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$f(x)$ is an odd function, and for $x > 0$, $f'(x) > 0$ and $f''(x) > 0$. Determine the behavior of $f(x)$ for $x < 0$:
A. A monotonically increasing convex function
B. A monotonically decreasing convex function
C. A monotonically increasing concave function
D. A monotonically decreasing concave function | To determine the properties of the function \( f(x) \) under the given conditions, let's analyze it step-by-step:
1. **Odd Function Property:**
Since \( f(x) \) is an odd function, it satisfies:
\[
f(-x) = -f(x) \text{ for any } x \in \mathbb{R}.
\]
This symmetry about the origin implies certain conditi... | null | null |
In the expansion of $(\frac{1}{x}+2)(1-x)^{4}$, the coefficient of $x^{2}$ is:
A: $16$
B: $12$
C: $8$
D: $4$ | To find the coefficient of $x^2$ in the expansion of $(\frac{1}{x}+2)(1-x)^{4}$, we first expand $(1-x)^{4}$ using the binomial theorem. The binomial expansion formula is given by $(a+b)^n = \sum_{k=0}^{n} {C}_{n}^{k} a^{n-k}b^{k}$, where ${C}_{n}^{k}$ is the binomial coefficient.
Applying this to $(1-x)^{4}$, we get:... | null | null |
How many ways are there to put 7 balls into 3 boxes if the balls are not distinguishable but the boxes are? | To solve this, we use the "stars and bars" theorem, which helps in finding the number of ways to distribute indistinguishable objects (balls) into distinguishable boxes. The formula for the number of ways to distribute \( n \) items into \( k \) boxes is given by \( \binom{n+k-1}{k-1} \).
Here, \( n = 7 \) (balls) and... | null | null |
Let $i$ be the imaginary unit. If the complex number $2- \frac{a}{2-i}$ (where $a \in \mathbb{R}$) is a pure imaginary number, then the value of $a$ is ( )
A: 5
B: 3
C: -5
D: -3 | First, we simplify $2- \frac{a}{2-i}$. To do this, we multiply the numerator and denominator by the conjugate of the denominator:
$$2- \frac{a}{2-i} = 2- \frac{a(2+i)}{(2-i)(2+i)} = 2- \frac{2a+ai}{5} = 2- \frac{2a}{5} - \frac{a}{5}i$$
Since the complex number $2- \frac{a}{2-i}$ (where $a \in \mathbb{R}$) is a pure i... | null | null |
The domain of the function $y = \sqrt{x}\ln(2-x)$ is $(\ )$
A: $(0,2)$
B: $[0,2)$
C: $(0,2]$
D: $[0,2]$ | **Analysis**
This question tests the domain of a function and the properties of the logarithmic function. The domain of the function is determined by setting up inequalities based on the properties of the function.
**Solution**
For the function $y= \sqrt{x}\ln(2-x)$,
we have $\begin{cases}x\geqslant 0 \\ 2-x > 0\en... | null | null |
Given the function $f(x)=2\sqrt{3}\sin \omega x\cdot \cos \omega x+2\cos^2 \omega x -1$ ($\omega > 0$) has the smallest positive period of $\frac{\pi}{2}$, then when $x\in [0,\frac{\pi}{4}]$, the range of the function $y=f(x)$ is ( )
A: $[-2,1]$
B: $[-2,2]$
C: $[-1,1]$
D: $[-1,2]$ | **Analysis**
This question examines the application of double angle formulas and auxiliary angle formulas, as well as the graph and properties of the function $y=A\sin (\omega x+\varphi)$.
Using the double angle formula and auxiliary angle formula, we get $f(x)=2\sin \left(2\omega x+ \frac{\pi}{6}\right)$. Then, by u... | null | null |
Arrange in ascending order all rational numbers between zero and one whose denominators do not exceed \( n \). Let \( \frac{a}{b} \) and \( \frac{c}{d} \) be two consecutive numbers in this sequence. Prove that \( |bc - ad| = 1 \). | 1. We start by recognizing that we need to show that for two adjacent rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ in the Farey sequence $F_n$ (where $a/b < c/d$), the condition $|bc - ad| = 1$ holds.
2. First, recall that the Farey sequence $F_n$ consists of all fractions in lowest terms between 0 and 1 with den... | null | null |
The lengths of the sides of a triangle with positive area are $\log_2 8$, $\log_2 48$, and $\log_2 n$, where $n$ is a positive integer. Find the number of possible values for $n$. | By the triangle inequality, a nondegenerate triangle with these side lengths exists if and only if:
\[\left\{
\begin{aligned}
\log_2 48 + \log_2 n &> \log_2 8, \\
\log_2 8 + \log_2 48 &> \log_2 n, \\
\log_2 8 + \log_2 n &> \log_2 48.
\end{aligned}
\right.\]
1. The first inequality simplifies to $\log_2(48n) > \log_2... | null | null |
A two-digit number has its unit digit greater than the tens digit by 2. If this two-digit number is less than 30, find the number. | Let the tens digit be $x$. Then the unit digit is $(x+2)$.
Thus, we have $10 < 10x + (x + 2) < 30$, which simplifies to $\frac{8}{11} < x < \frac{28}{11}$.
Since $x$ is an integer, we have $x = 1$ or $x = 2$.
When $x = 1$, the number is $13$.
When $x = 2$, the number is $24$.
Hence, the possible numbers are $\boxed{13... | null | null |
A pond begins with a small patch of algae that triples in size each day. By day 20 of the month, the pond is completely covered with algae. Determine on what day of that month was $75\%$ of the pond still free from algae. | 1. We know the pond is completely covered on day 20.
2. Since the algae triples each day, on day 19 the pond was $\frac{1}{3}$ covered.
3. On day 18, the pond was $\frac{1}{3^2} = \frac{1}{9}$ covered.
We need to determine when $75\%$ of the pond was still free from algae, meaning $25\%$ of it was covered:
4. We find ... | null | null |
Given an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 (a > b > 0)$, where the left vertex is $A$, the top vertex is $B$, and the right focus is $F$. Let the midpoint of segment $AB$ be $M$. If $2\overrightarrow{MA} \cdot \overrightarrow{MF} + |\overrightarrow{BF}|^2 \geqslant 0$, then the range of the eccentricity ... | **Analysis**
Represent the vectors in coordinates and use the dot product formula to obtain an inequality about $e$, from which the conclusion can be drawn.
**Solution**
Given that $A(-a,0)$, $B(0,b)$, $F(c,0)$,
then $M(-\dfrac{a}{2}, \dfrac{b}{2})$,
since $2\overrightarrow{MA} \cdot \overrightarrow{MF} + |\over... | null | null |
Given that the complex conjugate of a complex number $z$ is denoted as $\overline {z}$, and $(1-i) \overline {z}=2i$, find the complex number $z$.
A: $i$
B: $-i$
C: $-1+i$
D: $-1-i$ | From $(1-i) \overline {z}=2i$, we have: $\overline {z}= \frac {2i}{1-i} = \frac {2i(1+i)}{(1-i)(1+i)} = \frac {-2+2i}{2} = -1+i$.
Therefore, $z= \overline { \overline {z}} = \boxed{-1-i}$.
To solve this, multiply both sides of the given equation by $\frac {1}{1-i}$, then simplify using complex number division. After ... | null | null |
Let $I_1, I_2, I_3$ be three open intervals of $\mathbb{R}$ such that none is contained in another. If $I_1\cap I_2 \cap I_3$ is non-empty, then show that at least one of these intervals is contained in the union of the other two. | 1. Let the intervals be \( I_1 = (a_1, b_1) \), \( I_2 = (a_2, b_2) \), and \( I_3 = (a_3, b_3) \) with the assumption \( a_1 < a_2 < a_3 \). This assumption does not lose generality because we can always relabel the intervals to satisfy this ordering.
2. Since \( I_1 \cap I_2 \cap I_3 \) is non-empty, there exists so... | null | null |
If the function $f\left( x \right)={x}^{2}+ax+\frac{1}{x}$ is an increasing function in $\left( \frac{1}{2},{+}\infty \right)$, then the range of values for $a$ is $(\_ \ \ \_ \ \ )$.
A: $\[-1,0\]$
B: $\[-1,{+}\infty )$
C: $\[0,3\])$
D: $\[3,{+}\infty )$ | The problem tests the knowledge of the graph and properties of a quadratic function, and the use of derivatives to study the monotonicity of a function. It's a medium-difficulty question.
First, find the derivative of the function $f(x)$. By ensuring that the derivative is greater than or equal to $0$ in the interval ... | null | null |
The diagram shows a \(2 \times 4\) table in which the numbers in each column except the first column are the sum and the difference of the numbers in the previous column.
Carl completes a \(2 \times 7\) table in the same way and obtains the numbers 96 and 64 in the final column. What is the sum of the numbers in the fi... | 1. Consider a \(2 \times 7\) table where the numbers in the first column are \(a\) and \(b\). These will be the starting values.
2. According to the given rule, for each subsequent column, the entries are the sum and the difference of the numbers in the previous column. To understand this better, we will calculate thr... | null | null |
Observe the following equations: $5^{5}=3125$, $5^{6}=15625$, $5^{7}=78125$, $...$, the last four digits of $5^{2018}$ are $\_\_\_\_\_\_\_$. | **Analysis**
This problem primarily tests the student's observation skills and understanding of patterns, as well as their ability to apply these patterns flexibly. It can be categorized as a foundational question.
The last four digits of $5^{5}=3125$ are $3125$, the last four digits of $5^{6}=15625$ are $5625$, the l... | null | null |
Find the derivative.
$$
y=\frac{4^{x}(\ln 4 \cdot \sin 4 x-4 \cos 4 x)}{16+\ln ^{2} 4}
$$ | To find the derivative of the given function:
\[ y = \frac{4^x (\ln 4 \cdot \sin 4x - 4 \cos 4x)}{16 + (\ln 4)^2} \]
we will treat this as a quotient where the numerator and the denominator involve differentiable functions. Let us denote the numerator and denominator as follows for clarity:
\[ u(x) = 4^x (\ln 4 \cdo... | null | null |
The legs of a right triangle are 6 cm and 8 cm. A circle is drawn through the midpoint of the shorter leg and the midpoint of the hypotenuse, touching the hypotenuse. Find the area of the circle bounded by this circumference. | Given a right triangle \( \triangle ABC \) with \( \angle ABC = 90^\circ \), \( AB = 6 \) cm, \( BC = 8 \) cm. Let \( E \) and \( D \) be the midpoints of \( AB \) and \( AC \) respectively, and let \( O \) be the center of the circle that touches the hypotenuse \( AC \). We need to find the area of the circle.
1. **C... | null | null |
Alice needs to replace a ceiling fan switch located $15$ centimeters below her 3-meter-tall ceiling. Alice is $1.6$ meters tall and can reach $50$ centimeters above her head. If she uses a stack of books along with a stool to reach the switch, and the stack of books is $12$ centimeters tall, what is the minimum height... | 1. **Convert all measurements to centimeters**:
- Ceiling height is $300$ centimeters ($3$ meters).
- Alice's height is $160$ centimeters ($1.6$ meters).
- Alice can reach $50$ centimeters above her head.
2. **Calculate the total height Alice can reach with books**:
- Alice's total potential reach with the... | null | null |
Let the function $f(x) = \begin{cases} 3x-5, & \text{if } x\geq 6 \\ f(x+2), & \text{if } x<6 \end{cases}$. Then, find $f(3)$. | Given the function $f(x) = \begin{cases} 3x-5, & \text{if } x\geq 6 \\ f(x+2), & \text{if } x<6 \end{cases}$,
Therefore, $f(3) = f(5) = f(7) = 3 \times 7 - 5 = 16$.
Hence, the answer is $\boxed{16}$.
Since $3 < 6$, we have $f(3) = f(5) = f(7)$, which allows us to find the result.
This problem tests the method of fi... | null | null |
Given the function $f(x)=2\sqrt{3}\sin{x}\cos{x}+2\cos^2{x}-1$ where $x\in\mathbb{R}$.
$(1)$ Find the minimum positive period and the equation of the axis of symmetry of the function $f(x)$.
$(2)$ The graph of the function $f(x)$ is shifted to the left by $\frac{\pi}{12}$ units to obtain the graph of the function $... | ### Solution:
#### Part (1): Finding the Minimum Positive Period and the Equation of the Axis of Symmetry
Given the function $f(x)=2\sqrt{3}\sin{x}\cos{x}+2\cos^2{x}-1$, we can simplify it using trigonometric identities:
1. **Simplification using Trigonometric Identities**:
- We know that $\sin{2x} = 2\sin{x}\cos... | null | null |
The positive integers \(x\) and \(y\) are the two smallest positive integers such that the product of \(450\) and \(x\) is a square, and the product of \(450\) and \(y\) is a cube. What is the sum of \(x\) and \(y\)?
A) 48
B) 58
C) 62
D) 82
E) 100 | Firstly, factorize \(450\):
$$450 = 2 \cdot 3^2 \cdot 5^2.$$
#### Finding \(x\) such that \(450x\) is a square:
For \(450x\) to be a perfect square:
- The exponent of \(2\) is \(1\); we need one more \(2\) to make it \(2^2\).
- The exponents of \(3\) and \(5\) are both even, so they are fine.
So, \(x\) must include \... | null | null |
A mixture of 150 liters of wine and water contains a certain percentage of water. 30 liters more water should be added so that water becomes 25% of the new mixture. What is the initial percentage of water in the mixture? | Let's denote the initial percentage of water in the mixture as P%. This means that the amount of water in the initial 150-liter mixture is (P/100) * 150 liters.
According to the problem, we need to add 30 liters of water to the mixture so that water becomes 25% of the new mixture. The new total volume of the mixture a... | null | null |
Evaluate the expression \(8^{15} \div 64^5\). | \(8\) and \(64\) can both be expressed as powers of \(2\) because \(8 = 2^3\) and \(64 = 2^6\). So, \(8^{15}\) and \(64^5\) can be rewritten in terms of base \(2\):
\[ 8^{15} \div 64^5 = (2^3)^{15} \div (2^6)^{5}. \]
Apply the power of a power rule for exponents:
\[ (2^3)^{15} = 2^{45}, \quad (2^6)^{5} = 2^{30}. \]
N... | null | null |
The center of the ellipse is at the coordinate origin, and the ratio between the lengths of the major and minor axes is $2:1$. One focus point is $(0, -2)$. Find the eccentricity and the standard equation of the ellipse. | Given that one focus point of the ellipse is $(0, -2)$, we have that the distance from the center to the focus $c = 2$. Additionally, since the ratio between the lengths of the major and minor axes is given as $2:1$, for an ellipse with a major axis $2a$ and a minor axis $2b$, the equation is $\frac{2a}{2b} = \frac{2}{... | null | null |
A special school for deaf and blind students has a deaf student population three times the size of blind student population. If the number of deaf students is 180, how many students are there altogether? | If the number of deaf students is 180 and they are three times the size of the blind student population, we can find the number of blind students by dividing the number of deaf students by 3.
Number of blind students = Number of deaf students / 3
Number of blind students = 180 / 3
Number of blind students = 60
Now, t... | null | null |
A line passing through the focus F of the parabola $y^2=4x$ intersects the parabola at points A and B. Point O is the origin. If the area of $\triangle AOB$ is $\frac{3\sqrt{2}}{2}$, then the slope of line AB is ( )
A: $2\sqrt{2}$
B: $-2\sqrt{2}$
C: $\pm2\sqrt{2}$
D: $\frac{\sqrt{2}}{4}$ | Let the inclination angle of line AB be $\alpha$. Given that the area of $\triangle AOB$ is $S_{\triangle AOB} = \frac{2p}{2\sin\alpha} = \frac{3\sqrt{2}}{2}$, we find that $\sin\alpha = \frac{2\sqrt{2}}{3}$.
Therefore, $\cos\alpha = \pm\frac{1}{3}$, which leads to $\tan\alpha = \pm2\sqrt{2}$.
Hence, the slope of lin... | null | null |
Which of the following propositions are correct?
A: $\exists x\in R$ such that $|x| \gt x$
B: $\forall x\in R$ such that $x^{2}-3x-5 \gt 0$
C: $\forall x\in Q$ such that $x^{4}\in Q$
D: $\exists a$ and $b\in R$ such that $|a-2|+\left(b+1\right)^{2}\leqslant 0$ | **Solution in Step-by-Step Format:**
**Proposition A: $\exists x\in \mathbb{R}$ such that $|x| \gt x$**
- Consider the case when $x < 0$. For such $x$, $|x| = -x$ because $|x|$ is the absolute value, which makes it positive or zero.
- Therefore, $-x > x$ when $x < 0$.
- This shows that there indeed exists at least on... | null | null |
The area of a rectangle is 460 square meters. If the length is 15% more than the breadth, what is the breadth of the rectangular field? | Let the breadth of the rectangle be \( b \) meters. According to the problem, the length is 15% more than the breadth, so the length \( l \) can be expressed as:
\[ l = b + 0.15b \]
\[ l = 1.15b \]
The area \( A \) of a rectangle is given by the product of its length and breadth:
\[ A = l \times b \]
Given that the... | null | null |
Given the universal set $U=\{a, b, c, d, e\}$, and sets $M=\{a, b, c\}$, $N=\{a, c, e\}$, find the intersection of the complements $\complement_{U}M \cap \complement_{U}N$.
(A) $\varnothing$
(B) $\{d\}$
(C) $\{a, c\}$
(D) $\{b, e\}$ | Since the universal set is $U=\{a, b, c, d, e\}$, and the sets are $M=\{a, b, c\}$ and $N=\{a, c, e\}$, we can find:
1. The complement of $M$ with respect to $U$ is everything in $U$ that is not in $M$:
$$ \complement_{U}M = \{d, e\}. $$
2. Similarly, the complement of $N$ with respect to $U$ is everything in $U$ tha... | null | null |
Jim decides to open up a bike shop. He charges $20 for fixing bike tires and it cost him $5 in parts. In one month Jim does 300 of these repairs. He also does 2 more complex repairs for $300 each and cost $50 in parts. He also sells $2000 profit worth of things from his retail shop. If rent and other fixed expense for ... | First, let's calculate the profit from fixing bike tires.
Jim charges $20 per repair and it costs him $5 in parts. So, the profit per repair is $20 - $5 = $15.
He does 300 of these repairs in a month, so the total profit from tire repairs is 300 * $15 = $4500.
Next, let's calculate the profit from the more complex ... | null | null |
Calculate the probability of data falling outside the interval (-2, 2) for a normal distribution N(0, $\frac{4}{9}$).
A: 0.46
B: 0.997
C: 0.03
D: 0.0026 | Given the normal distribution N(0, $\frac{4}{9}$), we have a mean, $\mu = 0$, and a standard deviation, $\sigma = \frac{2}{3}$.
First, we calculate the probability of the data falling within the interval (-2, 2). By the empirical (68-95-99.7) rule, or the 3-sigma rule for normal distribution, we know that approximatel... | null | null |
Given vectors $\overrightarrow {a}$=(-2, 1, 3) and $\overrightarrow {b}$=(-1, 2, 1), if $\overrightarrow {a}$ is orthogonal to ($\overrightarrow {a}$-$λ \overrightarrow {b}$), find the value of the real number $λ$. | Since we have vectors $\overrightarrow {a}$=(-2, 1, 3) and $\overrightarrow {b}$=(-1, 2, 1),
then $\overrightarrow {a}$-$λ \overrightarrow {b}$=(-2+$λ$, 1-2$λ$, 3-$λ$).
Given that $\overrightarrow {a}$ is orthogonal to ($\overrightarrow {a}$-$λ \overrightarrow {b}$),
then the dot product of $\overrightarrow {a}$ and ($... | null | null |
Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. Calculate the sum $\sum_{i=1}^{1005} \frac{(2i-1)!!}{(2i)!!}$ and express it as a fraction in lowest terms. The denominator of this fraction has the form $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$. | First, we notice that $(2n)!! = 2^n \cdot n!$ and $(2n)!! \cdot (2n-1)!! = (2n)!$.
The fraction $\dfrac{(2i-1)!!}{(2i)!!}$ becomes $\dfrac{(2i)!}{(2i)!!^2} = \dfrac{(2i)!}{2^{2i}(i!)^2}$.
We recognize $\dfrac{(2i)!}{(i!)^2} = {2i \choose i}$, thus $\frac{{2i\choose i}}{2^{2i}}$. Our sum is $S=\sum_{i=1}^{1005} \dfrac{{... | null | null |
Given that $b$ is an odd multiple of $997$, find the greatest common divisor of $3b^2 + 17b + 31$ and $b + 7$. | We use the Euclidean Algorithm to solve this problem. Begin by simplifying the gcd expression:
\[
\text{gcd}(3b^2 + 17b + 31, b + 7)
\]
First, express $3b^2 + 17b + 31$ in terms of $(b + 7)$:
\[
3b^2 + 17b + 31 - 3(b + 7)^2 - 17(b + 7) = 3b^2 + 17b + 31 - (3b^2 + 42b + 147) - 17b - 119
\]
Simplify the above:
\[
3b^2 + ... | null | null |
A circle with center O inscribed in a convex quadrilateral ABCD is tangent to the lines
AB, BC, CD, DA at points K, L, M, N respectively. Assume that the lines KL and MN
are not parallel and intersect at the point S. Prove that BD is perpendicular OS.
I think it is very good and beautiful problem. I solved it with... | 1. **Lemma**: Let \( A \) and \( B \) be two arbitrary points (different from the center of the circle \( O \)). Then \( A \) is on the polar of \( B \) if and only if \( B \) is on the polar of \( A \).
2. **Proof of Lemma**:
- Consider the circle with center \( O \) and let \( P \) be a point on the circle. The ... | null | null |
Given the propositions $p$ and $q$, "$\neg p$ is false" is the (______) of "$p \lor q$ is true".
A. Necessary and sufficient condition
B. Necessary but not sufficient condition
C. Sufficient but not necessary condition
D. Neither necessary nor sufficient condition | First, let's translate the Chinese solution step by step:
1. If $\neg p$ is false, then $p$ is true, which implies $p \lor q$ is true. Thus, sufficiency holds.
2. When $p$ is false and $q$ is true, $p \lor q$ is true, but $\neg p$ is true. Therefore, necessity does not hold.
3. Hence, "$\neg p$ is false" is a sufficien... | null | null |
Determine $a, b, c$ such that the matrix
\[\mathbf{N} = \begin{pmatrix} a & b & c \\ b & -a & 0 \\ c & 0 & -a \end{pmatrix}\]
satisfies $\mathbf{N}^T \mathbf{N} = \mathbf{I}.$ Then find $a^2 + b^2 + c^2.$ | First, compute $\mathbf{N}^T$:
\[\mathbf{N}^T = \begin{pmatrix} a & b & c \\ b & -a & 0 \\ c & 0 & -a \end{pmatrix}^T = \begin{pmatrix} a & b & c \\ b & -a & 0 \\ c & 0 & -a \end{pmatrix}.\]
(Note: $\mathbf{N}$ is symmetric, hence $\mathbf{N}^T = \mathbf{N}$.)
Compute $\mathbf{N}^T \mathbf{N}$:
\[\mathbf{N}^T \mathbf... | null | null |
Regular polygons with $7, 8, 9,$ and $10$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?
A) 840
B) 852
C) 862
D) 872
E) 882 | 1. **Understanding the Intersection Rule**:
Each side of a smaller polygon intersects each side of a larger polygon twice — once upon entering and once upon exiting the side of the larger polygon.
2. **Combining Each Pair of Polygons**:
We'll consider the pairs $(7, 8), (7, 9), (7, 10), (8, 9), (8, 10), (9, 10)$... | null | null |
Let $\{\overrightarrow{{e}_{1}}, \overrightarrow{{e}_{2}}, \overrightarrow{{e}_{3}}\}$ be a unit orthogonal basis of space vectors, $\overrightarrow{a}=3\overrightarrow{{e}_{1}}-4\overrightarrow{{e}_{2}}+2\overrightarrow{{e}_{3}}$, $\overrightarrow{b}=-2\overrightarrow{{e}_{1}}-3\overrightarrow{{e}_{2}}+4\overrightarro... | To find the coordinates of $\overrightarrow{a} + \overrightarrow{b}$, we start by looking at the given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ in terms of the unit orthogonal basis $\{\overrightarrow{{e}_{1}}, \overrightarrow{{e}_{2}}, \overrightarrow{{e}_{3}}\}$:
- The vector $\overrightarrow{a}$ is giv... | null | null |
Mai loves books and recently started reading "Theoretical Mathematics Unveiled". On the first day, Mai read $1/6$ of the pages plus $10$ more, and on the second day, she read $1/5$ of the remaining pages plus $20$ pages. On the third day, she read $1/4$ of the remaining pages plus $25$ pages. She then discovered that t... | Let $x$ be the total number of pages in the book.
1. **Reading on the first day:**
Mai reads $\frac{1}{6}x + 10$ pages. The remaining pages are:
\[
x - \left(\frac{1}{6}x + 10\right) = \frac{5}{6}x - 10
\]
2. **Reading on the second day:**
From the remaining pages, Mai reads $\frac{1}{5}(\frac{5}{6}x -... | null | null |
Calculate $\frac{2}{3} \cdot \frac{4}{7} \cdot \frac{9}{11}$. | We have
\[
\frac{2}{3} \cdot \frac{4}{7} \cdot \frac{9}{11} = \frac{2 \cdot 4 \cdot 9}{3 \cdot 7 \cdot 11}.
\]
This simplifies to
\[
\frac{72}{231}.
\]
To simplify $\frac{72}{231}$, we find the greatest common divisor (GCD) of 72 and 231. The GCD is 3, so we divide both numerator and denominator by 3:
\[
\frac{72}{23... | null | null |
Class 1 of Senior Year 3 has 52 students. Now, all students are randomly numbered, and a sample of size 4 is drawn using the systematic sampling method. It is known that students numbered 5, 31, and 44 are in the sample. Therefore, the number of another student in the sample is ( )
A: 8
B: 13
C: 15
D: 18 | According to the definition and method of systematic sampling, the numbers of the 4 individuals drawn form an arithmetic sequence.
Given that three of the individuals' numbers are 5, 31, and 44, thus the number of another drawn individual must be 18.
Therefore, the correct choice is: $\boxed{\text{D}}$.
**Analys... | null | null |
Determine the angles of a triangle with sides 3, $\sqrt{11}$, and $2 + \sqrt{5}$, giving the angles in degrees separated by commas. | First, check that the sides form a triangle using the triangle inequality:
- $3 + \sqrt{11} > 2 + \sqrt{5}$
- $3 + 2 + \sqrt{5} > \sqrt{11}$
- $\sqrt{11} + 2 + \sqrt{5} > 3$
All these inequalities hold, so it's a valid triangle.
Next, apply the Law of Cosines:
1. For the angle opposite side $3$, using sides $3$, $\sq... | null | null |
If you add the allowance of Yul-gi and Ga-yeong have, you get 6,000 won. If you subtract the difference between their allowance from the sum of their allowance, you get 4,800 won. If Yul-gi's allowance is more than Ga-yeong's, find out how much Yul-gi's allowance is. | Let's denote Yul-gi's allowance as Y and Ga-yeong's allowance as G.
According to the problem, we have two equations:
1) Y + G = 6,000 (the sum of their allowances)
2) (Y + G) - (Y - G) = 4,800 (the sum of their allowances minus the difference between their allowances)
From equation 2, we can simplify:
Y + G - Y + G... | null | null |
Given that the 5th term ($a_5$) of an arithmetic sequence is 10, and the sum of the first three terms ($a_1 + a_2 + a_3$) is 3, determine the correct option for the first term ($a_1$) and the common difference ($d$).
( )
A: $a_1 = -2$, $d = 3$
B: $a_1 = 2$, $d = -3$
C: $a_1 = -3$, $d = 2$
D: $a_1 = 3$, $d = -2$ | **Analysis:**
Utilize the properties of an arithmetic sequence along with the given equation ($a_1 + a_2 + a_3 = 3$) to find $a_2$. Then, use the known value of $a_5$ and the arithmetic sequence formula to find the common difference $d$, and subsequently, the first term $a_1$.
**Step-by-step solution:**
In the arithme... | null | null |
Find how many positive integer factors of $2550$ have more than $4$ factors.
A) 2
B) 3
C) 4
D) 5 | First, we need to determine the prime factorization of $2550$:
\[ 2550 = 2 \times 3 \times 5^2 \times 17 \]
Using the formula for the number of divisors from the prime factorization, we find the total divisors for $2550$:
\[ d(2550) = (1+1)(1+1)(2+1)(1+1) = 2 \times 2 \times 3 \times 2 = 24 \]
Next, we evaluate the d... | null | null |
The area of a rectangle is 432 square centimeters. What would the new area be if the length of the rectangle was increased by $20\%$ and the width of the rectangle was decreased by $10\%$? | 1. **Initial area:** The original area of the rectangle is given as 432 square cm.
2. **Change in dimensions:**
- Length is increased by 20%, so the new length factor is 1.2.
- Width is decreased by 10%, so the new width factor is 0.9.
3. **New area calculation:**
- New area = original area × (factor for len... | null | null |
A service hotline has the following probabilities of being answered: 0.1 on the first ring, 0.2 on the second ring, 0.3 on the third ring, and 0.35 on the fourth ring.
1. What is the probability that a call will be answered before the fifth ring?
2. What is the probability that a call will not be answered after four r... | Let's denote the event "the call is answered on the k-th ring" as A<sub>k</sub> (k ∈ N). These events are mutually exclusive. Let's denote the event "the call is answered before the fifth ring" as event A.
According to the addition rule of probabilities for mutually exclusive events, we have:
P(A) = P(A<sub>1</sub> ∪... | null | null |
Donna is trying to figure out if her truck is over a particular bridge's weight limit. The bridge can hold 20,000 pounds. Donna's empty truck weighs 12,000 pounds. She's carrying 20 crates of soda that each weigh 50 pounds, 3 dryers that each weigh 3000 pounds, and some weight in fresh produce. Her fully loaded truck w... | First, let's calculate the total weight of the soda and the dryers.
The soda crates: 20 crates * 50 pounds each = 1000 pounds
The dryers: 3 dryers * 3000 pounds each = 9000 pounds
Now, let's add the weight of the soda and the dryers to the weight of the empty truck to find out how much weight is accounted for before ... | null | null |
Two trains of equal length are running on parallel lines in the same direction at 75 km/hr and 60 km/hr. The faster train passes the slower train in 45 seconds. What is the length of each train (in meters)? | Let's denote the length of each train as L meters.
Since the trains are of equal length and the faster train is overtaking the slower one, we can consider the relative speed between the two trains. The relative speed is the difference in their speeds because they are moving in the same direction.
Relative speed = Spe... | null | null |
In a school of 650 boys, a certain percentage are Muslims, 28% are Hindus, 10% are Sikhs, and the remaining belong to other communities. There are 117 boys from other communities. What percentage of the boys are Muslims? | Let's denote the percentage of Muslim boys as M%.
We know that the total number of boys is 650. According to the information given, 28% are Hindus, 10% are Sikhs, and the remaining are from other communities or Muslims.
First, let's calculate the number of Hindu and Sikh boys:
Hindu boys = 28% of 650
Hindu boys = 0.... | null | null |
Given the quadratic function $f(x) = ax^2 - 2x + c$ whose range is $[0, +\infty)$, find the minimum value of $\frac{9}{a} + \frac{1}{c}$. | For the quadratic function $f(x) = ax^2 - 2x + c$ with a range of $[0, +\infty)$, it must mean that the vertex of the parabola lies on the x-axis and the parabola opens upwards. Therefore, the discriminant $\Delta = 4 - 4ac$ should be equal to zero in order for the parabola to have a single point (the vertex) touching ... | null | null |
Evaluate the complex number $\dfrac{2}{i(3-i)}=($ $)$,
A: $\dfrac{1}{5}-\dfrac{3}{5}i$,
B: $\dfrac{1}{5}+\dfrac{3}{5}i$,
C: $\dfrac{3}{5}+\dfrac{1}{5}i$,
D: $\dfrac{3}{5}-\dfrac{1}{5}i$, | **Step 1**: Multiply the numerator and the denominator by the conjugate of the denominator.
$\dfrac{2}{i(3-i)} = \dfrac{2}{3i - i^2} = \dfrac{2}{3i + 1}$
**Step 2**: Multiply the numerator and the denominator by the conjugate of the new denominator.
$\dfrac{2}{3i + 1} \times \dfrac{1 - 3i}{1 - 3i} = \dfrac{2 - 6i}{1 -... | null | null |
Let \( f(x, y) \) be a polynomial in \( x \) and \( y \), satisfying the conditions:
1. \( f(1, 2) = 2 \);
2. \( y f(x, y) = x f(f(x, y), y) = (f(x, y))^2 \).
Determine all possible functions \( f(x, y) \). | We are given the following conditions for the polynomial function \( f(x, y) \):
1. \( f(1, 2) = 2 \)
2. \( y f\left(x, y\right) = x f(f(x, y), y) = (f(x, y))^2 \)
We are to determine all possible functions \( f(x, y) \).
1. **Substitute the first condition**:
\[
f(1, 2) = 2
\]
2. **Analyze the second condi... | null | null |
Fill in the blanks with appropriate numbers.
216+35+84=35+（\_\_\_\_\_\_+\_\_\_\_\_\_）
298-35-165=298-（\_\_\_\_\_\_+\_\_\_\_\_\_）
400÷25÷4=400÷（\_\_\_\_\_\_×\_\_\_\_\_\_）
a×6+6×15=\_\_\_\_\_\_×（\_\_\_\_\_\_+\_\_\_\_\_\_） | Solution:
(1) 216+35+84=35+(216+84),
(2) 298-35-165=298-(35+165),
(3) 400÷25÷4=400÷(25×4),
(4) a×6+6×15=6×(a+15),
Therefore, the answers in order are: 216, 84, 35, 165, 25, 4, a, 15.
(1) This is solved based on the associative property of addition,
(2) This is solved using a simplified method for subtract... | null | null |
Consider a sphere with radius 2 centered at $(2,0,2)$. A light source is located at $P = (2,-2,6)$. Find the equation of the boundary of the shadow of this sphere on the $xy$-plane, expressed as $y = f(x)$. | Let $O = (2,0,2)$ be the center of the sphere, and let $X = (x,y,0)$ be a point on the boundary of the shadow. Since $X$ is on the boundary, $\overline{PX}$ is tangent to the sphere, and thus, point $T$ of tangency satisfies $\angle PTO = 90^\circ$.
Define the vector from $P$ to $O$ as $\overrightarrow{PO} = (-0, 2, ... | null | null |
Among the following shapes, the one with the most axes of symmetry is ( )
A: Isosceles triangle
B: Square
C: Circle
D: Line segment | A: Since an isosceles triangle can be folded along the median line of its base, and the two parts after folding can completely coincide, the isosceles triangle is an axisymmetric figure. The median line of its base is its axis of symmetry, so the isosceles triangle has $1$ axis of symmetry.
B: Since a square can be fo... | null | null |
Lanie has been sick and was only able to work 4/5 of her usual 40-hour week. If her hourly rate is $15, how much was her salary for that week? | To calculate Lanie's salary for the week, we first need to determine how many hours she worked. Since she worked 4/5 of her usual 40-hour week, we can calculate the hours worked as follows:
Hours worked = (4/5) * 40 hours
Now, let's calculate the hours worked:
Hours worked = (0.8) * 40 hours
Hours worked = 32 hours
... | null | null |
In a regular tetrahedron \(ABCD\), \(M\) and \(N\) are the midpoints of \(AB\) and \(CD\) respectively. What is the degree measure of the angle between \(MN\) and \(AD\)?
A. \(30^{\circ}\)
B. \(45^{\circ}\)
C. \(60^{\circ}\)
D. \(90^{\circ}\) | 1. Consider the regular tetrahedron \(ABCD\). Let \(M\) and \(N\) be the midpoints of the edges \(AB\) and \(CD\) respectively.
2. Place the tetrahedron in a way that fits inside a cube. Assume all edges of the cube have unit length. Thus, every edge of the regular tetrahedron within this cube equals \(\sqrt{2}\) as de... | null | null |
There is one black ball and one white ball in a bag. Each time a ball is taken from the bag, it is not returned, but instead, another black ball is added. What is the probability of drawing a black ball on the $n$-th draw? | We are given a problem where we have one black ball and one white ball in a bag. Each time we draw a ball at random from the bag, we do not put the drawn ball back in the bag, but we add an additional black ball to the bag. We need to find the probability of drawing the black ball on the $n$-th draw.
1. **Step 1: Init... | null | null |
Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{3} = 1$ ($a > 0$) has an eccentricity of $2$, then the real number $a$ is ( ).
A: $2$
B: $\frac{\sqrt{6}}{2}$
C: $\frac{\sqrt{5}}{2}$
D: $1$ | From the given information, we have the eccentricity formula:
\[e= \frac{c}{a}= \frac{\sqrt{a^2+3}}{a}=2\],
Solving this, we get $a=1$.
Therefore, the correct choice is $\boxed{D}$.
This problem involves finding $a$, $b$, and $c$ from the equation of the hyperbola, and then substituting them into the formula for eccent... | null | null |
Which of the following equations is transformed correctly?
A: If $S=\frac{1}{2}ab$, then $b=\frac{S}{2a}$
B: If $x-3=y-3$, then $x+y=0$
C: If $mx=my$, then $x=y$
D: If $\frac{x}{m}=\frac{y}{m}$, then $x=y$ | Let's analyze each option step by step:
**Option A:**
Given: $S=\frac{1}{2}ab$
To isolate $b$, we multiply both sides by $2$ to get:
\[2S = ab\]
Then, to solve for $b$, we divide both sides by $a$:
\[b = \frac{2S}{a}\]
The given transformation was $b=\frac{S}{2a}$, which is incorrect because the correct transformation... | null | null |
Given that $a=\int_{0}^{1}{{{x}^{-{\frac{1}{3}}}}dx}$, $b=1-\int_{0}^{1}{{{x}^{\frac{1}{2}}}}dx$, and $c=\int_{0}^{1}{{{x}^{3}}}dx$, determine the relationship among $a$, $b$, and $c$.
A: $a > b > c$
B: $b > a > c$
C: $a > c > b$
D: $b > c > a$ | **Analysis**
This problem requires us to perform integration operations and compare the values of $a$, $b$, and $c$. We will compute each integral separately using the fundamental theorem of calculus and then determine their relationship.
**Step-by-Step Solution**
1. Compute the value of $a$:
$a=\int_{0}^{1}{x}^{-{\... | null | null |
Evaluate: \(6 - 8 \left( 9 - 4^2 \right) 3.\) | Following the order of operations (PEMDAS/BODMAS), we first calculate the expression inside the parentheses:
\[
4^2 = 16
\]
Subtract from 9:
\[
9 - 16 = -7
\]
Next, multiply by the values outside the parentheses:
\[
8 \times (-7) \times 3 = -168
\]
Finally, subtract this result from 6:
\[
6 - (-168) = 6 + 168 = \boxed{... | null | null |
What is the smallest base-10 integer that can be represented as $21_a$ in one base and $12_b$ in another base, where $a$ and $b$ are any bases larger than 2? | Convert $21_a$ and $12_b$ into base-10:
\[ 21_a = 2a + 1 \]
\[ 12_b = 1b + 2 \]
Set these equal since they represent the same number:
\[ 2a + 1 = 1b + 2 \]
Solving for one variable:
\[ 2a - 1b = 1 \]
For the smallest base-10 integer, choose the smallest values for $a$ and $b$ greater than 2. Let's start with $b = 3$... | null | null |
The volume of a rectangular prism is $8 \mathrm{~cm}^{3}$. If each edge of the rectangular prism is increased by 1 centimeter, then the resulting rectangular prism has a volume of $27 \mathrm{~cm}^{3}$. What will be the volume of the rectangular prism if the edges are increased by 1 additional centimeter each? | 1. Given that the volume of a cuboid is $8 \, \mathrm{cm}^3$, let the edges of the cuboid be $a$, $b$, and $c$ such that:
\[
V = abc = 8
\]
2. When each edge of the cuboid is increased by $1 \, \mathrm{cm}$, the new volume is $27 \, \mathrm{cm}^3$. Thus we have:
\[
(a+1)(b+1)(c+1) = 27
\]
3. Expand ... | null | null |
A quadratic expression in the form $4x^2 - 20x - 32$ is rewritten as $(px + q)^2 + r$. Find the product $pq$ where $p$, $q$, and $r$ are integers. | Following the format $(px + q)^2 + r = p^2x^2 + 2pqx + q^2 + r$, equate this to $4x^2 - 20x - 32$.
1. From $p^2x^2 = 4x^2$, we deduce $p^2 = 4$. Therefore, $p = 2$ or $p = -2$.
2. Plugging $p = 2$ into $2pqx = -20x$, we get $2 \cdot 2 \cdot q = -20$, thus $q = -5$.
3. Substituting $p = 2$ and $q = -5$ into $q^2 + r = ... | null | null |
The number $a+\sqrt{b}$ and its radical conjugate have a sum of $-6$ and a product of $4$. Find $a+b$. | 1. **Finding a**: The sum of $a + \sqrt{b}$ and $a - \sqrt{b}$ is $2a$. Given that their sum is $-6$, we have:
\[
2a = -6 \implies a = -3.
\]
2. **Finding b**: The product of $a+\sqrt{b}$ and $a-\sqrt{b}$ is $a^2 - b$. Given the product is $4$:
\[
(-3)^2 - b = 4 \implies 9 - b = 4 \implies b = 5.
\]
... | null | null |
Given $f(x)=2^{|x-a|}$, the graph of $f(x)$ is symmetric about the line $x=1$. Then, the value of the real number $a$ is $\boxed{1}$. | **Method 1:** Since $y=|x-a|$ is symmetric about $x=a$,
it follows that $f(x)=2^{|x-a|}$ is symmetric about $x=a$.
Therefore, the axis of symmetry is $x=a=1$, which means $a=1$.
**Method 2:** Since the graph of $f(x)=2^{|x-a|}$ is symmetric about the line $x=1$,
we have $f(1+x)=f(1-x)$.
This implies $2^{|1+x-a... | null | null |
Five fair six-sided dice are rolled. What is the probability that at least three of the five dice show the same value?
$\textbf {(A) } \frac{113}{648} \qquad \textbf {(B) } \frac{125}{1296} \qquad \textbf {(C) } \frac{1}{6} \qquad \textbf {(D) } \frac{100}{648} \qquad \textbf {(E) } \frac{7}{36}$ | We need to calculate the probability of two scenarios:
1. Exactly three dice show the same number.
2. Four or five dice show the same number.
#### Case 1: Exactly three dice show the same number
- The probability that any three dice show the same number (and the other two do not match this number or each other) is cal... | null | null |
The area of trapezoid $ABCD$ is 90. The diagonals intersect at point $O$. Segments connecting the midpoint $P$ of the base $AD$ with vertices $B$ and $C$ intersect with the diagonals of the trapezoid at points $M$ and $N$. Find the area of quadrilateral $OMPN$ if one of the bases of the trapezoid is twice as long as th... | Let's denote the trapezoid by $ABCD$ with bases $AD$ and $BC$, where $AD = 2BC$. Let $O$ be the point where the diagonals $AC$ and $BD$ intersect, and let $P$ be the midpoint of $AD$. Lines $BP$ and $CP$ intersect the diagonals $AC$ and $BD$ at points $M$ and $N$ respectively. We need to find the area of the quadrilate... | null | null |
In acute-angled triangle $ABC$ , $AH$ is an altitude and $AM$ is a median. Points $X$ and $Y$ on lines $AB$ and $AC$ respectively are such that $AX=XC$ and $AY=YB$ . Prove that the midpoint of $XY$ is equidistant from $H$ and $M$ .
*Proposed by Danylo Khilko* | 1. **Assume the unit circle and express points in complex form:**
Let the vertices of the triangle \(ABC\) be on the unit circle. We can represent the points \(A\), \(B\), and \(C\) as complex numbers \(a\), \(b\), and \(c\) respectively, where \(|a| = |b| = |c| = 1\).
2. **Express the orthocenter \(H\) and the med... | null | null |
Among the following propositions:
\\(①\\) The equation \\(x^{2}+(a-3)x+a=0\\) has one positive real root and one negative real root, then \\(a < 0\\);
\\(②\\) The function \\(y= \sqrt {x^{2}-1}+ \sqrt {1-x^{2}}\\) is both an even function and an odd function;
\\(③\\) If the range of the function \\(f(x)\\) is \\(\[-2,2... | \\(①\\) If the equation \\(x^{2}+(a-3)x+a=0\\) has one positive real root and one negative real root, then the product of the two roots is negative, hence \\(a < 0\\), which is correct.
\\(②\\) The function \\(y= \sqrt {x^{2}-1}+ \sqrt {1-x^{2}}\\) is defined only for \\(x \in \{-1, 1\}\\), and its values are the same ... | null | null |
Given the function $f(x)= \begin{cases} f(x-4), & x > 2 \\ e^{x}, & -2\leqslant x\leqslant 2 \\ f(-x), & x < -2\end{cases}$, then $f(-2017)=$ ()
A: $1$
B: $e$
C: $\dfrac {1}{e}$
D: $e^{2}$ | Since the function $f(x)= \begin{cases} f(x-4), & x > 2 \\ e^{x}, & -2\leqslant x\leqslant 2 \\ f(-x), & x < -2\end{cases}$,
we have $f(-2017)=f(2017)=f(1)=e$.
Therefore, the correct choice is $\boxed{B}$.
By the properties of the function, we get $f(-2017)=f(2017)=f(1)$, from which we can find the result.
This... | null | null |
Find the domain of the expression $\frac{\sqrt{x-2}}{\sqrt{5-x}}$. | To find the domain of the expression $\frac{\sqrt{x-2}}{\sqrt{5-x}}$, we need to ensure two things:
1. The expressions inside each square root must be non-negative.
2. The denominator cannot be equal to zero.
For the numerator $\sqrt{x-2}$ to be defined, we need $x-2 \ge 0$. Solving this inequality:
\[x - 2 \ge 0 \]
... | null | null |
Anna, Bridgit, and Carol run in a $100 \mathrm{~m}$ race. When Anna finishes, Bridgit is $16 \mathrm{~m}$ behind her, and when Bridgit finishes, Carol is $25 \mathrm{~m}$ behind her. The girls run at constant speeds throughout the race. How far behind was Carol when Anna finished?
A $37 \mathrm{~m}$
B $41 \mathrm{~m}$
... | 1. Given:
- When Anna finishes the $100 \, \mathrm{m}$ race, Bridgit is $16 \, \mathrm{m}$ behind her.
- When Bridgit finishes the $100 \, \mathrm{m}$ race, Carol is $25 \, \mathrm{m}$ behind her.
- The girls run at constant speeds.
2. Determine Carol's distance traveled when Bridgit finishes:
- When Bridg... | null | null |
Evaluate the expression $3 + 2\sqrt{3} + \frac{1}{3 + 2\sqrt{3}} + \frac{1}{2\sqrt{3} - 3}$.
A. $\frac{10\sqrt{3}}{3}\qquad$ B. $3 + \frac{10\sqrt{3}}{3}\qquad$ C. $3\sqrt{3}\qquad$ D. $10\qquad$ E. $6 + \frac{10\sqrt{3}}{3}$ | 1. **Simplify $\frac{1}{3 + 2\sqrt{3}}$:**
Multiply the numerator and denominator by the conjugate of the denominator:
\[
\frac{1}{3 + 2\sqrt{3}} \cdot \frac{3 - 2\sqrt{3}}{3 - 2\sqrt{3}} = \frac{3 - 2\sqrt{3}}{(3 + 2\sqrt{3})(3 - 2\sqrt{3})}.
\]
Using the difference of squares formula:
\[
(3 + 2\s... | null | null |
Prove that if 21 people collected 200 nuts, then there are two people who collected the same number of nuts. | To solve this problem, we will use the Pigeonhole Principle. We need to prove that if 21 people have collectively gathered 200 nuts, then at least two of them must have gathered the same number of nuts.
1. **Understanding the Problem:**
We have 21 people and 200 nuts. If every person gathered a different amount of ... | null | null |
Three positive integers have a sum of 90 and are in the ratio 1:3:5. What is the least of these three integers? | Let's denote the smallest integer as $x$. According to the ratio, the other integers are $3x$ and $5x$. The sum of these integers is given as:
\[ x + 3x + 5x = 9x \]
According to the problem statement, their total sum is 90. Therefore, we can set up the equation:
\[ 9x = 90 \]
Solving for $x$:
\[ x = \frac{90}{9} = 10... | null | null |
In a right triangle XYZ, where angle XYZ is right, find $\tan X$ if the lengths of side XY is 40 and the hypotenuse YZ is 41. | 1. Apply the Pythagorean theorem $XZ^2 + XY^2 = YZ^2$ to find $XZ$.
Given $XY = 40$, $YZ = 41$,
\[
XZ^2 = YZ^2 - XY^2 = 41^2 - 40^2 = 1681 - 1600 = 81
\]
Therefore, $XZ = \sqrt{81} = 9$.
2. Calculate $\tan X$ using $\tan X = \frac{opposite}{adjacent} = \frac{XZ}{XY}$.
\[
\tan X = \frac{XZ}{XY}... | null | null |
Two lines with slopes $\frac{3}{4}$ and $\frac{1}{3}$ intersect at $(3,3)$. What is the area of the triangle enclosed by these two lines and the line $x+y=8$?
$\textbf{(A) } 0.2 \qquad\textbf{(B) } 0.36 \qquad\textbf{(C) } 0.5 \qquad\textbf{(D) } 0.6 \qquad\textbf{(E) } 0.8$ | 1. **Identify the equations of the lines**:
- The line with slope $\frac{3}{4}$ passing through $(3,3)$ has the equation $y - 3 = \frac{3}{4}(x - 3)$, which simplifies to $y = \frac{3}{4}x + \frac{3}{4}$.
- The line with slope $\frac{1}{3}$ passing through $(3,3)$ has the equation $y - 3 = \frac{1}{3}(x - 3)$, w... | null | null |
Given the points \( A(0,3) \), \( B(0,6) \), \( C(0,-2) \), and \( D(0,2) \), point \( P \) is a moving point such that \(\frac{|PA|}{|PB|}=\frac{1}{2}\). Let line \( l \) be the perpendicular bisector of \( PC \), which intersects line \( PD \) at point \( Q \). Find the equation of the trajectory of point \( Q \) as ... | 1. Given points \(A(0, 3)\), \(B(0, 6)\), \(C(0, -2)\), and \(D(0, 2)\), and a point \(P\) moving such that \(\frac{|PA|}{|PB|} = \frac{1}{2}\).
2. We need to find the locus equation for point \(Q\), where line \(l\) is the perpendicular bisector of \(PC\), and \(l\) intersects with line \(PD\).
3. From the given con... | null | null |
I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different marbles at random. What is the expected value of the sum of the numbers on the marbles? | To find the expected value of the sum of the numbers on two marbles drawn at random from a bag containing marbles numbered 1 through 5, we first calculate the total number of different pairs of marbles that can be drawn. This is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n$ is the tot... | null | null |
On a plane, there are \( n \) points arranged such that any triangle with vertices at these points has an area of no more than 1. Prove that all these points can be placed within a triangle of area 4. | 1. We start by choosing the triangle with the largest area among all triangles that can be formed using the given $n$ points on the plane.
2. Let the area of this largest triangle be $s$, such that $s \leqslant 1$.
3. Denote the vertices of this triangle as \(A\), \(B\), and \(C\).
4. Draw lines through each vertex ... | null | null |
Let \(\theta\) be the angle between the line
\[\frac{x - 3}{3} = \frac{y + 4}{-1} = \frac{z - 2}{4}\]
and the plane \(4x + 5y - 7z = 10.\)
Find \(\sin \theta.\) | First, identify the direction vector of the line, \(\mathbf{d} = \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}\), and the normal vector of the plane, \(\mathbf{n} = \begin{pmatrix} 4 \\ 5 \\ -7 \end{pmatrix}\).
The cosine of the angle \(90^\circ - \theta\) between the line and the vector normal to the plane is given by
... | null | null |
Find two sets of consecutive integers: one pair and one triplet, such that the product of the pair and the product of the triplet both equal 360. What is the sum of all five integers? | First, factorize $360 = 2^3 \cdot 3^2 \cdot 5$.
**For the pair of consecutive integers**:
Suppose the integers are $n$ and $n+1$. Then, we have:
\[ n(n+1) = 360 \]
\[ n^2 + n - 360 = 0 \]
Using the quadratic formula, $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1, b=1$, and $c=-360$:
\[ n = \frac{-1 \pm \sqrt{1 ... | null | null |
For the one-variable quadratic equation $x^{2}-5x+m=0$ with two equal real roots in terms of $x$, then $m=$____. | To solve for $m$ in the quadratic equation $x^{2}-5x+m=0$ with two equal real roots, we follow these steps:
1. Recall that for a quadratic equation $ax^2 + bx + c = 0$, the discriminant $\Delta$ is given by $\Delta = b^2 - 4ac$. For the equation to have two equal real roots, $\Delta$ must equal $0$.
2. In our equatio... | null | null |
A square of paper is folded in half four times to obtain a smaller square. Then a corner is removed. Which of the following could be the paper after it is unfolded? | 1. **Understanding the problem and the folding process**:
- We start with a square piece of paper.
- The paper is folded in half four times.
- Folding in half each time doubles the number of layers of paper.
2. **Calculating the number of layers of paper after folding**:
- Initially, the paper is a sin... | null | null |
The diameter of a certain coronavirus is $120$ nanometers. Given that $1$ nanometer $=10^{-9}$ meters, what is the diameter of this coronavirus in centimeters?
A: $120\times 10^{-9}$
B: $1.2\times 10^{-7}$
C: $1.2\times 10^{-6}$
D: $1.2\times 10^{-5}$ | To convert the diameter of the coronavirus from nanometers to centimeters, we follow these steps:
1. Convert the diameter from nanometers to meters.
- The diameter is $120$ nanometers.
- Since $1$ nanometer $= 10^{-9}$ meters, we multiply the diameter by $10^{-9}$ to convert it to meters:
\[120 \times 10^{-... | null | null |
Find the monotonic intervals and extreme values of the function $y=2\ln x\cdot x^{2}$. | From the given problem, we know that the domain of the function is: $(0,+\infty)$
The derivative of the function is: $f'(x)=4x\cdot\ln x+2\cdot x^{2}\cdot\dfrac{1}{x}=4x\cdot\ln x+2x$.
By setting $f'(x)\leqslant 0$, we have $2x\cdot\ln x+x\leqslant 0$. Solving for $x$, we get $0\leqslant x\leqslant e^{-\frac{1}{2}}$. ... | null | null |
If $\tan \alpha = 3$, then $\cos^{2}(\alpha + \frac{\pi}{4}) - \cos^{2}(\alpha - \frac{\pi}{4}) =$ ?
A: $- \frac{3}{5}$
B: $- \frac{4}{5}$
C: $\frac{3}{5}$
D: $\frac{4}{5}$ | We have $\cos^{2}(\alpha + \frac{\pi}{4}) - \cos^{2}(\alpha - \frac{\pi}{4}) = \cos^{2}(\frac{\pi}{4} + \alpha) - \sin^{2}(\frac{\pi}{4} + \alpha) = \cos(2\alpha + \frac{\pi}{2})$
$= -\sin 2\alpha = - \frac{2\sin \alpha \cos \alpha}{\sin^{2}\alpha + \cos^{2}\alpha} = - \frac{2\tan \alpha}{\tan^{2}\alpha + 1} = - \fra... | null | null |
A car sales company has adopted a more flexible payment method for promoting sales. For a sedan priced at 100,000 yuan, under the premise that the payment is completed within a year, customers can choose from the following two installment payment plans:
Plan 1: Pay in 3 installments, with the first payment made 4 mont... | **Answer**
If payment is made according to Plan 1, let each payment be $a$ (in ten thousand yuan).
Then we have $a + a(1+0.8\%)^4 + a(1+0.8\%)^8 = 10 \times (1+0.8\%)^{12}$
Thus, $a = \frac{10 \times 1.008^{12}}{3}$, and the total payment amount $S_1 = 3a = 10 \times 1.008^{12}$
If payment is made according to Plan... | null | null |
Suppose we have 12 dogs and we want to divide them into three groups, one with 4 dogs, one with 5 dogs, and one with 3 dogs. How many ways can we form the groups such that Fluffy is in the 4-dog group and Nipper is in the 5-dog group? | 1. **Placing Fluffy and Nipper**: Place Fluffy in the 4-dog group and Nipper in the 5-dog group. This leaves 10 dogs to be placed.
2. **Filling Fluffy's group**: Fluffy's 4-dog group needs 3 more dogs from the remaining 10. The number of ways to choose 3 dogs out of 10 is $\binom{10}{3}$.
3. **Filling Nipper's group**:... | null | null |
If the fractional equation $\frac{{2-m}}{{1-x}}=1$ has no solution with respect to $x$, then $m=$____. | To solve the problem, we start by analyzing the given fractional equation:
\[
\frac{2-m}{1-x} = 1
\]
**Step 1:** Remove the denominator to simplify the equation. This is done by multiplying both sides of the equation by the denominator $(1-x)$, resulting in:
\[
2-m = 1-x
\]
**Step 2:** Rearrange the equation to exp... | null | null |
Find the coordinates of the point symmetric to point $(1, -2, 3)$ with respect to the $x$-axis. | To find the coordinates of the point symmetric to a given point with respect to the $x$-axis, we need to change the signs of the $y$ and $z$ coordinates, while keeping the $x$ coordinate unchanged. This is because the reflection across the $x$-axis only affects the $y$ and $z$ coordinates.
Given the point $(1, -2, 3)... | null | null |
Expanding $(1+0.1)^{2000}$ by the binomial theorem and doing no further manipulation gives
\[{2000 \choose 0}(0.1)^0+{2000 \choose 1}(0.1)^1+{2000 \choose 2}(0.1)^2+\cdots+{2000 \choose 2000}(0.1)^{2000}= B_0 + B_1 + B_2 + \cdots + B_{2000},\]where $B_k = {2000 \choose k}(0.1)^k$ for $k = 0,1,2,\ldots,2000.$ For which ... | To compare different values of $B_k,$ we look at the ratio $B_k/B_{k-1},$ which equals
\[\frac{B_k}{B_{k-1}} = \frac{\binom{2000}{k} (0.1)^k}{\binom{2000}{k-1} (0.1)^{k-1}} = \frac{\frac{2000!}{k!(2000-k)!} (0.1)^k}{\frac{2000!}{(k-1)!(2001-k)!} (0.1)^{k-1}} = \frac{2001-k}{10k}.\]
The inequality
\[\frac{B_k}{B_{k-1}}... | null | null |
Let $a, b, c$ be real numbers such that $a + b + c = 1$. Find the set of all possible values of $ab + ac + bc$. | Starting with the equation $a + b + c = 1$, we square both sides to obtain:
\[(a + b + c)^2 = 1^2 = 1.\]
Expanding the left side,
\[a^2 + b^2 + c^2 + 2(ab + ac + bc) = 1.\]
Rearranging to solve for $ab + ac + bc$,
\[2(ab + ac + bc) = 1 - (a^2 + b^2 + c^2).\]
Thus,
\[ab + ac + bc = \frac{1 - (a^2 + b^2 + c^2)}{2}.\]
Sin... | null | null |
Jasmine swims 12 laps every afternoon, Monday through Friday. How many laps does she swim in five weeks? | To calculate the total number of laps Jasmine swims in five weeks, we follow these steps:
1. Determine the number of laps Jasmine swims per day, which is given as 12 laps.
2. Since Jasmine swims 5 days a week, we calculate the total laps per week as follows:
\[
12 \, \text{laps/day} \times 5 \, \text{days/week} ... | null | null |
Given functions $f\left(x\right)=2^{x}$ and $g\left(x\right)=\log _{a}x$, if for any point $P$ on the graph of $f\left(x\right)$, there always exists a point $Q$ on the graph of $g\left(x\right)$ such that $OP\bot OQ$ and $|OP|=|OQ|$, then the real number $a=\_\_\_\_\_\_$. | To solve this problem, we start by defining the points on the graphs of the given functions $f(x) = 2^x$ and $g(x) = \log_a x$. Let's denote a point $P$ on the graph of $f(x)$ by $P(x, 2^x)$ and a point $Q$ on the graph of $g(x)$ by $Q(b, y)$.
1. **Case when $x=0$:**
- For $x=0$, the point on $f(x)$ is $P(0,1)$. Ac... | null | null |
End of preview. Expand in Data Studio
NuminaMath Worked Solutions
Competition and school maths problems with step-by-step solutions.
| Rows | 24,000 |
| Domain | mathematics |
| Format | data.parquet, one row per example |
| Licence | apache-2.0 |
| Built for | supervised fine-tuning (SFT) experiments on Adaption AutoScientist |
Columns
| Column | Description |
|---|---|
original_prompt |
The prompt (user turn) as uploaded. |
original_completion |
The target response as uploaded. |
enhanced_prompt |
Empty in this dataset. |
enhanced_completion |
Empty in this dataset. |
How it was built
Sampled from NuminaMath-CoT, balanced across its sources and decontaminated against MATH-500, GSM8K test and the NuminaMath test split.
Sources and licence
- AI-MO/NuminaMath-CoT (Apache-2.0)
Notes
- Columns
enhanced_prompt,enhanced_completionare empty in this dataset (the Adaption export reserves them for rewritten text).
Loading
from datasets import load_dataset
ds = load_dataset("rodriguescarson/adaption-math-numina-worked-24k", split="train")
import pandas as pd
df = pd.read_parquet("hf://datasets/rodriguescarson/adaption-math-numina-worked-24k/data.parquet")
Published by Carson Rodrigues (Hugging Face, Kaggle).
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