instruction_seed stringlengths 2 10.7k | solution stringlengths 0 10.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
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Given the proposition $p$: $\exists x 1$
B: $\exists x 1$
C: $\forall x 1$
D: $\exists x \geqslant 1$, $x^{2} > 1$ | Solution: Since the negation of an existential proposition is a universal proposition, we can derive that
the negation of proposition $p$: $\exists x 1$,
Therefore, the correct choice is $\boxed{C}$.
By applying the negation of an existential proposition as a universal proposition, along with the changes in quan... | C | Logic and Puzzles | MCQ | Incomplete | Incomplete | cn_k12 | false |
\begin{aligned} & \text { The Law of Sines } \\ & {[\text { Area of a triangle (using two sides and the included angle).] }}\end{aligned}
Two circles with centers $O_{1}$ and $O_{2}$, lying on the side $M N$ of triangle $M P N$, touch each other and intersect sides $M P$ and $P N$ at points $M, D$, and $N, C$ respecti... | Compose a system of trigonometric equations with respect to angles $P M N$ and $P N M$.
## Solution
Let $\angle P M N=\alpha, \angle P N M=\beta$. Then $\angle D O_{1} N=2 \alpha, \angle C O_{2} M=2 \beta$. Since
$$
\frac{S_{\triangle M C O_{2}}}{S_{\triangle O_{1} D N}}=\frac{O_{2} C \cdot O_{2} M \sin 2 \beta}{O_{... | \frac{81(\sqrt{3}-1)}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A lucky number is one whose sum of digits is divisible by 7. For example, 7, 25, and 849 are lucky numbers. The smallest pair of lucky numbers is 7 and 16.
(a) Find eight consecutive numbers, of which two are lucky numbers.
(b) Find 12 consecutive numbers, such that none of them are lucky.
(c) Show that any sequence... | (a) The sequence of eight consecutive numbers from 52 to 59 has exactly two lucky numbers: 52 and 59. Another example is any sequence of 8 numbers that contains 59 and 61, for example: $55,56,57,58,59,60,61,62$.
(b) Two examples: $994, \ldots, 1005$ and $7994, \ldots, 8005$. There are more examples, find some.
(c) Le... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Among the propositions in the form of "If $p$, then $q$", $q$ is a necessary condition of $p$ in which option?
A: If $x=1$, then $x^{2}=1$
B: If $ac=bc$, then $a=b$
C: If $mn$ is an irrational number, then $m$ and $n$ are irrational numbers
D: If the diagonals of a quadrilateral are perpendicular, then the quadrila... | To analyze the propositions given in the form of "If $p$, then $q$" and determine where $q$ is a necessary condition of $p$, let's evaluate each option step by step:
- **Option A**: If $x=1$, then $x^{2}=1$.
- Forward implication: If we have $x=1$, squaring both sides gives us $x^{2}=1^2=1$. So, $p$ leads to $q$.
... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false |
21-4 (G43, USA) Given a plane $\pi$, a point $P$ on $\pi$, and a point $Q$ outside $\pi$, find a point $R$ on $\pi$ such that the ratio $\frac{Q P+P R}{Q R}$ is maximized. | 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 | Yes | Problem not solved | olympiads | false |
The average mark of the students of a class in a particular exam is 80. If 5 students whose average mark in that exam is a certain value are excluded, the average mark of the remaining will be 90. There were 10 students who wrote the exam. What was the average mark of the excluded students? | Let's denote the total number of students as \( N \), the average mark of all students as \( A \), the average mark of the excluded students as \( E \), and the average mark of the remaining students as \( R \). We also know the number of excluded students, which we'll denote as \( X \).
From the information given:
- ... | 70 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
21. Grandma has two granddaughters, Grandma's age is a two-digit number, the two digits that make up this two-digit number are exactly the ages of the two granddaughters. The younger granddaughter is 5 years old, the sum of the three ages is 69 years, Grandma $\qquad$ years old. | $57$ | 57 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A trapezoid is given in which one base is twice as large as the other. Use one ruler (no divisions) to draw the midline of this trapezoid. | 1. **Constructing the midpoint of base \( AD \)**
- Let the trapezoid be \(ABCD\) with bases \(BC\) and \(AD\) where \(AD > BC\).
- Construct point \(P\) as the intersection of lines \(AB\) and \(CD\).
- Construct point \(Q\) as the intersection of lines \(AC\) and \(BD\).
- It is a well-known fact that th... | null | Geometry | other | Yes | Yes | aops_forum | false |
Solve the equation \[ \frac{x+6}{x-3} = 4 \] for $x$. | Starting with the given equation:
\[ \frac{x+6}{x-3} = 4 \]
Perform cross-multiplication to eliminate the fraction:
\[ x + 6 = 4(x - 3) \]
Expanding and simplifying the right-hand side:
\[ x + 6 = 4x - 12 \]
Bring all terms involving \( x \) to one side and constants to the other:
\[ x - 4x = -12 - 6 \]
\[ -3x = -18... | 6 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Let $f(x)$ and $g(x)$ be odd and even functions defined on $\mathbb{R}$, respectively. When $x 0$, and $g(-3) = 0$. Then, the solution set of the inequality $f(x)g(x) < 0$ is ( )
A: $\left( -3,0 \right)\cup \left( 3,+\infty \right)$
B: $\left( -3,0 \right)\cup \left( 0,3 \right)$
C: $\left( -\infty ,-3 \right)\c... | **Analysis**
This problem examines the odd and even properties of functions, constructing the function model $h(x)=f(x)·g(x)$, and uses derivatives to study monotonicity to obtain the solution set of the inequality.
**Solution**
Let $h(x)=f(x)·g(x)$,
Then $h(-x)=f(-x)·g(-x)=-f(x)·g(x)$,
So $h(x)$ is an odd functio... | \text{D} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
in country z , 10 % of the people do not have a university diploma but have the job of their choice , and 25 % of the people who do not have the job of their choice have a university diploma . if 20 % of the people have the job of their choice , what percent of the people have a university diploma ? | Let's denote the total population of Country Z as 100% for simplicity.
According to the information given:
- 10% of the people do not have a university diploma but have the job of their choice.
- 20% of the people have the job of their choice.
Since 10% of the people have the job of their choice without a university... | 30\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A batsman has an average of 58 runs from 46 innings in a cricket tournament. The difference between his highest and lowest score is 150 runs. After excluding his highest and lowest scores, his average remains the same at 58 runs. Additionally, he scores exactly a century (100 runs) in 5 of his innings and the rest of h... | Let's denote the batsman's highest score as H and his lowest score as L. We are given that the difference between the highest and lowest score is 150 runs, so we can write:
H - L = 150 (1)
We are also given that the batsman's average remains the same after excluding his highest and lowest scores. This means that the... | 143 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Let be given three quadratic polynomials:
$P_1(x) = x^2 + p_1x+q_1, P_2(x) = x^2+ p_2x+q_2, P_3(x) = x^2 + p_3x+q_3$.
Prove that the equation $|P_1(x)|+|P_2(x)| = |P_3(x)|$ has at most eight real roots. | null | null | Algebra | proof | Yes | Yes | aops_forum | false |
6. Solve the system of equations for positive real numbers:
$$
\frac{1}{x y}=\frac{x}{z}+1, \quad \frac{1}{y z}=\frac{y}{x}+1, \quad \frac{1}{z x}=\frac{z}{y}+1
$$
|
Solution: The given system reduces to
$$
z=x^{2} y+x y z, x=y^{z}+x y z, y=z^{2} x+x y z
$$
Hence
$$
z-x^{2} y=x-y^{2} z=y-z^{2} x
$$
If $x=y$, then $y^{2} z=z^{2} x$ and hence $x^{2} z=z^{2} x$. This implies that $z=x=y$. Similarly, $x=z$ implies that $x=z=y$. Hence if any two of $x, y, z$ are equal, then all are... | \frac{1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The following propositions are given: ① Vertical angles are equal; ② Complementary angles of equal angles are equal; ③ Corresponding angles are equal; ④ If $b$ is parallel to $a$ and $c$ is parallel to $a$, then $b$ is parallel to $c$; ⑤ If an angle has its sides parallel to the sides of another angle, then the two ang... | To determine the true propositions among the given, we analyze each one step by step:
1. **Proposition ①: Vertical angles are equal.** This is a fundamental theorem in geometry stating that when two lines intersect, the opposite (or vertical) angles formed are equal. Thus, proposition ① is $\text{true}$.
2. **Proposi... | C | Geometry | MCQ | Yes | Yes | cn_k12 | false |
Two trains are moving in opposite directions at 100 kmph and 120 kmph. Their lengths are 500 m and 700 m. How long does it take for the slower train to cross the faster train in seconds? | When two trains are moving in opposite directions, their relative speed is the sum of their individual speeds.
So, the relative speed of the two trains is:
100 kmph + 120 kmph = 220 kmph
To find the time it takes for the slower train to cross the faster train, we need to consider the total distance to be covered, wh... | 19.63 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
The roots of the polynomial $P(x) = x^4 -10x^3 + 35x^2 -51x + 26$ form the side lengths of a quadrilateral. The maximum area of such a quadrilateral is of the form $\sqrt{n}$ for some positive integer $n$. Find $n$. | 1. **Identify the roots of the polynomial:**
The polynomial given is \( P(x) = x^4 - 10x^3 + 35x^2 - 51x + 26 \). Let the roots of this polynomial be \( a, b, c, \) and \( d \).
2. **Sum and product of the roots:**
By Vieta's formulas, we know:
\[
a + b + c + d = 10,
\]
\[
ab + ac + ad + bc + bd +... | 224.5 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
If the sum of interior angles of a polygon is $1260^{\circ}$, then the number of sides of the polygon is ( )
A: $8$
B: $9$
C: $10$
D: $11$ | To find the number of sides of a polygon given the sum of its interior angles, we use the formula for the sum of interior angles of a polygon, which is $\left(n-2\right)\cdot 180^{\circ}$, where $n$ is the number of sides. Given that the sum of interior angles is $1260^{\circ}$, we set up the equation:
\[
\left(n-2\ri... | \text{B} | Geometry | MCQ | Yes | Yes | cn_k12 | false |
Let the angle between vectors **a** and **b** be $\theta$, **a** = (2,1), and **a** + **2b** = (4,5). Then $\cos \theta$ equals
A.
B.
C.
D. | Given that **a** = (2,1) and **a** + **2b** = (4,5), we can find vector **b** by solving the equation **a** + **2b** = (4,5). Subtracting **a** from both sides gives us **2b** = (2,4), so **b** = (1,2).
The cosine of the angle between two vectors **a** and **b** can be calculated using the dot product formula: $\cos \... | \text{D} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ satisfy $|\overrightarrow{a}|=1$ and $\overrightarrow{a}\cdot \overrightarrow{b}=-1$, then $\overrightarrow{a}\cdot (2\overrightarrow{a}-\overrightarrow{b})$ equals $\left(\ \ \right)$
A: $4$
B: $3$
C: $2$
D: $0$ | To solve the problem, we start with the expression given and apply the properties of dot product and the given conditions:
\[
\overrightarrow{a}\cdot (2\overrightarrow{a}-\overrightarrow{b}) = 2\overrightarrow{a}\cdot \overrightarrow{a} - \overrightarrow{a}\cdot \overrightarrow{b}
\]
Given that $|\overrightarrow{a}|=... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false |
$$
61609
$$
99 children stand in a circle, each initially having a ball. Every minute, each child with a ball throws their ball to one of the two neighbors; if two balls land with the same child, one of these balls is lost irretrievably. What is the minimum time after which only one ball may remain with the children? | Number the children and balls clockwise from 1 to 99.
Example. Suppose children 1 and 2 toss the first ball to each other. The other balls with odd numbers are always thrown counterclockwise until they reach the second child, who discards them (this happens after an odd number of minutes, and at that moment he also re... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Given the function $y=\cos^2x+ \frac{\sqrt{3}}{2}\sin 2x- \frac{1}{2}$, with $x \in \left(0, \frac{\pi}{2}\right)$, find the range of this function. | Let's rewrite the function using the double angle identity for cosine.
\[
y=\cos^2x + \frac{\sqrt{3}}{2}\sin 2x - \frac{1}{2} = \frac{1+\cos 2x}{2} + \frac{\sqrt{3}}{2}\sin 2x - \frac{1}{2}
\]
Now, we can simplify the expression by combining like terms:
\[
y = \frac{\sqrt{3}}{2}\sin 2x + \frac{1}{2}\cos 2x
\]
At th... | \left(-\frac{1}{2}, 1\right] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Define the characteristic function of a subset $A$ of the universal set $U$ as $f_A(x) = \begin{cases} 1, & x \in A \\ 0, & x \in \complement_U A \end{cases}$, where $\complement_U A$ denotes the complement of set $A$ in the universal set $U$. Given $A \subseteq U$, $B \subseteq U$, consider the following statements: ... | For statement ①, $f_A(x) = \begin{cases} 1, & x \in A \\ 0, & x \in \complement_U A \end{cases}$, $f_B(x) = \begin{cases} 1, & x \in B \\ 0, & x \in \complement_U B \end{cases}$,
since $\complement_U A$ may contain elements of $B$, but $\complement_U B$ cannot contain elements of $A$
∴ $f_A(x) \leq f_B(x)$,
i.e.,... | ①②③ | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given the function $f(x)= \sqrt {3}(\sin ^{2}x-\cos ^{2}x)+2\sin x\cos x$.
(I) Find the smallest positive period of $f(x)$;
(II) If $x\in\[- \frac {π}{3}, \frac {π}{3}\]$, find the range of $f(x)$ and the interval where it is strictly increasing. | (I) First, we simplify the function $f(x)$. Using the double angle identities for sine and cosine, we have:
$$\begin{align}
f(x)&=\sqrt {3}(\sin ^{2}x-\cos ^{2}x)+2\sin x\cos x \\
&=-\sqrt {3}(\cos ^{2}x-\sin ^{2}x)+2\sin x\cos x \\
&=\sin 2x-\sqrt {3}\cos 2x \\
&=2\sin \left(2x-\frac{\pi}{3}\right)
\end{align}$$
The s... | \left[-\frac{\pi}{12},\frac{\pi}{3}\right] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
The function $f(x)$, for any natural number $x$, satisfies $f(x+1) = f(x) + 1$, and $f(0) = 1$. Then, $f(10) =$ （ ）
A: 11
B: 12
C: 13
D: 14 | Let $x = n$, where $n \in \mathbb{N}^{*}$,
Since $f(x+1) = f(x) + 1$ and $f(0) = 1$,
Therefore, $f(n+1) = f(n) + 1$, and $f(1) = 2$,
Therefore, the sequence $\{f(n)\}$ is an arithmetic sequence with the first term as 2 and common difference as 1,
Therefore, $f(10) = 2 + (10 - 1) = 11$,
Hence, the correct choice is... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false |
# Problem 10. (2 points)
The sum of a geometric progression consisting of four natural numbers is 1417. Find the second largest of these numbers.
If there are multiple possible answers, list them in ascending order separated by commas or semicolons. | # Problem 10. (2 points)
The sum of a geometric progression consisting of four natural numbers is 1417. Find the second largest of these numbers.
If there are multiple possible answers, list them in ascending order separated by commas or semicolons.
Answer: 300 | 300 | Number Theory | math-word-problem | Yes | Problem not solved | olympiads | false |
If $x$ is real, calculate the maximum integer value of
\[
\frac{4x^2 + 12x + 20}{4x^2 + 12x + 8}.
\] | First, start by simplifying the expression:
\[
\frac{4x^2 + 12x + 20}{4x^2 + 12x + 8} = \frac{(4x^2 + 12x + 8) + 12}{4x^2 + 12x + 8} = 1 + \frac{12}{4x^2 + 12x + 8}.
\]
Our goal is now to minimize $4x^2 + 12x + 8$ to maximize the fraction.
Completing the square:
\[
4x^2 + 12x + 8 = 4(x^2 + 3x + 2) = 4(x^2 + 3x + \frac... | 12001 | Algebra | math-word-problem | Yes | Incomplete | synthetic_math | true |
13.044. Sugar and granulated sugar were delivered to the store in 63 bags, totaling 4.8 tons, with the number of bags of granulated sugar being $25 \%$ more than those of sugar. The mass of each bag of sugar was $3 / 4$ of the mass of a bag of granulated sugar. How much sugar and how much granulated sugar were delivere... | Solution.
Let $x$ be the number of bags of sugar; $1.25 x$ be the number of bags of sand. According to the condition, $x + 1.25 x = 63$, from which $x = 28$ bags of sugar; $1.25 \cdot 28 = 35$ bags of sand. Now let the total be $y$ tons of sand; $4.8 - y$ tons of sugar. Then $\frac{y}{35}$ tons is the weight of one ba... | 1.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
a) Prove that for every $\epsilon>0$ there is a positive integer $n$ and real
numbers $\lambda_{1},\dots,\lambda_{n}$ such that
$$\max_{x\in [-1,1]}|x-\sum_{k=1}^{n}\lambda_{k}x^{2k+1}|<\epsilon.$$
b) Prove that for every odd continuous function $f$ on $[-1,1]$ and for every $\epsilon>0$ there is a positive integer $n... | ### Part (a)
1. **Define the function to approximate:**
We need to approximate the function \( f(x) = x \) on the interval \([-1, 1]\) using a sum of the form \( \sum_{k=1}^{n} \lambda_k x^{2k+1} \).
2. **Construct the approximation:**
Consider the function \( g_n(x) = x - x(1 - x^2)^n \). We can rewrite this a... | null | Calculus | proof | Yes | Yes | aops_forum | false |
In the rectangular coordinate system $(xOy)$, the coordinate origin is the pole, and the positive direction of the $x$-axis is the polar axis. Establish a polar coordinate system. The polar equation of the curve $C$ is $({{\rho }^{2}}-4\sqrt{2}\cos ({\theta -\dfrac{{\pi }}{4}})+6=0)$.
(I) Find the parametric equation o... | (I) The given polar equation of curve $C$ can be rewritten as $ρ^{2}-4ρ\cos θ-4ρ\sin θ+6=0$.
Thus, the rectangular coordinate equation of $C$ is $x^{2}+y^{2}-4x-4y+6=0$.
This can be further simplified to $(x-2)^{2}+(y-2)^{2}=2$.
Hence, the parametric equation of $C$ is $ \begin{cases}x=2+ \sqrt {2}\cos φ \\ y=2+ \sqrt ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
The owner of a furniture shop charges his customer 24% more than the cost price. A customer paid Rs. 7967 for a computer table. What was the cost price of the computer table? | Let's denote the cost price of the computer table as \( CP \).
The owner charges 24% more than the cost price, so the selling price (SP) that the customer paid is:
\( SP = CP + \frac{24}{100} \times CP \)
We know that the customer paid Rs. 7967, so:
\( 7967 = CP + \frac{24}{100} \times CP \)
\( 7967 = CP + 0.24 \t... | 6424 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Determine how many integers between 1 and 1200 inclusive can be expressed as the difference of the squares of two nonnegative integers. | We know that the difference of squares formula is given by $(a+b)(a-b)$. For odd numbers, setting $a-b = 1$ results in $(a+1)^2 - a^2 = 2a+1$, which can generate all odd numbers as $a$ varies among nonnegative integers. The highest odd number below 1200 is 1199. Therefore, there are $\frac{1199+1}{2} = 600$ odd numbers... | 900 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
## Task 1 - V00901
The sum of two numbers is 20, and the sum of their squares is 202. Solve the problem mathematically. | Let $x$ and $y$ be the two numbers we are looking for. Then we get the system of equations
$$
x+y=20 \quad(1) \quad ; \quad x^{2}+y^{2}=202
$$
Rearranging (1) for $y$ and substituting into (2) yields
$$
x^{2}+(20-x)^{2}=202 \Rightarrow 2 x^{2}-40 x+198=0 \Rightarrow x^{2}-20 x+99=0
$$
This quadratic equation has th... | 911 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Two divers, A and B, scored as shown in the table below. Which option is correct?
| | 1st Dive | 2nd Dive | 3rd Dive | 4th Dive | 5th Dive |
|--------|----------|----------|----------|----------|----------|
| A | $85.5$ | $96$ | $86.4$ | $75.9$ | $94.4$ |
| B | $79.5$ | $80$ | $95.7$... | To solve this problem, we first need to organize the scores of divers A and B in ascending order and then calculate the median, average, and variance for each diver.
**Diver A:**
- Scores in ascending order: $75.9$, $85.5$, $86.4$, $94.4$, $96$
- Median: The third score in the ordered list is $86.4$.
- Average: $\frac... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Among the following four shapes, the one with the most number of axes of symmetry is ( )
A: Equilateral triangle
B: Circle
C: Rectangle
D: Square | To determine the shape with the most number of axes of symmetry among the given options, let's analyze each option:
- **A: Equilateral Triangle**
- An equilateral triangle has equal sides and angles, which allows it to have 3 axes of symmetry. Each axis passes through a vertex and the midpoint of the opposite side.
... | \text{B. Circle} | Geometry | MCQ | Yes | Yes | cn_k12 | false |
A certain company is preparing to purchase two types of materials, $A$ and $B$, to make two types of crafts, $X$ and $Y$. It is known that for every one unit of material $A$ purchased, it costs $40$ yuan less than one unit of material $B$. Additionally, purchasing $2$ units of material $A$ and $3$ units of material $B$... | ### Solution:
#### Part 1: Finding the Cost of Materials $A$ and $B$
Let's denote the cost of one unit of material $A$ as $a$ yuan and the cost of one unit of material $B$ as $b$ yuan. We are given two equations based on the problem statement:
1. The cost difference between materials $B$ and $A$ is $40$ yuan, so $b ... | \text{The maximum profit is } 44600 \text{ yuan, obtained by producing } 180 \text{ units of craft } X \text{ and } 380 \text{ units of craft } Y. | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A goods train runs at a speed of 72 kmph and crosses a 290 m long platform in a certain amount of time. The length of the goods train is 230.0416 meters. How many seconds does it take for the train to cross the platform? | To calculate the time it takes for the goods train to cross the platform, we need to determine the total distance covered by the train when it crosses the platform. This distance is the sum of the length of the train and the length of the platform.
Length of the train = 230.0416 meters
Length of the platform = 290 met... | 26.00208 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
In acute $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are respectively $a$, $b$, and $c$, with $a= \sqrt {7}$, $b=3$, and $\sqrt {7}\sin B+\sin A=2 \sqrt {3}$.
(Ⅰ) Find the magnitude of angle $A$;
(Ⅱ) Find the value of $\sin (2B+ \frac {π}{6})$. | Solution:
(Ⅰ) In acute $\triangle ABC$, by the Law of Sines, we have $\frac {a}{\sin A}= \frac {b}{\sin B}$, $\therefore \sin B= \frac {b\sin A}{a}= \frac {3 \sqrt {7}\sin A}{7}$.
Since $\sqrt {7}\sin B+\sin A=2 \sqrt {3}$, $\therefore 4\sin A=2 \sqrt {3}$.
$\therefore \sin A= \frac { \sqrt {3}}{2}$.
Given $0 <... | - \frac {1}{7} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
4. In trapezoid $ABCD$, $BC \parallel DA$, diagonals $AC$ and $BD$ intersect at $P$. Let the inradii of $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, and $\triangle PDA$ be $r_{1}$, $r_{2}$, $r_{3}$, and $r_{4}$ respectively, and $\frac{1}{r_{1}} + \frac{1}{r_{3}} = \frac{1}{r_{2}} + \frac{1}{r_{4}}$. Prove: $AB +... | 4. Note that $S_{\triangle P A B}=S_{\triangle P C D}=\sqrt{S_{\triangle P B C} \cdot S_{\triangle P D M}}$, it is only necessary to prove that the sum of the perimeters of $\triangle P A B$ and $\triangle P C D$ equals the sum of the perimeters of $\triangle P B C$ and $\triangle P D A$.
| proof | Geometry | proof | Yes | Yes | olympiads | false |
In measuring the sides of a triangle, the base is measured 10% in excess, the height is taken 8% in deficit and the third side is taken 7% in excess. Find the error percent in calculating the area of triangle from these measurements. | Let's denote the actual base of the triangle as \( b \), the actual height as \( h \), and the actual third side as \( s \).
According to the problem, the measured base is 10% in excess, so the measured base \( b_m \) is:
\[ b_m = b + 0.10b = 1.10b \]
The height is taken 8% in deficit, so the measured height \( h_m \... | 1.2\% | Geometry | math-word-problem | Yes | Yes | orca_math | true |
40-3 (C50, Belarus) Let $n$ be a fixed positive even number, and consider an $n \times n$ square board, which is divided into $n^2$ unit squares. Two different squares on the board are called adjacent if they share a common edge.
Mark $N$ unit squares on the board such that every square on the board (marked or unmarke... | 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 | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
Given the real numbers $a$, $b$, and $c$ that satisfy the equation $\frac{1}{4}a^{2}+ \frac{1}{4}b^{2}+c^{2}=1$, determine the range of possible values for $ab+2bc+2ca$.
A: $(-\infty, 4]$
B: $[-4, 4]$
C: $[-2, 4]$
D: $[-1, 4]$ | From the given equation $\frac{1}{4}a^{2}+ \frac{1}{4}b^{2}+c^{2}=1$, we have $a^{2}+b^{2}+4c^{2}=4$, which can also be written as $2a^{2}+2b^{2}+8c^{2}=8$.
Now, observe that:
$8=2a^{2}+2b^{2}+8c^{2}=(a^{2}+b^{2})+(a^{2}+4c^{2})+(b^{2}+4c^{2})$
$\geqslant 2ab+4ac+4bc$ (by applying the basic inequality $x^2+y^2 \geq ... | \text{C} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
The monotonic decreasing interval of the function $y=2+3\sin x$ is \_\_\_\_. | **Analysis**
This question examines the application of the monotonicity and monotonic intervals of the sine function, which is a basic problem. Based on the monotonicity of the sine function, we write the monotonic decreasing interval of the function $y=2+3\sin x$.
**Solution**
The monotonic decreasing interval of t... | \left[ \frac{\pi}{2}+2k\pi, \frac{3\pi}{2}+2k\pi \right], k\in\mathbb{Z} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
Given the function $f(x) = x - a\ln{x}$ ($a \in \mathbb{R}$)
(Ⅰ) When $a=2$, find the equation of the tangent line to the curve $y=f(x)$ at point $A(1, f(1))$.
(Ⅱ) Discuss the monotonic intervals of the function $f(x)$. | Solution:
(Ⅰ) When $a=2$, $y=f(x)=x-2\ln{x}$, thus $f(1)=1-2\ln{1}=1$, which means $A(1, 1)$.
$f'(x)=1- \frac{2}{x}$, thus $f'(1)=1-2=-1$,
According to the geometric meaning of the derivative, the slope of the tangent line we are looking for is $k=f'(1)=-1$,
Therefore, the equation of the tangent line is $y-1=-... | (a, +\infty) | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
Given that the sequence $\{a_n\}$ is an arithmetic sequence, and $a_1 - a_9 + a_{17} = 7$, then $a_3 + a_{15} =$ （ ）
A: 7
B: 14
C: 21
D: 7$(n-1)$ | Since $\{a_n\}$ is an arithmetic sequence and $a_1 - a_9 + a_{17} = 7$,
we have $2a_9 - a_9 = 7$, which gives $a_9 = 7$.
Therefore, $a_3 + a_{15} = 2a_9 = 14$.
Hence, the correct answer is $\boxed{B}$.
**Analysis:** This can be easily derived using the properties of an arithmetic sequence. | B | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Given a function $f(x)$ defined on $\mathbb{R}$ that is symmetric about the origin, when $x > 0$, we have $f(x) = 2^{x} - \log_{3}(x^{2} - 3x + 5)$. Find the value of $f(-2)$. | Since the function $f(x)$ is symmetric about the origin and defined on $\mathbb{R}$, we can infer that $f(x)$ is an odd function. This implies that $f(-x) = -f(x)$.
For $x>0$, we know that $f(x) = 2^{x} - \log_{3}(x^{2} - 3x + 5)$, as given.
For $x0$, we apply the property of odd functions to get $f(-x) = -f(x)$ whic... | -3 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
The root of the equation $\lg x = x - 5$ that is greater than 1 lies in the interval $(n, n+1)$. Then, the positive integer $n = \_\_\_\_\_\_$. | First, plot the graphs of $y = \ln x$ and $y = x - 5$. It is easy to see that the root lies in the interval $(5, 6)$. Therefore, $n = \boxed{5}$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
In triangle $\triangle ABC$, it is known that $A+B=3C$ and $2\sin \left(A-C\right)=\sin B$. Find:
1. $\sin A$;
2. Given $AB=5$, find the height on side $AB$. | 1. **Finding $\sin A$**
Given that $A + B = 3C$ and $A + B + C = \pi$, we can combine these equations to find $C$:
\[A + B = 3C \quad \text{and} \quad A + B + C = \pi\]
\[4C = \pi \quad \Rightarrow \quad C = \frac{\pi}{4}\]
Next, using the given $2\sin(A-C) = \sin B$ and the identity $\sin B = \sin[\pi - (A + C)] = \... | 6 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
A certain unit has 1000 employees, each creating an average profit of 100,000 yuan per year. In order to increase the competitiveness of the enterprise, it is decided to optimize the industrial structure. $x$ ($x\in N^{*}$) employees will be adjusted to engage in the tertiary industry. After the adjustment, the average... | Let's break down the solution step by step, adhering to the requirements:
### For Part (1):
Given that each of the 1000 employees creates an average profit of 100,000 yuan per year, if $x$ employees are adjusted to engage in the tertiary industry, we have the following conditions:
The new average profit per person f... | (0,5] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given that $17^{-1} \equiv 53 \pmod{89}$, find $72^{-1} \pmod{89}$. | Since $72 \equiv -17 \pmod{89}$, we use the property of modular inverses:
\[
72^{-1} \equiv (-17)^{-1} \equiv (-1)^{-1} \cdot 17^{-1}.
\]
Here, $(-1)^{-1} \equiv -1$. So, continuing:
\[
72^{-1} \equiv -1 \cdot 53 \equiv -53 \equiv 36 \pmod{89}.
\]
Thus, we find that $72^{-1} \equiv \boxed{36} \pmod{89}$. | 36 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
The sum of $3x + 15$ is one-third of the value of $6x + 45$. Find the value of $x$. | The problem can be translated to the equation:
\[ 3x + 15 = \frac{1}{3}(6x + 45) \]
To solve,
1. Multiply both sides by 3 to clear the fraction:
\[ 3 \times (3x + 15) = 6x + 45 \]
\[ 9x + 45 = 6x + 45 \]
2. Subtract $6x$ and $45$ from both sides to isolate $x$:
\[ 9x - 6x = 45 - 45 \]
\[ 3x = 0 \]
3. Divide by 3 to ... | 0 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Let $r = 3^s - 2s + 1$ and $s = 2^{n+1} + 2$. What is the value of $r$ when $n = 3$? | 1. Substitute $n = 3$ into the expression for $s$:
\[
s = 2^{3+1} + 2 = 2^4 + 2 = 16 + 2 = 18
\]
2. Substitute $s = 18$ into the expression for $r$:
\[
r = 3^{18} - 2 \times 18 + 1 = 3^{18} - 36 + 1 = 3^{18} - 35
\]
3. The expression $3^{18}$ can be calculated as:
\[
3^{18} = (3^9)^2 = (19683)^2... | 387420454 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
2. In an exam with 50 questions, answering a question correctly earns 3 points, answering incorrectly deducts 1 point, and not answering a question scores no points. Xiao Long scored 120 points. How many questions did Xiao Long answer correctly at most?
A. 40
B. 42
C. 48
D. 50 | 【Answer】B
【Analysis】Since it is required to find out the maximum number of questions Xiao Long answered correctly, and because Xiao Long finally scored 120 points with 3 points for each correct answer, the number of questions Xiao Long got wrong is a multiple of 3. Therefore, the total number of questions answered corr... | 42 | Algebra | MCQ | Yes | Yes | olympiads | false |
In the chapter "Equations" of the ancient Chinese mathematical work "Nine Chapters on the Mathematical Art," negative numbers were formally introduced for the first time in the history of mathematics. If spending $100$ yuan is denoted as $-100$ yuan, then $+80$ yuan represents ( )
A: spending $80$ yuan
B: earning $80... | In the context of the ancient Chinese mathematical work "Nine Chapters on the Mathematical Art," where negative numbers were introduced, we interpret the transactions as follows:
- Spending money is represented by a negative number, as shown by spending $100$ yuan being denoted as $-100$ yuan.
- Conversely, earning mo... | B: \text{earning } 80 \text{ yuan} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Two right triangles share a side as follows: Triangle ABC and triangle ABD have AB as their common side. AB = 8 units, AC = 12 units, and BD = 8 units. There is a rectangle BCEF where point E is on line segment BD and point F is directly above E such that CF is parallel to AB. What is the area of triangle ACF? | Since AB = BD and both are sides of right triangles, triangle ABD is an isosceles right triangle. Therefore, $\angle DAB = 45^\circ$. This implies that diagonal AF of rectangle BCEF, which extends through triangle ACF, bisects $\angle CAB$.
By the properties of the rectangle, BE = CF and $\angle EBC = \angle FCB = 90^... | 24 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
The distance from the right focus of the ellipse $\frac {x^{2}}{4} + \frac {y^{2}}{3} = 1$ to the line $y= \sqrt {3}x$ is
A: $\frac {1}{2}$
B: $\frac { \sqrt {3}}{2}$
C: 1
D: $\sqrt {3}$ | Let's denote the ellipse as $\frac {x^{2}}{a^{2}} + \frac {y^{2}}{b^{2}} = 1$. From the given equation $\frac {x^{2}}{4} + \frac {y^{2}}{3} = 1$, we can identify $a^{2} = 4$ and $b^{2} = 3$. The focal length of the ellipse, $c$, can be found using the formula $c^2 = a^2 - b^2$. Now we calculate it:
$$
c^2 = a^2 - b^2 ... | \frac{\sqrt {3}}{2} | Geometry | MCQ | Yes | Yes | cn_k12 | false |
Which fraction should be added to 1 / (2 ã — 3) + 1 / (3 ã — 4) + 1 / (4 ã — 5) + . . . . . . . + 1 / (23 ã — 24) so that the result is equal to 1? | To find the fraction that should be added to the given series to make the sum equal to 1, we first need to simplify the series.
The series is a telescoping series, where each fraction can be broken down into partial fractions that will cancel out with adjacent terms. Let's look at the pattern:
1 / (n(n - 1)) = 1 / (n... | 1 | Algebra | math-word-problem | Incomplete | Yes | orca_math | true |
Given the graph of $y = f(x)$ as described below:
[asy]
unitsize(0.5 cm);
real func(real x) {
real y;
if (x >= -3 && x = 0 && x = 2 && x = -3 && x = 0 && x = 2 && x <= 3) {y = 2*(x - 2);}
return(y);
}
real func_mod(real x) {
return(func(x + 3));
}
for (n = 1; n <= 5; ++n) {
graf[n] = new picture;
for (i... | The graph of $y = f(x - 3)$ is obtained by shifting the graph of $y = f(x)$ three units to the right. This transformation moves every point of the original graph rightwards by 3 units on the x-axis.
By examining the provided options, graph C correctly shows this transformation. None of the other graphs correspond to t... | \text{C} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Determine the values of a, b, and c that satisfy the equation:
( a / √(18*b) ) * ( c / √(72*b) ) = 1 | To solve for a, b, and c, we need to simplify the equation and then find the values that satisfy the equation.
First, let's simplify the left side of the equation:
( a / √(18*b) ) * ( c / √(72*b) )
We can combine the two square roots in the denominator by multiplying them:
√(18*b) * √(72*b) = √(18 * 72 * b^2)
Now,... | 36 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Given lines $l_{1}$: $x+ay-2a-2=0$, $l_{2}$: $ax+y-1-a=0$.
(1) Find the value of $a$ when $l_{1}$ is parallel to $l_{2}$;
(2) Find the value of $a$ when $l_{1}$ is perpendicular to $l_{2}$. | (1) For line $l_{1}$: $x+ay-2a-2=0$, we can rewrite it as $y=-\frac{1}{a}x+\frac{2a+2}{a}$. The slope of line $l_{1}$ is $k_{1}=-\frac{1}{a}$.
For line $l_{2}$: $ax+y-1-a=0$, we can rewrite it as $y=-ax+1+a$. The slope of line $l_{2}$ is $k_{2}=-a$.
(1) Since $l_{1}$ is parallel to $l_{2}$, their slopes are equal. Th... | 0 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Diana works 10 hours per day on Monday, Wednesday, and Friday, and some hours per day on Tuesday and Thursday. She does not work on Saturday and Sunday. She earns $1800 per week and $30 per hour. How many hours does she work on Tuesday and Thursday? | First, let's calculate the total number of hours Diana works on Monday, Wednesday, and Friday:
She works 10 hours per day on each of these days, so:
10 hours/day * 3 days = 30 hours
Now, let's calculate her total earnings per hour for the entire week. She earns $1800 per week and $30 per hour, so:
Total earnings per... | 30 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
10CT2 ** In the convex quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $P, \angle DBC=60^{\circ}$, $\angle ACB=50^{\circ}, \angle ABD=20^{\circ}, \angle ACD=30^{\circ}$, find $\angle ADB$. | Parse 1 $\frac{A P}{A C}=\frac{S_{\triangle A B P}}{S_{\triangle A B C}}=\frac{B P \sin 20^{\circ}}{B C \sin 80^{\circ}}=\frac{\sin 20^{\circ} \sin 50^{\circ}}{\sin 70^{\circ} \sin 80^{\circ}}$,
Ifij $\frac{D P}{C D}=\frac{\sin 30^{\circ}}{\sin 70^{\circ}}$, since $\frac{\sin 20^{\circ} \sin 50^{\circ} \sin 70^{\circ}}... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
For positive integers $n$, let $f(n)$ denote the number of integers $1 \leq a \leq 130$ for which there exists some integer $b$ such that $a^b-n$ is divisible by $131$, and let $g(n)$ denote the sum of all such $a$. Find the remainder when$$\sum_{n = 1}^{130} [f(n) \cdot g(n)]$$is divided by $131$.
[i]Proposed by [b]... | 1. **Understanding the Problem:**
We need to find the sum of \( f(n) \cdot g(n) \) for \( n \) from 1 to 130, and then find the remainder when this sum is divided by 131. Here, \( f(n) \) is the number of integers \( 1 \leq a \leq 130 \) such that there exists an integer \( b \) for which \( a^b - n \) is divisible ... | 54 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Find all triplets $x, y, z$ such that
\[
\begin{aligned}
x+y+z & =5 \\
x^{2}+y^{2}+z^{2} & =19 \\
x^{3}+y^{3}+z^{3} & =53
\end{aligned}
\] | We once again use the recurrence formula for symmetric polynomials: here $S_{1}=5, S_{2}=19$ and $S_{3}=53$, so $\sigma_{1}=S_{1}=5, \sigma_{2}=\frac{1}{2}\left(\sigma_{1} S_{1}-S_{2}\right)=3$ and $\sigma_{3}=\frac{1}{3}\left(S_{3}-\sigma_{1} S_{2}+\sigma_{2} S_{1}\right)=-9$. The real numbers $x, y, z$ we are looking... | (-1,\sqrt{3}i,-\sqrt{3}i) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Suppose the function $g$ has all real numbers in its domain and range and is invertible. Some values of $g$ are given by the following table:
$$\begin{array}{c || c | c | c | c | c}
x & 1 & 2 & 3 & 4 & 5 \\
\hline
g(x) & 4 & 5 & 7 & 9 & 10
\end{array}$$
What is the value of $g(g(2)) + g(g^{-1}(9)) + g^{-1}(g^{-1}(7))?$... | From the table, we read that $g(2) = 5$, so $g(g(2)) = g(5) = 10.$
Since $g(4) = 9$, we have $g^{-1}(9) = 4$, thus $g(g^{-1}(9)) = g(4) = 9$, because $g(g^{-1}(x)) = x$ for all $x$ by the definition of an inverse function.
Furthermore, since $g(3) = 7$, we find that $g^{-1}(7) = 3$ and consequently, $g^{-1}(g^{-1}(7)... | 21 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
7.240. $\lg ^{4}(x-1)^{2}+\lg ^{2}(x-1)^{3}=25$. | ## Solution.
Domain of definition: $x>1$.
From the condition, we have $16 \lg ^{4}(x-1)+9 \lg ^{2}(x-1)-25=0$. Solving this equation as a biquadratic equation in terms of $\lg (x-1)$, we get $\lg ^{2}(x-1)=1 \Rightarrow$ $\Rightarrow \lg (x-1)=-1$ or $\lg (x-1)=1$, from which $x_{1}=1.1, x_{2}=11$.
Answer: 1,$1 ; 11... | 1.1;11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Given the function $f(x)=1-e^{-x}$ and the function $g(x)=\frac{x}{ax+1}$ (where $a\in R$, $e$ is the base of the natural logarithm),
(1) When $a=0$, find the extreme value of the function $h(x)=f′(x)⋅g(x)$;
(2) If $f(x)\leqslant g(x)$ holds true on $[0,+\infty)$, find the range of values for the real number $a$. | (I) Given $f(x)=1-e^{-x}$, hence $f′(x)=-e^{-x}\cdot (-1)=e^{-x}$,
The function $h(x)=f′(x)⋅g(x)=xe^{-x}$,
Hence $h′(x)=(1-x)\cdot e^{-x}$, when $x 0$; when $x > 1$, $h′(x) 0$, then $a > -\frac{1}{x}$ always holds, hence $a\geqslant 0$.
The inequality $1-e^{-x}\leqslant \frac{x}{ax+1}$ always holds if and only if $((... | \text{Case 1:} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
Example 7 Given three points $A, B, C$ on $\odot O$ satisfying $\angle A B C>90^{\circ}$, draw a perpendicular line to $A C$ through point $C$ intersecting the extension of $A B$ at point $D$. Draw a perpendicular line to $A O$ through $D$, intersecting $A C$ and $\odot O$ at points $E$ and $F$ respectively, with $F$ b... | Prove as shown in Figure 8, extend $AO$, intersecting $\odot O$ at point $G$, and connect $CG$, $BG$.
Since $AG$ is the diameter, $\angle ACG=90^{\circ}$.
Also, $DC \perp AC$, so points $D$, $C$, and $G$ are collinear.
Therefore, $E$ is the orthocenter of $\triangle DAG$.
By $\angle CDF=\angle GAC=\angle GFC$ and the c... | proof | Geometry | proof | Yes | Yes | olympiads | false |
Let $f(x) = x^3 - 2$ and $g(x) = 3x^2 + 3x + 2$. What is $g(f(3))$? | First, evaluate $f(x)$ at $x = 3$:
$$f(3) = 3^3 - 2 = 27 - 2 = 25.$$
Next, substitute $f(3) = 25$ into $g(x)$:
$$g(25) = 3 \cdot 25^2 + 3 \cdot 25 + 2 = 3 \cdot 625 + 75 + 2 = 1875 + 75 + 2 = 1952.$$
Therefore, the value of $g(f(3))$ is $\boxed{1952}$. | 1952 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
There is a rectangular prism made of 1 in cubes that has been covered in tin foil. There are some cubes that are not touching any tin foil on any of their sides. The width of the figure created by these cubes is twice the length and twice the height. The width of the foil-covered prism is 10 inches. How many cubes are ... | Let's denote the length of the figure created by the cubes not touching any tin foil as \( L \), the width as \( W \), and the height as \( H \). According to the problem, the width \( W \) is twice the length \( L \) and twice the height \( H \), so we can write:
\[ W = 2L \]
\[ W = 2H \]
Since the width of the foil... | 128 | Geometry | math-word-problem | Yes | Yes | orca_math | true |
The train left the first station at 9:45 am and traveled at a constant speed of 60 km/h until it reached the second station, 30 km away. At the second station, it stopped for 10 minutes before continuing to the third station, 40 km away, at a reduced speed of 40 km/h. After a 5-minute stop at the third station, the tra... | To calculate the total travel time, we need to find the travel time for each leg of the journey and add the stop times.
**First Leg:**
Distance = 30 km
Speed = 60 km/h
Time = Distance / Speed = 30 km / 60 km/h = 0.5 hours
**Stop at Second Station:**
Time = 10 minutes = 10/60 hours = 1/6 hours
**Second Leg:**
Distanc... | 2 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
On a 6 by 6 grid of points, what fraction of the larger square's area is inside the shaded rhombus? The vertices of the rhombus are located at points (2,2), (4,2), (3,3), and (3,1). | 1. **Calculate the side length of the shaded rhombus**: The vertices indicate that the rhombus is formed by vertical and horizontal shifts of 1 unit from the center point (3,2). Each side of the rhombus is the hypotenuse of a right triangle with legs of length 1, thus each side is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
2. **A... | \frac{1}{18} | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
For how many integer values of $a$ does the equation $$x^2 + ax + a^2 = 0$$ have integer solutions for $x$? | Suppose $m$ and $n$ are the integer roots of the quadratic equation. According to Vieta's formulas:
\begin{align*}
m + n &= -a \\
mn &= a^2
\end{align*}
These equations imply that:
\[ m+n = -a \quad \text{and} \quad mn = a^2 \]
By substituting $m+n = -a$ into $mn = a^2$, we get:
\[ mn = (-m-n)^2 \]
Let's factorize t... | 1 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Determine the fifth element in Row 20 of Pascal's triangle. Additionally, confirm whether this element is the same as the fifth element from the end of Row 20. | In Pascal's triangle, the \(k^{\text{th}}\) element in row \(n\) is given by \(\binom{n}{k-1}\). In Row 20, to find the fifth element, we calculate \(\binom{20}{4}\):
\[
\binom{20}{4} = \frac{20!}{4!(20-4)!} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = \frac{116280}{24} = 4845
\]
To confi... | 4845 | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
The quadratic equation $ax^2 + 30x + c = 0$ has exactly one solution. If $a+c=35$, and $a < c$, find the ordered pair $(a, c)$. | Since the quadratic equation has only one solution, its discriminant must be zero:
\[ b^2 - 4ac = 30^2 - 4ac = 0 \]
\[ 900 - 4ac = 0 \]
\[ ac = \frac{900}{4} = 225 \]
Given $a + c = 35$, we can use the identity $(a+c)^2 = a^2 + 2ac + c^2$:
\[ 35^2 = 1225 \]
\[ a^2 + 2ac + c^2 = 1225 \]
Using $2ac = 450$ from $ac = 225$... | \left(\frac{35 - 5\sqrt{13}}{2}, \frac{35 + 5\sqrt{13}}{2}\right) | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
a salesman ’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1300 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ? | Let's denote the total sales by S.
Under the old scheme, the salesman's remuneration would be 5% of S, which is 0.05S.
Under the new scheme, the salesman's remuneration is a fixed salary of Rs. 1300 plus 2.5% commission on sales exceeding Rs. 4000. So, if S is greater than Rs. 4000, the commission is 2.5% of (S - 400... | 24000 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
The distance from city A to city B is 120 miles. While driving from city A to city B, Cara drives at a constant speed of 30 miles per hour. Dan leaves city A 90 minutes after Cara. What is the minimum constant speed in miles per hour that Dan must exceed in order to arrive in city B before Cara? | First, let's calculate the time it takes for Cara to drive from city A to city B at a constant speed of 30 miles per hour.
Time = Distance / Speed
Time = 120 miles / 30 mph = 4 hours
Cara will take 4 hours to reach city B.
Dan leaves 90 minutes (which is 1.5 hours) after Cara. Therefore, to arrive before Cara, Dan m... | 48 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Compute the least positive value of $t$ such that
\[\arcsin (\sin \alpha), \ \arcsin (\sin 3 \alpha), \ \arcsin (\sin 5 \alpha), \ \arcsin (\sin t \alpha)\]
is a geometric progression for some $\alpha$ with \(0 < \alpha < \frac{\pi}{2}\). | Let $r$ be the common ratio of the geometric progression. Given $0 < \alpha < \frac{\pi}{2}$, then:
- $\arcsin (\sin \alpha) = \alpha$
- $\arcsin (\sin 3 \alpha)$ varies depending on $\alpha$, but for $0 < \alpha \leq \frac{\pi}{6}$, $\arcsin (\sin 3 \alpha) = 3\alpha$
- $\arcsin (\sin 5 \alpha)$ similarly depends on t... | 27 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
2. Answer. The sign is positive. | Solution. We will prove that negative numbers are in odd positions, and positive numbers are in even positions. Then the product of all numbers will be negative, since we are multiplying 20 negative numbers and 19 positive numbers.
Let the numbers $a_{1}, a_{2}, \ldots, a_{39}$ be written down. Note that if we remove ... | proof | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
8. 141 The sequence $\left\{a_{n} \mid n \in N\right\}$ is defined by the following formula: $a_{1}=1994$, $a_{n+1}=\frac{a_{n}^{2}}{2\left[a_{n}\right]+21}, n \in N$.
Here $\left[a_{n}\right]$ is the greatest integer not exceeding $a_{n}$.
(1) Prove that $a_{12}<1$.
(2) Prove that this sequence is convergent, and find... | [Solution] (1) Clearly, $a_{n}>0(n \in \mathbb{N})$. Using $\left[a_{n}\right]>a_{n-1}$, we get
$$
\begin{aligned}
a_{n+1} & =\frac{a_{n}^{2}}{2\left[a_{n}\right]+21}>0, \text{ thus it is sufficient that }
\end{aligned}
$$
$$
\begin{aligned}
0 &< a_{n+1} = \frac{a_{n}^{2}}{2\left[a_{n}\right]+21}, \\
s+\sqrt{s^{2}+19 s... | 10 | Algebra | proof | Yes | Yes | olympiads | false |
The owner of the Most Amazing Circus was counting the money from one day 's ticket sales. He knew that a total of 522 tickets were sold. Adult tickets cost 15 dollars each and children 's tickets cost 8 dollars each. 130 adult tickets were sold. How much were the total receipts for the day? | To calculate the total receipts for the day, we need to find out how much money was made from the adult tickets and the children's tickets.
First, let's calculate the money made from adult tickets:
Number of adult tickets sold = 130
Cost of each adult ticket = $15
Total money from adult tickets = Number of adult ticke... | \$5086 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Example As shown in Figure 1, three circles with centers at $A$, $B$, and $C$ are externally tangent to each other, and all are tangent to the line $l$. Let the radii of the three circles be $a$, $b$, and $c$ $(0<c<a<b)$. Prove:
$$
\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}} \text {. }
$$ | $$
\begin{array}{l}
A_{1} B_{1}=2 \sqrt{a b}, B_{1} C_{1}=2 \sqrt{b c}, A_{1} C_{1}=2 \sqrt{a c}, \\
A_{1} C_{1}+B_{1} C_{1}=A_{1} B_{1} .
\end{array}
$$
Therefore, equation (1) holds.
Proof In fact, if four circles in a plane are pairwise tangent, they either all touch each other externally, or three of them touch ea... | \frac{1}{\sqrt{}}=\frac{1}{\sqrt{}}+\frac{1}{\sqrt{b}} | Geometry | proof | Yes | Yes | olympiads | false |
Complex number $\frac{{2i}}{{1-i}}=\left(\ \ \right)$
A: $1+i$
B: $1-i$
C: $-1+i$
D: $-1-i$ | To solve the given problem, we start by multiplying both the numerator and the denominator of the complex fraction by the conjugate of the denominator. The conjugate of $1-i$ is $1+i$. This step is crucial as it helps to eliminate the imaginary part in the denominator, making the complex number easier to handle. Here's... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
One, (20 points) Find all positive integer solutions $x, y$ that satisfy the equation
$$
x^{2}+y^{2}=2(x+y)+x y
$$ | $$
\begin{array}{l}
x^{2}-(y+2) x+y^{2}-2 y=0 . \\
\text { By } \Delta=(y+2)^{2}-4\left(y^{2}-2 y\right) \\
=-3 y^{2}+12 y+4 \\
=16-3(y-2)^{2} \geqslant 0 \\
\Rightarrow(y-2)^{2} \leqslant \frac{16}{3}<9 \\
\Rightarrow-3<y-2<3 \\
\Rightarrow-1<y<5 .
\end{array}
$$
Since $y$ is a positive integer, thus,
$$
y=1,2,3,4 \t... | {\begin{pmatrix}2,\\4\end{pmatrix}\text{or}{\begin{pmatrix}4,\\2\end{pmatrix}\text{or}{\begin{pmatrix}4,\\4\end{pmatrix}...}} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In rectangle $ABCD$, $AB = 8$ cm, $BC = 10$ cm, and $DE = DF$. The area of triangle $DEF$ is one-third the area of rectangle $ABCD$. What is the length in centimeters of segment $EF$? Express your answer in simplest radical form. | - First, calculate the area of rectangle $ABCD$:
\[
\text{Area} = AB \times BC = 8 \times 10 = 80 \text{ cm}^2.
\]
- The area of triangle $DEF$ is one-third of the area of rectangle $ABCD$, so:
\[
\text{Area of } \triangle DEF = \frac{80}{3} \text{ cm}^2.
\]
- Since $DE = DF$, the area of $\triangle DEF$ c... | \frac{8\sqrt{15}}{3} \text{ cm} | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
Given that $w$ is a complex number such that $w+\frac{1}{w}=2\cos 45^\circ$, find $w^{12}+\frac{1}{w^{12}}$. | From the given equation, $w + \frac{1}{w} = 2 \cos 45^\circ$. This implies:
\[ w^2 + 1 = 2w \cos 45^\circ \]
\[ w^2 - 2w \cos 45^\circ + 1 = 0 \]
We solve for $w$ using the quadratic formula:
\[ w = \frac{2 \cos 45^\circ \pm \sqrt{(2 \cos 45^\circ)^2 - 4}}{2} = \cos 45^\circ \pm i \sin 45^\circ \]
Using DeMoivre's Theo... | -2 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
A carpenter is building a house. He needs 80 woodblocks to build it. He has 8 logs that can make five woodblocks each. How many logs does the carpenter still need? | The carpenter has 8 logs, and each log can make 5 woodblocks. So, the total number of woodblocks he can make from the logs he has is:
8 logs * 5 woodblocks/log = 40 woodblocks
He needs 80 woodblocks in total, so the number of woodblocks he still needs is:
80 woodblocks - 40 woodblocks = 40 woodblocks
Since each log... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | orca_math | true |
A teacher and her 30 students play a game on an infinite cell grid. The teacher starts first, then each of the 30 students makes a move, then the teacher and so on. On one move the person can color one unit segment on the grid. A segment cannot be colored twice. The teacher wins if, after the move of one of the 31 play... | 1. **Define the Problem and Strategy:**
- The teacher and 30 students play a game on an infinite cell grid.
- The teacher wins if, after any move, there is a $1 \times 2$ or $2 \times 1$ rectangle such that each segment from its border is colored, but the segment between the two adjacent squares isn't colored.
... | null | Logic and Puzzles | proof | Yes | Yes | aops_forum | false |
2、Let $a, b, c>0$, prove:
$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \geq \frac{9}{1+a b c}$ (Wa 1 ther Janous)
2、Suppose $a, b, c>0$, prove:
$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \geq \frac{9... | Proof: By the rearrangement inequality, we have
$$\frac{1}{a(1+a)}+\frac{1}{b(1+b)}+\frac{1}{c(1+c)} \geq \frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{1}{a(1+b)}$$
Therefore, if the following inequalities hold, then the original inequality also holds.
$$\sum_{c y c} \frac{1}{b(1+c)} \geq \frac{3}{1+a b c} ; \quad \sum_{c y... | proof | Inequalities | proof | Yes | Yes | inequalities | false |
Let $p \ge 5$ be a prime integer. Prove that amongst the numbers $2,3,...,p-2$, there exists at least two distinct prime numbers, $q$ and $r$, such that $q^{p-1}\not\equiv 1$ ($mod$ $p^2$) and $r^{p-1}\not\equiv 1$ ($mod$ $p^2$). | null | null | Number Theory | proof | Yes | Yes | aops_forum | false |
In triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, and $\frac{{\sin A + \sin B}}{{\sin C}} = \frac{{c + b}}{{a - b}}$.
$(1)$ If $a = 2\sqrt{3}$ and $b = 2$, find angle $B$.
$(2)$ Let $AD$ be the angle bisector of $\angle BAC$ intersecting $BC$ at point $D... | ### Solution:
#### Part (1):
Given $\frac{{\sin A + \sin B}}{{\sin C}} = \frac{{c + b}}{{a - b}}$, by applying the Law of Sines, we can rewrite the equation as $\frac{{a + b}}{c} = \frac{{c + b}}{{a - b}}$. This leads to the equation $a^2 - b^2 = bc + c^2$. Rearranging the terms, we get:
\[
b^2 + c^2 - a^2 = -bc
\]
... | 1 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
Find the projection of $\begin{pmatrix} 4 \\ -1 \\ 2 \end{pmatrix}$ onto the plane $x + 2y - 3z = 0.$ | Let $\mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$ be the normal vector to the plane defined by $x + 2y - 3z = 0.$
Let $\mathbf{v} = \begin{pmatrix} 4 \\ -1 \\ 2 \end{pmatrix}$, and $\mathbf{p}$ is the projection of $\mathbf{v}$ onto the plane.
As $\mathbf{v} - \mathbf{p}$ must be parallel to $\mathbf{n}$,... | \begin{pmatrix} 30/7 \\ -3/7 \\ 8/7 \end{pmatrix} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
I3.4 In the expansion of $(a x+b)^{2012}$, where $a$ and $b$ are relatively prime positive integers.
If the coefficients of $x^{\gamma}$ and $x^{\gamma+1}$ are equal, find the value of $\delta=a+b$. | Coefficient of $x^{23}=C_{23}^{2012} \cdot a^{23} \cdot b^{1989}$; coefficient of $x^{24}=C_{24}^{2012} \cdot a^{24} \cdot b^{1988}$
$$
\begin{array}{l}
C_{23}^{2012} \cdot a^{23} \cdot b^{1989}=C_{24}^{2012} \cdot a^{24} \cdot b^{1988} \\
b=\frac{C_{24}^{2012}}{C_{23}^{2012}} \cdot a \\
b=\frac{2012-24+1}{24} \cdot a ... | 671 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the average of the first 15 even numbers? | The first 15 even numbers are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, and 30.
To find the average, we add all these numbers together and then divide by the number of values, which is 15.
Sum of the first 15 even numbers = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30
= 240
Now, we... | 16 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Tanika is selling boxes of crackers for her scout troop's fund-raiser. On Thursday, she sold 60 boxes at a 10% discount. On Friday, she sold 50% more boxes than on Thursday at a 15% discount. On Saturday, she sold 80% more boxes than on Friday at a 20% discount. On Sunday, she sold 30% less boxes than on Saturday at a ... | Let's calculate the number of boxes sold each day and then sum them up.
Thursday:
She sold 60 boxes.
Friday:
She sold 50% more than on Thursday, so:
50% of 60 boxes = 0.50 * 60 = 30 boxes more
Total for Friday = 60 boxes + 30 boxes = 90 boxes
Saturday:
She sold 80% more than on Friday, so:
80% of 90 boxes = 0.80 * 9... | 425 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Circles \( C_1 \) and \( C_2 \) are externally tangent and both internally tangent to a larger circle \( C_3 \). The radii of \( C_1 \) and \( C_2 \) are 6 and 9, respectively. The centers of all three circles lie on a straight line. A common external tangent to \( C_1 \) and \( C_2 \) touches them at points \( T_1 \) ... | 1. Let the centers of \( C_1, C_2, \) and \( C_3 \) be \( O_1, O_2, \text{ and } O_3 \) with radii \( r_1 = 6, r_2 = 9, \text{ and } r_3 \) respectively.
2. Define \( H \) as the meeting point of the line passing through \( O_1 \) and \( O_2 \) extended, and the tangent line touching \( C_3 \) at \( T \).
3. Set up eq... | 11 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
If the solution set of the inequality system $\left\{{\begin{array}{l}{\frac{{x-1}}{2}≥\frac{{x-2}}{3}}\\{2x-m≥x}\end{array}}\right.$ is $x\geqslant m$, then the range of values for $m$ is ______. | To solve the given system of inequalities $\left\{{\begin{array}{l}{\frac{{x-1}}{2}≥\frac{{x-2}}{3}}\\{2x-m≥x}\end{array}}\right.$ and find the range of values for $m$ such that the solution set is $x \geqslant m$, we proceed as follows:
1. **Solve the first inequality:**
Starting with the first inequality, $\frac{{x... | m \geqslant -1 | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false |
On the sides $AB, BC$ and $CA$ of triangle $ABC$, points $L, M$ and $N$ are chosen, respectively, such that the lines $CL, AM$ and $BN$ intersect at a common point O inside the triangle and the quadrilaterals $ALON, BMOL$ and $CNOM$ have incircles. Prove that
$$\frac{1}{AL\cdot BM} +\frac{1}{BM\cdot CN} +\frac{1}{CN \c... | null | null | Geometry | proof | Yes | Yes | aops_forum | false |
3. The sum of the first $n$ terms of an arithmetic sequence is 305, the common difference is 2, and the first term is
(A) 570
(B) 5 e.
(C) 5 ?s
(D) 498 | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Incomplete | Yes | cn_contest | false |
73. What is greater: the length of the large semicircle or the sum of the lengths of the small ones (see Fig. 2)?
Fig. 2
 | 73. Since the diameter of the large semicircle is equal to the sum of the diameters of the small semicircles, the length of the large semicircle is also equal to the sum of the lengths of the small ones. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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