
(The 1st Zu Chongzhi Cup Middle School Mathematics Invitational Competition) | Solution: (1) When $AB$ is the leg (right angle side) of the required $\therefore$ shape, we obtain the four points $C_{1}, C_{2}, C_{3}, C_{4}$ in the figure:
(2) When $AB$ is the hypotenuse of the required $\vdots$ angle $C_{6}$.
In summary, there are 6 points that satisfy the conditions of the problem.
Note: This pr... | 6 | Geometry | math-word-problem | Incomplete | Yes | cn_contest | false |
Let the ordered triples $(x,y,z)$ of complex numbers satisfy
\begin{align*}
x + yz &= 9, \\
y + xz &= 12, \\
z + xy &= 12.
\end{align*}
Find the sum of all distinct $x$ values from the solutions $(x_1, y_1, z_1), (x_2, y_2, z_2), \dots, (x_n, y_n, z_n)$. | Subtract the second equation from the third:
\[y + xz - z - xy = 0\]
This simplifies to:
\[y - z + x(z - y) = 0 \Rightarrow (y - z)(1-x) = 0\]
Thus, either $y = z$ or $x = 1$.
Case 1: $x = 1$
\[yz = 8, \quad y + z = 12 \]
The roots $y$ and $z$ of $t^2 - 12t + 8 = 0$ happen to be $(11, 1)$ and $(1, 11)$ (solving by qua... | 5 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
## Task 2 - 250932
Determine all pairs $(a, b)$ of two-digit numbers $a$ and $b$ for which the following holds: If one forms a four-digit number $z$ by writing $a$ and $b$ consecutively in this order, then $z=(a+b)^{2}$. | According to the problem statement, $z=100a+b=(a+b)^2=a^2+2ab+b^2$ or $a^2+(2b-100)a+b(b-1)=0$. For $b \geq 26$, however,
$0=a^2+(2b-100)a+b(b-1) \geq a^2-48a+26 \cdot 25=(a-24)^2-24^2+650=(a-24)^2+650-576>0$ which is a contradiction. Therefore, $b \leq 25$ must hold.
$b$ is the number formed by the last two digits o... | (20,25),(30,25) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The condition for the square root $\sqrt{x-1}$ to be meaningful is ______. | To ensure the square root $\sqrt{x-1}$ is meaningful, the expression inside the square root must be greater than or equal to zero. This leads us through the following steps:
1. Start with the condition inside the square root: $x - 1$.
2. Set it greater than or equal to zero: $x - 1 \geqslant 0$.
3. Solve for $x$ by ad... | x \geqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given the function $f(x) = \sin x + \cos x$, where $x \in \mathbb{R}$,
- (I) Find the value of $f\left(\frac{\pi}{2}\right)$.
- (II) Determine the smallest positive period of the function $f(x)$.
- (III) Calculate the minimum value of the function $g(x) = f\left(x + \frac{\pi}{4}\right) + f\left(x + \frac{3\pi}{4}\righ... | - (I) Since $f(x) = \sin x + \cos x$, we have $f\left(\frac{\pi}{2}\right) = \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1$.
Therefore, $f\left(\frac{\pi}{2}\right) = \boxed{1}$.
- (II) The function $f(x) = \sin x + \cos x$ can be rewritten using the sum to product identity as $f(x) = \sqrt{2} \sin \left(x + \... | -2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
For two non-coincident planes $\alpha$ and $\beta$, given the following conditions:
① There exists a plane $\gamma$ such that both $\alpha$ and $\beta$ are parallel to $\gamma$;
② There exists a plane $\gamma$ such that both $\alpha$ and $\beta$ are perpendicular to $\gamma$;
③ There are three non-collinear points in p... | For condition ①, if planes $\alpha$ and $\beta$ are not parallel, then there does not exist a plane $\gamma$ such that both $\alpha$ and $\beta$ are parallel to $\gamma$. Therefore, condition ① is incorrect.
For condition ②, if planes $\alpha$ and $\beta$ are parallel, then a plane $\gamma$ perpendicular to $\alpha$ m... | \text{B} | Geometry | MCQ | Yes | Yes | cn_k12 | false |
5. Given real numbers $a, b, c$ satisfy
$$
a+b+c=2, ab+bc+ca=0 \text{, and } abc=-1 \text{.}
$$
Then the set of real numbers $a, b, c$ is
$\qquad$ | 5. $\left\{\frac{1-\sqrt{5}}{2}, 1, \frac{1+\sqrt{5}}{2}\right\}$.
From the problem, we know
$$
a+b=2-c, a b+c(a+b)=0, a b=-\frac{1}{c},
$$
we get
$$
\begin{array}{l}
-\frac{1}{c}+c(2-c)=0 \\
\Rightarrow c^{3}-2 c^{2}+1=0 \\
\Rightarrow(c-1)\left(c^{2}-c-1\right)=0 \\
\Rightarrow c=\frac{1 \pm \sqrt{5}}{2} \text { or... | \left\{\frac{1-\sqrt{5}}{2}, 1, \frac{1+\sqrt{5}}{2}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For every natural number $n$ we define \[f(n)=\left\lfloor n+\sqrt{n}+\frac{1}{2}\right\rfloor\] Show that for every integer $k \geq 1$ the equation \[f(f(n))-f(n)=k\] has exactly $2k-1$ solutions. | 1. **Define the function \( f(n) \):**
\[
f(n) = \left\lfloor n + \sqrt{n} + \frac{1}{2} \right\rfloor
\]
This function essentially rounds down the value of \( n + \sqrt{n} + \frac{1}{2} \) to the nearest integer.
2. **Analyze the behavior of \( f(n) \):**
For large \( n \), \( \sqrt{n} \) is the domina... | 2k - 1 | Number Theory | proof | Yes | Yes | aops_forum | false |
Jungkook has 3 red bikes and Yoongi has 4 blue bikes. How many bikes are there in total? |
Jungkook has 3 red bikes and Yoongi has 4 blue bikes. In total, there are 3 + 4 = $\boxed{7}$ bikes. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | orca_math | true |
Find all positive integers $n$ for which all positive divisors of $n$ can be put into the cells of a rectangular table under the following constraints:
[list]
[*]each cell contains a distinct divisor;
[*]the sums of all rows are equal; and
[*]the sums of all columns are equal.
[/list] |
To solve the problem, we need to find all positive integers \( n \) for which all positive divisors of \( n \) can be arranged into a rectangular table such that each cell contains a distinct divisor, the sums of all rows are equal, and the sums of all columns are equal. This implies that the sum of the divisors of \(... | 1 | NaN | NaN | NaN | NaN | aops_forum | false |
10th Putnam 1950 Problem B2 An ellipse with semi-axes a and b has perimeter length p(a, b). For b/a near 1, is π(a + b) or 2π √(ab) the better approximation to p(a, b)? Solution | : π(a + b). Put b 2 = a 2 (1 - ε). The perimeter is 4 ∫ 0 π/2 (a 2 sin 2 t + b 2 cos 2 t) 1/2 dt = 4a ∫ (1 - ε cos 2 t) 1/2 dt = 4a ∫ ( 1 - 1/2 ε cos 2 t - 1/8 ε 2 cos 4 t + O(ε 4 ) dt = 2π a ( 1 - ε/4 - 3ε 2 /64 + O(ε 4 ) ). b = a(1 - ε) 1/2 = a(1 - ε/2 - ε 2 /8 + O(ε 3 ) ). Hence π (a + b) = 2π a (1 - ε/4 - 4ε 2 /64 ... | \pi(b) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. (10 points) $n$ pirates divide gold coins. The 1st pirate takes 1 coin first, then takes $1 \%$ of the remaining coins; then, the 2nd pirate takes 2 coins, then takes $1 \%$ of the remaining coins; the 3rd pirate takes 3 coins, then takes $1 \%$ of the remaining coins; ... the $n$-th pirate takes $n$ coins, then ta... | 【Solution】Solution: According to the problem, we know:
The $n$-th pirate takes $n$ coins first, then takes 1% of the remaining, and the coins are all taken. This means the remaining coins are 0. The $n$-th pirate actually took $n$ coins.
The $n$-1-th pirate takes $n-1$ coins first, then takes 1% of the remaining. Sinc... | 9801 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Prove: The equation $x\left(z^{2}-y^{2}\right)=z\left(x^{2}+y^{2}\right)$ has no positive integer solutions $(x, y, z)$. | Assume the original equation has positive integer solutions. Transform the original equation into
$$
x z^{2}-\left(x^{2}+y^{2}\right) z-x y^{2}=0 .
$$
Then the discriminant
$$
\Delta=\left(x^{2}+y^{2}\right)^{2}+(2 x y)^{2}
$$
is a perfect square.
Let $a=x^{2}+y^{2}, b=2 x y$. Then
$a>b$, and $a^{2}-b^{2}=\left(x^{2}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false |
In the Cartesian coordinate system, the coordinates of point $P(-2, \sqrt{5})$ with respect to the $x$-axis are ______. | To find the coordinates of point $P(-2, \sqrt{5})$ with respect to the $x$-axis, we need to reflect the point across the $x$-axis. The reflection of a point across the $x$-axis changes the sign of the $y$-coordinate while keeping the $x$-coordinate the same.
Given the point $P(-2, \sqrt{5})$, its reflection across the... | (-2, -\sqrt{5}) | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
Simplify $\dfrac {\cos 2\alpha}{4\sin ^{2}\left( \dfrac {\pi}{4}+\alpha\right)\tan \left( \dfrac {\pi}{4}-\alpha\right)}=$ ()
A: $\cos \alpha$
B: $\sin \alpha$
C: $1$
D: $\dfrac {1}{2}$ | First, consider the denominator: $4\sin ^{2}\left( \dfrac {\pi}{4}+\alpha\right)\tan \left( \dfrac {\pi}{4}-\alpha\right)=4 \dfrac {1-\cos \left( \dfrac {\pi}{2}+2\alpha\right)}{2}\cdot \dfrac {1-\tan \alpha}{1+\tan \alpha}$
$=2(1+\sin 2\alpha)\cdot \dfrac {\cos \alpha-\sin \alpha}{\cos \alpha+\sin \alpha}=2(\cos ^{2... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Let the function $f(x)$ have an inverse function denoted as $f^{-1}(x)$, and let $y=f^{-1}(-x+2)$ pass through the point $(-1, 2)$. Determine the fixed point through which $y=f(x-1)$ passes. | Since the function $y=f^{-1}(-x+2)$ passes through the point $(-1, 2)$, it implies that the function $y=f^{-1}(-x)$ passes through the point $(-3, 2)$.
Therefore, $y=f^{-1}(x)$ passes through the point $(-3, -2)$. Consequently, $y=f(x)$ passes through the point $(-2, -3)$.
Thus, $y=f(x-1)$ passes through the fixed po... | (-3, -3) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given the function $f(x) = \ln x + \frac{1}{x} + ax$ where $x \in (0, +\infty)$ and $a$ is a real constant,
(1) If the function $f(x)$ has an extremum at $x=1$, find the minimum value of $f(x)$ at this point;
(2) If the function $f(x)$ has an extremum in the interval $(2, 3)$, determine the range of values for the real... | (1) Firstly, we differentiate the function $f(x)$ with respect to $x$ and set the derivative equal to zero to find the extremum point:
$$ f'(x) = \frac{d}{dx} (\ln x + \frac{1}{x} + ax) = \frac{1}{x} - \frac{1}{x^2} + a $$
Setting $f'(1) = 0$, we get:
$$ f'(1) = \frac{1}{1} - \frac{1}{1^2} + a = 0 \Rightarrow a = 0 $$
... | x_1 \leq 1 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
The non-decreasing functions $f,g: \mathbb{R}\rightarrow \mathbb{R}$ are such that $f(r)\leq g(r)$ for $\forall$ rational numbers $r$. Is it true that $f(x)\leq g(x)$ for $\forall$ real numbers $x$? | 1. **Define the functions \( f \) and \( g \):**
- Let \( f(x) \) be defined as:
\[
f(x) =
\begin{cases}
x & \text{if } x < \pi \\
\pi + 4 & \text{if } x = \pi \\
x + 5 & \text{if } x > \pi
\end{cases}
\]
- Let \( g(x) \) be defined as:
\[
g(x) =
\begin{c... | null | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
If $a$, $b$, $c$, $d$, $e$, and $f$ are integers for which $1728x^4 + 64 = (ax^3 + bx^2 + c)(dx^2 + ex + f)$ for all $x$, then what is $a^2+b^2+c^2+d^2+e^2+f^2$? | Use the identity $1728x^4 + 64 = (12x^2)^2 + 8^2$ to factorize the polynomial. Recognizing this as the sum of two squares, we can apply the factorization:
\[
1728x^4 + 64 = (12x^2 + 8i)(12x^2 - 8i),
\]
where $i$ is the imaginary unit. However, this factorization involves complex numbers. Since we are given that $a$, $b... | 416 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
In trapezoid $K L M N$, the base $L M$ is 17, and angle $L K N$ is acute and twice the angle $K N M$.
A circle with its center on line $L M$ is tangent to lines $K M$, $K N$, and segment $M N... | See the solution to problem 54334.
## Answer
$$
84+5 \sqrt{34}
$$ | 84+5\sqrt{34} | Geometry | math-word-problem | Incomplete | Incomplete | olympiads | false |
A regular unit $7$-simplex is a polytope in $7$-dimensional space with $8$ vertices that are all exactly a distance of $ 1$ apart. (It is the $7$-dimensional analogue to the triangle and the tetrahedron.) In this $7$-dimensional space, there exists a point that is equidistant from all $8$ vertices, at a distance $d$. D... | null | null | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let $b_n = 7^n + 9^n$. Determine the remainder upon dividing $b_{96}$ by $50$. | First, observe that using Euler's totient theorem, $\phi(50) = 20$, implies $x^{20} \equiv 1 \pmod{50}$ for $x$ co-prime with $50$. Since $7$ and $9$ both have $\gcd(7,50) = \gcd(9,50) = 1$, it follows:
1. Calculate $7^{96} \pmod{50}$
- Since $96 \equiv 16 \pmod{20}$, we have $7^{96} \equiv 7^{16} \pmod{50}$.
- ... | 2 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
32. There are $\qquad$ triangles in the figure below.
The text above has been translated into English, preserving the original text's line breaks and format. | Answer: 48 | 48 | Geometry | math-word-problem | Incomplete | Problem not solved | olympiads | false |
6. Suppose the axial section of an inverted conical container is an equilateral triangle. Water is poured into this container, and a solid sphere with radius $r$ is placed inside. If the sphere is submerged in the water and is tangent to the container walls and the water surface, then the height of the water level afte... | 6. D.
Figure 3 shows the axial section $P A B$ of a cone, with the center of the sphere being O. It is easy to see that,
$$
\begin{array}{c}
P C=O C+P O \\
=r+2 r=3 r, \\
A C=\sqrt{3} r, \\
V_{P A B}=\frac{1}{3} \pi(\sqrt{3} r)^{2} \cdot 3 r=3 \pi r^{3} .
\end{array}
$$
Let the water surface after removing the sphere... | D | Geometry | MCQ | Yes | Yes | cn_contest | false |
2. Given quadrilateral $A B C D$, where $A D+D B+B C=16$, find the maximum value of the area $S$ of quadrilateral $A B C D$. | 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 | math-word-problem | Incomplete | Incomplete | olympiads | false |
1. The Haier Brothers go on an adventure together, the elder brother has 17 more energy bars than the younger brother, after the elder brother gives 3 energy bars to the younger brother, the elder brother still has $\qquad$ more energy bars than the younger brother. | \begin{tabular}{c}
1 \\
\hline 11
\end{tabular} | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false | |
Triangle \( \triangle PQR \) has side-lengths \( PQ = 15, QR = 30, \) and \( PR = 20. \) The line through the incenter of \( \triangle PQR \) parallel to \( \overline{QR} \) intersects \( \overline{PQ} \) at \( X \) and \( \overline{PR} \) at \( Y. \) What is the perimeter of \( \triangle PXY? \) | Let \( I \) be the incenter of \( \triangle PQR \). Since \( \overline{XI} \parallel \overline{QR} \) and \( \overline{PI} \) is the angle bisector of \( \angle QPR \), we find:
\[
\angle PXI = \angle PRI = \angle PIQ = \frac{1}{2}\angle QPR
\]
This implies by alternate interior angles and the base angles of isosceles ... | 35 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
24.8*. The origin is the center of symmetry of a convex figure with an area greater than 4. Prove that this figure contains at least one point with integer coordinates, different from the origin (Minkowski). | 24.8. Consider all convex figures obtained from a given one by translations by vectors, both coordinates of which are even. We will prove that at least two of these figures intersect. The original figure can be enclosed in a circle of radius $R$ centered at the origin, and $R$ can be chosen as an integer. Let's take th... | proof | Geometry | proof | Yes | Yes | olympiads | false |
In a game, there are two identical, uniform tetrahedral dice, each with the numbers $1$, $2$, $3$, $5$ on their faces. If both dice are rolled simultaneously and $x$ represents the sum of the numbers on the two faces facing down, what is the probability that $x$ is not less than $6$?
A: $\dfrac {1}{8}$
B: $\dfrac {1}{... | In a game, there are two identical, uniform tetrahedral dice,
Each face of the dice has the numbers $1$, $2$, $3$, $5$. When both dice are rolled simultaneously, let $x$ represent the sum of the numbers on the two faces facing down.
The total number of basic events is $n=4 \times 4 = 16$.
The basic events where $x$ ... | D | Combinatorics | MCQ | Yes | Yes | cn_k12 | false |
24.6*. The vertices of triangle $A B C$ are located at the nodes of an integer lattice, and there are no other nodes on its sides, while there is exactly one node $O$ inside it. Prove that $O$ is the point of intersection of the medians of triangle $A B C$.
See also problem 23.32.
## §2. Various Problems | 24.6. According to Pick's formula $S_{A O B}=S_{B O C}=S_{C O A}=1 / 2$, which means that $O-$ is the point of intersection of the medians of triangle $A B C$ (see problem 4.2). | proof | Geometry | proof | Yes | Yes | olympiads | false |
If the sum of the monomials $\dfrac{1}{3}a^{m}b^{3}$ and $-2a^{2}b^{n}$ is still a monomial, then the solution to the equation $\dfrac{x-7}{n}-\dfrac{1+x}{m}=1$ is ( ).
A: $-23$
B: $23$
C: $-29$
D: $29$ | Given that the sum of the monomials $\dfrac{1}{3}a^{m}b^{3}$ and $-2a^{2}b^{n}$ is still a monomial, we conclude that these monomials must be like terms. For monomials to be like terms, their variables and corresponding exponents must match. Therefore, we have:
1. For the exponents of $a$, it must be that $m = 2$.
2. ... | \text{A} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Example 8 Let $a_{1}, a_{2}, \cdots, a_{k}, \cdots$ be distinct positive integers. Prove that for any positive integer $n$, we have
$$
\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} \sum_{k-1}^{n} \frac{1}{k} \text {. }
$$
(20th IMO) | $\begin{array}{l}\text { Prove: } \because\left(\sum_{k=1}^{n} \frac{1}{k}\right)^{2}=\left(\sum_{k=1}^{n} \frac{\sqrt{a_{k}}}{k} \cdot \frac{1}{\sqrt{a_{k}}}\right)^{2} \\ \leqslant\left(\sum_{k=1}^{n} \frac{a_{k}}{k^{2}}\right)\left(\sum_{k=1}^{n} \frac{1}{a_{k}}\right) \leqslant\left(\sum_{k=1}^{n} \frac{a_{k}}{k^{2... | proof | Algebra | proof | Incomplete | Incomplete | cn_contest | false |
Shaypak I.A.
Let $a$ be the positive root of the equation $x^{2017}-x-1=0$, and $b$ be the positive root of the equation $y^{4034}-y=$ $3a$.
a) Compare $a$ and $b$.
b) Find the tenth digit after the decimal point of the number $|a-b|$. | Let $n=2017$.
a) Note that $b>1$, since $b^{2n}-b=3a>0$. Similarly, $a>1$. Therefore, $1+a+a^2>3a$, as this inequality is equivalent to the inequality $(1-a)^2>0$. Thus, $a^{2n}-a=1+a+a^2>3a=b^{2n}-b$. Since the function $f(t)=t^{2n}-t$ is strictly increasing for $t>1$ (for these $t$
$f'(t)=2nt^{2n-1}-1>0$), then $a>... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Distribute four students, named A, B, C, and D, into three different classes, with each class having at least one student. Students A and B cannot be in the same class. The number of different ways to distribute them is ______. | We use the exclusion method. First, we ignore the condition that students A and B cannot be in the same class. The number of ways to divide the four students into three groups is $C_4^2 = 6$. Then, assigning these three groups to three different classes can be done in $A_3^3 = 6$ ways. However, we also need to consider... | 30 | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false |
3. [4 points] Solve the inequality $\left(\sqrt{x^{3}-18 x-5}+2\right) \cdot\left|x^{3}-4 x^{2}-5 x+18\right| \leqslant 0$. | Answer: $x=1-\sqrt{10}$.
Solution. The first factor on the left side is positive for any admissible value of $x$. Since the second factor is always non-negative, the inequality on the domain of definition (i.e., under the condition $x^{3}-18 x-5 \geqslant 0$) is equivalent to the equation $x^{3}-4 x^{2}-5 x+18=0$. One... | 1-\sqrt{10} | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
What is the composite rate percent when the compound interest on Rs. 4000 amounts to Rs. 5500 in 3 years, with variable interest rates of X% for the first year, Y% for the second year, and Z% for the third year? | To find the composite rate percent, we need to calculate the effective interest rate over the 3 years that would result in the same final amount as the variable interest rates.
Let's denote the principal amount as P, the final amount as A, and the interest rates for the first, second, and third years as X%, Y%, and Z%... | 11.1\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
2. If the function
$$
f(x)=x^{2}-2 a x-2 a|x-a|+1
$$
has exactly three zeros, then the value of the real number $a$ is $\qquad$. | 2. 1.
Let $t=|x-a|(t \geqslant 0)$.
Then the original problem is equivalent to the equation $t^{2}-2 a t+1-a^{2}=0$ having two roots $t_{1}=0, t_{2}>0$.
Upon verification, $a=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Problem 5. Prove that for any two real numbers $a$ and $b$ there exists a real number $c \in(0,1)$, such that
$$
\left|a c+b+\frac{1}{c+1}\right|>\frac{1}{24}
$$
|
Solution: Consider $f(x)=a x+\frac{1}{x+1}$ and let $m$ and $M$ be respectively the minimum and the maximum value of the function $f$ in the interval $[0,1]$. Since $f(0)=1, f(1)=a+\frac{1}{2}$ and $f^{\prime}(x)=a-\frac{1}{(x+1)^{2}}$ there are four cases for the difference $M-m$.
1. $a \leq \frac{1}{4}$. Thus $f^{\... | proof | Algebra | proof | Yes | Yes | olympiads | false |
In the expansion of \\({(1+x)}^{2}+{(1+x)}^{3}+\cdots+{(1+x)}^{9}\\), the coefficient of the \\({x}^{2}\\) term is
A: \\(119\\)
B: \\(120\\)
C: \\(121\\)
D: \\(720\\) | **Analysis**
This question examines the specific term of the binomial expansion, using the general term formula of the expansion combined with the properties of binomial coefficients to obtain the result.
**Solution**
The coefficient of the \\({x}^{2}\\) term in the expansion of \\({(1+x)}^{2}+{(1+x)}^{3}+\cdots+{(1... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Fran's school just instituted a school uniform policy. Each student needs to buy five complete uniforms, each consisting of pants, shirt, tie, socks, jacket, and shoes. The pants cost $20, the shirt costs twice as much as the pants, the tie costs 1/5 as much as the shirt, the socks cost $3/pair, the jacket costs three ... | First, let's calculate the cost of one complete uniform without any discounts:
- Pants: $20
- Shirt: $20 * 2 = $40
- Tie: $40 / 5 = $8
- Socks: $3
- Jacket: $40 * 3 = $120
- Shoes: $40
Now, let's add up the cost of one complete uniform:
Total cost of one uniform = $20 (pants) + $40 (shirt) + $8 (tie) + $3 (socks) + ... | \$1117.77 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
## Task 1
Riddle: I am younger than 13 and older than 9 years old, how old can I be? | I can be 10, 11 or 12 years old. | 10,11,12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Given the sequence ${{a_n}}$ that satisfies ${{a_2}=102,{{a_{n+1}}-{{a_n}}=4n,n\in {\mathbb{N}^*}}}$, find the minimum value of the sequence ${\{\frac{{{a_n}}}{n}\}}$ ( ).
A: $25$
B: $26$
C: $27$
D: $28$ | This problem involves finding the general term of a sequence from a given recursive formula and using basic inequalities to find the minimum value. It tests the ability to synthesize and apply knowledge to solve problems.
Using cumulative sum, we can find $a_n$ and then express $\frac{{a_n}}{n}$. After that, we can us... | 26 | Algebra | MCQ | Yes | Yes | cn_k12 | false |
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