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Which of the following sets of three line segments can form a right triangle?
A: $2cm$, $3cm$, $4cm$
B: $3cm$, $4cm$, $5cm$
C: $4cm$, $5cm$, $6cm$
D: $5cm$, $6cm$, $7cm$ | To determine which set of three line segments can form a right triangle, we apply the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Let's examine ... | \text{B} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,076 |
Calculate.$(1)\sqrt[3]{{-8}}×|{-3}|+{({-\frac{1}{2}})^{-2}}÷{5^0}+{({-1})^{2023}}$;$(2)\left(b+a\right)\left(b-a\right)+\left(a+2b\right)^{2}$. | ### Solution:
#### For Problem (1):
We start by breaking down the expression into simpler parts:
1. Calculate the cube root of $-8$: $\sqrt[3]{{-8}} = -2$.
2. Take the absolute value of $-3$: $|-3| = 3$.
3. Evaluate the exponent: $\left(-\frac{1}{2}\right)^{-2} = 4$.
4. Note that any number to the power of $0$ is $1... | 5b^2 + 4ab | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,077 |
In $\triangle ABC$, the lengths of the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given that $a=3$, $b=\sqrt{3}$, and $A=\dfrac{\pi}{3}$, find the measure of angle $B$. | **Analysis**
This problem involves the application of the Law of Sines and the values of trigonometric functions for special angles. Pay attention to the range of the internal angles, which is a fundamental aspect of the problem.
We can solve the problem using the Law of Sines.
**Step-by-Step Solution**
1. Accordin... | \frac{\pi}{6} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,078 |
The rectangular coordinates of point $M$ with polar coordinates $(4, \frac{5\pi}{6})$ are $\_\_\_\_\_\_$. | We have $x=4\cos{\frac{5\pi}{6}}=-2\sqrt{3}$ and $y=4\sin{\frac{5\pi}{6}}=2$.
Thus, the rectangular coordinates of point $M$ are $(-2\sqrt{3}, 2)$.
Here's a step-by-step solution:
1. Use the formulas for converting polar coordinates to rectangular coordinates:
$$x=r\cos{\theta} \quad \text{and} \quad y=r\sin{\thet... | (-2\sqrt{3}, 2) | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,080 |
Among the following four propositions, the true proposition is ( )
A: $x>3$ is a sufficient condition for $x>5$
B: $x^2=1$ is a sufficient condition for $x=1$
C: $a>b$ is a necessary condition for $ac^2>bc^2$
D: $α= \frac {π}{2}$ is a necessary condition for $\sinα=1$ | Solution: A. $x>3$ is a necessary but not sufficient condition, so it is incorrect;
B. $x^2=1$ is a necessary but not sufficient condition;
C. $a>b$ is a necessary but not sufficient condition for $ac^2>bc^2$;
D. $α= \frac {π}{2}$ implies $\sinα=1$, but the converse is not true.
Therefore, the correct choice is... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,081 |
Given that the inequality $(ax-20) \lg \frac{2a}{x} \leqslant 0$ holds for any positive real number $x$, determine the range of values for $a$.
A: $[-10,10]$
B: $[-\sqrt{10}, \sqrt{10}]$
C: $(-\infty, \sqrt{10}]$
D: $\{\sqrt{10}\}$ | The inequality $(ax-20) \lg \frac{2a}{x} \leqslant 0$ is equivalent to either of the following systems:
$$
\begin{cases}
ax-20 \geqslant 0 \\
\lg \frac{2a}{x} \leqslant 0 \\
x > 0
\end{cases} \text{ or }
\begin{cases}
ax-20 \leqslant 0 \\
\lg \frac{2a}{x} \geqslant 0 \\
x > 0
\end{cases}
$$
Considering the first syste... | \{\sqrt{10}\} | Inequalities | MCQ | Yes | Yes | cn_k12 | false | 549,082 |
If $n\left(n\neq 0\right)$ is a root of the equation $x^{2}+mx+3n=0$ with respect to $x$, then the value of $m+n$ is ______. | Given that $n$ (where $n\neq 0$) is a root of the equation $x^{2}+mx+3n=0$, we substitute $x=n$ into the equation to find the relationship between $m$ and $n$.
Starting with the equation:
\[x^{2}+mx+3n=0\]
Substituting $x=n$ gives:
\[n^{2}+mn+3n=0\]
Since $n\neq 0$, we can simplify this equation to find a relations... | -3 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,083 |
Given the function $f(x) = -f'(0)e^x + 2x$, point $P$ lies on the tangent line $l$ to the curve $y=f(x)$ at the point $(0, f(0))$, and point $Q$ lies on the curve $y=e^x$. Find the minimum value of the distance $|PQ|$. | Firstly, let's consider the function $f(x) = -f'(0)e^x + 2x$. We need to find its derivative with respect to $x$:
$$ f'(x) = -f'(0)e^x + 2. $$
Since $f'(0) = -f'(0)e^0 + 2$, we can solve for $f'(0)$:
$$ f'(0) = -f'(0) + 2. $$
From this equation, we determine that $f'(0) = 1$.
After finding $f'(0)$, we can write the f... | \sqrt{2} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,084 |
A cylinder with a base radius of $R$ is intersected by a plane that forms an angle of $\theta(0^\circ < \theta < 90^\circ)$ with its base. The cross-section is an ellipse. When $\theta$ is $30^\circ$, the eccentricity of this ellipse is ( ).
A: $\dfrac{1}{2}$
B: $\dfrac{\sqrt{3}}{2}$
C: $\dfrac{\sqrt{3}}{3}$
D: $\dfra... | 1. **Understand the Problem**: We are given a cylinder with a base radius of $R$ and an intersecting plane that forms an angle of $\theta$ with the base. We need to find the eccentricity of the resulting elliptical cross-section when $\theta$ is $30^\circ$.
2. **Formula for Eccentricity**: The eccentricity ($e$) of an... | \frac{1}{2} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,085 |
Given that the domain of the function $f(x)$ is $(0,+\infty)$, and $f(xy)=f(x)+f(y)$, when $x>1$, $f(x)<0$, and $f(2)=-1$, which of the following statements is correct?
A: $f(1)=0$
B: The function $f(x)$ is decreasing on $(0,+\infty)$
C: $f(\frac{1}{2022})+f(\frac{1}{2021})+\cdots+f(\frac{1}{3})+f(\frac{1}{2})+f(2)+... | To analyze the given statements about the function $f(x)$, we proceed as follows:
**Statement A: $f(1)=0$**
- Let's consider $x=y=1$. Then, by the given property $f(xy)=f(x)+f(y)$, we have:
\[
f(1\cdot1)=f(1)+f(1) \implies f(1)=2f(1)
\]
Simplifying this equation gives us:
\[
0=f(1) \implies \boxed{f(1)=0}... | ABD | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,087 |
If $2|x-1|+|x-a| \geq 2$ holds for any real number $x$, then the range of the real number $a$ is. | Given that when $|x-1| \geq 1$, which means $x \geq 2$ or $x \leq 0$, we have $2|x-1| \geq 2$.
Therefore, $2|x-1|+|x-a| \geq 2$ holds for any real number $x$.
Thus, the original inequality holds for any real number $a$.
Therefore, $2|x-1|+|x-a| \geq 2$ holds for any real number $x$ if and only if $2|x-1|+|x-a| \geq ... | (-\infty, -1] \cup [3, +\infty) | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false | 549,088 |
Given $$\cos\left( \frac {3\pi}{14}-\theta\right)= \frac {1}{3}$$, then $$\sin\left( \frac {2\pi}{7}+\theta\right)$$ equals ( )
A: $$\frac {1}{3}$$
B: $$- \frac {1}{3}$$
C: $$\frac {2 \sqrt {2}}{3}$$
D: $$- \frac {2 \sqrt {2}}{3}$$ | Since $$\cos\left( \frac {3\pi}{14}-\theta\right)= \frac {1}{3}$$,
it follows that $$\cos\left( \frac {3\pi}{14}-\theta\right)$$ equals $$\sin\left( \frac {\pi}{2}- \frac {3\pi}{14}+\theta\right)$$ which equals $$\sin\left( \frac {2\pi}{7}+\theta\right)$$ equals $$\frac {1}{3}$$.
Therefore, the correct choice is: A.
... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,089 |
(I) Simplify and evaluate: $\log \_{ \frac {1}{3}} \sqrt {27}+\lg 25+\lg 4+7^{-\log \_{7}2}+(-0.98)^{0}$
(II) Given a point $P( \sqrt {2},- \sqrt {6})$ on the terminal side of angle $α$, evaluate: $\frac{\cos ( \frac {π}{2}+α)\cos (2π-α)+\sin (-α- \frac {π}{2})\cos (π-α)}{\sin (π+α)\cos ( \frac {π}{2}-α)}$. | (I) $\log \_{ \frac {1}{3}} \sqrt {27}+\lg 25+\lg 4+7^{-\log \_{7}2}+(-0.98)^{0}$
$=\log \_{ \frac {1}{3}}3^{ \frac {3}{2}}+\lg 100+7^{\log \_{7} \frac {1}{2}}+1$
$=- \frac {3}{2}+2+ \frac {1}{2}+1$
$=\boxed{2}$.
(II) Given a point $P( \sqrt {2},- \sqrt {6})$ on the terminal side of angle $α$,
we have $\tan α= \frac {... | \frac { \sqrt {3}+1}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,090 |
In the triangular prism $ABC-A_{1}B_{1}C_{1}$, $\overrightarrow{AB}=\overrightarrow{a}$, $\overrightarrow{AC}=\overrightarrow{b}$, $\overrightarrow{A{A_1}}=\overrightarrow{c}$. If point $D$ is the midpoint of $B_{1}C_{1}$, then $\overrightarrow{CD}=$
A: $\frac{1}{2}\overrightarrow{a}-\frac{1}{2}\overrightarrow{b}+\ove... | To solve for $\overrightarrow{CD}$ in the triangular prism $ABC-A_{1}B_{1}C_{1}$, we start by noting the given vectors and the midpoint condition:
1. Given that $\overrightarrow{AB}=\overrightarrow{a}$, $\overrightarrow{AC}=\overrightarrow{b}$, and $\overrightarrow{A{A_1}}=\overrightarrow{c}$.
2. Point $D$ is the midp... | \text{A: } \frac{1}{2}\overrightarrow{a} - \frac{1}{2}\overrightarrow{b} + \overrightarrow{c} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,091 |
Given the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ with an eccentricity of $2$, and one of its focal points having coordinates $(4, 0)$, find the values of $a$ and $b$. | Since the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ has an eccentricity of $2$ and one of its focal points is at $(4, 0)$,
We know that $\frac{c}{a} = 2$ and $c = 4$,
Thus, $a = 2$ and $b = 2\sqrt{3}$.
Therefore, the answer is: $\boxed{a = 2}$ and $\boxed{b = 2\sqrt{3}}$.
Using the hyperbola's stand... | b = 2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,092 |
Given the function $f\left(x\right)=ax^{5}+b\sin x+c$, if $f\left(-1\right)+f\left(1\right)=2$, then $c=\left(\ \ \right)$
A: $-1$
B: $0$
C: $1$
D: $\frac{2}{3}$ | Given the function $f\left(x\right)=ax^{5}+b\sin x+c$, we are asked to find the value of $c$ given that $f\left(-1\right)+f\left(1\right)=2$. Let's solve this step by step:
1. First, we substitute $x=-1$ and $x=1$ into the function $f(x)$:
- For $x=-1$: $f(-1)=a(-1)^{5}+b\sin(-1)+c=-a-b\sin(1)+c$.
- For $x=1$: $... | 1 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,093 |
Given the curve C: $x^2+y^2+2x+m=0$ ($m \in \mathbb{R}$)
(1) If the trajectory of curve C is a circle, find the range of values for $m$;
(2) If $m=-7$, and a line passing through point P(1, 1) intersects curve C at points A and B, with $|AB|=4$, find the equation of line AB. | (1) Completing the square for the original equation yields: $(x+1)^2+y^2=1-m$,
Since the trajectory of curve C is a circle,
we have $1-m>0$, thus $m<1$.
So, the range of values for $m$ is $\boxed{m<1}$.
(2) When $m=-7$, we have $(x+1)^2+y^2=8$, with the center of the circle at (-1, 0) and radius $2\sqrt{2}$.
W... | 3x+4y-7=0 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,094 |
Which of the following real numbers is irrational?
A: $0.3$
B: $3.14$
C: $\sqrt{9}$
D: $\sqrt{3}$ | To determine which of the given real numbers is irrational, we analyze each option step by step:
- **Option A**: $0.3$ can be expressed as the fraction $\frac{3}{10}$, which means it is a rational number. Therefore, $0.3$ does not meet the criteria for being irrational.
- **Option B**: $3.14$ can also be expressed as... | D | Number Theory | MCQ | Yes | Yes | cn_k12 | false | 549,095 |
Given $(a-2i)i=b-i$, where $a, b \in \mathbb{R}$, and $i$ is the imaginary unit, the complex number $a+bi= \_\_\_\_\_\_$. | From $(a-2i)i=b-i$, we get $2+ai=b-i$.
Therefore, $a=-1$ and $b=2$.
Thus, $a+bi=-1+2i$.
Hence, the answer is $\boxed{-1+2i}$.
By expanding the left side of the given equation through the multiplication of a monomial by a polynomial, and then using the condition of equality of complex numbers to set up equations... | -1+2i | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,096 |
Given the function $y= \frac {\sin x}{x} + \sqrt {x} + 2$, then $y′= $ ______. | Since $y= \frac {\sin x}{x} + \sqrt {x} + 2$,
then $y′= \frac {x\cos x-\sin x}{x^{2}} + \frac {1}{2} \cdot x^{- \frac {1}{2}} = \frac {x\cos x-\sin x}{x^{2}} + \frac {1}{2 \sqrt {x}}$.
Therefore, the answer is: $\boxed{\frac {x\cos x-\sin x}{x^{2}} + \frac {1}{2 \sqrt {x}}}$.
This problem is solved directly using... | \frac {x\cos x-\sin x}{x^{2}} + \frac {1}{2 \sqrt {x}} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,097 |
When $x=$______, the value of the fraction $\frac{2{x}^{2}-6x}{x-3}$ is $0$. | To solve the given problem, we start by setting the numerator of the fraction equal to zero because a fraction is zero if and only if its numerator is zero, and the denominator is not zero. The fraction given is $\frac{2x^2 - 6x}{x - 3}$.
1. Set the numerator equal to zero: $2x^2 - 6x = 0$.
2. Factor out the common fa... | 0 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,098 |
In the interval \\((- \dfrac {π}{2}, \dfrac {π}{2})\\), a number \\(x\\) is randomly selected. The probability that \\(\tan x > 1\\) is \_\_\_\_\_\_. | Given that \\(\tan x > 1 = \tan \dfrac {π}{4}\\) and \\(x \in (- \dfrac {π}{2}, \dfrac {π}{2})\\),
We can infer that \\(\dfrac {π}{4} 1\\) within the given interval, and then use the length as the measure to calculate the probability. This problem tests geometric understanding and computational skills. | null | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,099 |
Given that $f(x)=x^{2}+3ax+4$, and $b-3 \leqslant x \leqslant 2b$ is an even function, find the value of $a-b$. | Since the function $f(x)=x^{2}+3ax+4$, and $b-3 \leqslant x \leqslant 2b$ is an even function,
We know that $f(-x)=f(x)$, which means $x^{2}-3ax+4=x^{2}+3ax+4$, and $b-3+2b=0$
Solving these equations, we get $a=0$, and $b=1$,
Hence, $a-b=-1$.
So, the answer is $\boxed{-1}$.
By the definition of an even function, w... | -1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,100 |
Given the function $f(x) = x^2 - bx + c$, the axis of symmetry of $f(x)$ is $x = 1$ and $f(0) = -1$.
(1) Find the values of $b$ and $c$.
(2) Find the range of $f(x)$ when $x \in [0, 3]$.
(3) If the inequality $f(\log_2 k) > f(2)$ holds, find the range of the real number $k$. | (1) Since the axis of symmetry of $f(x)$ is $x = 1$ and $f(0) = -1$, we know that
$$
\frac{b}{2a} = 1 \quad \text{and} \quad f(0) = c = -1.
$$
Hence, for the axis of symmetry $x=\frac{-b}{2a}$ to be equal to 1, we have
$$
\frac{-b}{2 \cdot 1} = 1 \quad \Rightarrow \quad -b = 2 \quad \Rightarrow \quad b = -2.
$$
Usi... | k > 4 \quad \text{or} \quad 0 < k < 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,101 |
A certain agricultural machinery factory produced 500,000 parts in April and a total of 1,820,000 parts in the second quarter. If the average monthly growth rate of the factory in May and June is $x$, then the equation satisfied by $x$ is:
A: $50(1+x)^{2}=182$
B: $50+50(1+x)+50(1+x)^{2}=182$
C: $50(1+2x)=182$
D: $5... | To solve the problem, we start by understanding the given information and the question. The factory produced 500,000 parts in April and a total of 1,820,000 parts in the second quarter (April, May, and June). The question introduces a variable $x$, which represents the average monthly growth rate in May and June.
Give... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,102 |
Given point $Q$ on the ellipse $C$: $\frac{x^{2}}{16} + \frac{y^{2}}{10} = 1$, point $P$ satisfies $\overrightarrow{OP} = \frac{1}{2}(\overrightarrow{OF_{1}} + \overrightarrow{OQ})$, where $O$ is the coordinate origin and $F_{1}$ is the left focus of ellipse $C$. Determine the trajectory of point $P$.
A: Circle
B: Par... | Since point $P$ satisfies $\overrightarrow{OP} = \frac{1}{2}(\overrightarrow{OF_{1}} + \overrightarrow{OQ})$, we can deduce that $P$ is the midpoint of line segment $QF_{1}$.
Let $P(a, b)$.
Given that $F_{1}$ is the left focus of ellipse $C$: $\frac{x^{2}}{16} + \frac{y^{2}}{10} = 1$, it follows that $F_{1}(-\sqrt{6}... | D | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,103 |
If $A=\{x|1＜{(\frac{1}{2})}^{x-a}＜4\}$, $B=\{x|\frac{1}{2-x}≤1\}$,
$(1)$ When $a=\frac{3}{2}$, find $A∩\overline{B}$;
$(2)$ If $x\in A$ is a sufficient condition for $x\in B$, find the range of real number $a$. | ### Solution:
#### Part (1):
Given $1＜{(\frac{1}{2})}^{x-a}＜4$, we can rewrite this inequality as $2^{0} \lt 2^{-x+a} \lt 2^{2}$, which implies:
\[0 \lt -x+a \lt 2.\]
Solving for $x$, we get:
\[a-2 \lt x \lt a.\]
Thus, the set $A$ can be represented as $A=\left(a-2,a\right)$.
For the set $B$, given $\frac{1}{2-x}≤1$... | \{a\left|\right.a\leqslant 1 \text{ or } a\geqslant 4\} | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false | 549,105 |
Which of the following statements is correct?
A: If $a$, $b$, and $c$ are lines, $a$∥$b$, $b\bot c$, then $a$∥$c$
B: If $a$, $b$, and $c$ are lines, $a\bot b$, $b\bot c$, then $a\bot c$
C: If $a$, $b$, and $c$ are lines, $a$∥$b$, $b$∥$c$, then $a$∥$c$
D: If $a$, $b$, and $c$ are lines, $a$∥$b$, $b$∥$c$, then $a\bot... | Let's analyze each statement step by step, following the given conditions and geometrical properties:
**Statement A**: Given $a \parallel b$ and $b \perp c$, we infer that $a$ cannot be parallel to $c$ because $a$ would also have to be perpendicular to $c$ due to the transitive relation of perpendicular lines. This co... | C | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,106 |
The complex number $z$ satisfies the equation $\overline{z}(1+2i)=4+3i$. Determine the value of $z$ from the options below:
A: $2-i$
B: $2+i$
C: $1+2i$
D: $1-2i$ | **Analysis**
This problem primarily tests the understanding of basic complex number arithmetic and the concept of complex conjugates. It is a fundamental question.
**Step-by-Step Solution**
1. We start with the given equation:
$$
\overline{z}(1+2i)=4+3i
$$
Our goal is to find the value of $z$.
2. To solve for $z$, w... | 2+i | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,107 |
If the circle $x^2+y^2+Dx+Ey+F=0$ intersects the y-axis at two points on either side of the origin, then ( )
A: $D\neq0$, $F>0$
B: $E=0$, $F>0$
C: $E\neq0$, $D=0$
D: $F<0$ | Let $x=0$, then the equation of the circle becomes $y^2+Ey+F=0$.
When $E^2>4F$, i.e., the equation has two solutions,
then the roots of this equation are the y-coordinates of the points where the circle intersects the y-axis.
According to the problem, the two intersection points with the y-axis are required to be on... | D | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,108 |
Three manufacturers, A, B, and C, have market shares of $25\%$, $35\%$, and $40%$ for their mobile phone chargers, respectively. The qualification rates of their chargers are $70\%$, $75\%$, and $80%$, respectively.
$(1)$ The local industrial and commercial quality inspection department randomly selects $3$ mobile phon... | Let's break down the solution into detailed steps:
### Part 1: Probability Distribution, Expectation, and Variance of $X$
Given:
- Manufacturer A has a market share of $25\%$, so $P(A) = 0.25$.
- The possible values of $X$ (chargers produced by Manufacturer A) are $0$, $1$, $2$, and $3$.
**Probability Distribution:*... | \frac{30}{97} | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false | 549,109 |
The distance from the vertex of the hyperbola $C: \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ to its asymptote is _______. | The coordinates of one vertex of the hyperbola $C: \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ are (4, 0), and the equation of one of its asymptotes is 3x + 4y = 0.
Hence, the required distance is: $$\frac{12}{\sqrt{3^{2} + 4^{2}}} = \frac{12}{5}$$.
Thus, the answer is: $\boxed{\frac{12}{5}}$.
The solution involves find... | \frac{12}{5} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,110 |
Given lines $l$, $m$, $n$ are different, and planes $\alpha$, $\beta$ are different, there are the following four propositions. The number of correct propositions is ( )
① If $\alpha \perp \beta$, $l \perp \alpha$, then $l \parallel \beta$
② If $\alpha \perp \beta$, $l \subset \alpha$, then $l \perp \beta$
③ If... | ① If $\alpha \perp \beta$, $l \perp \alpha$, then $l \parallel \beta$ or $l \subset \beta$, hence proposition ① is incorrect,
② If $\alpha \perp \beta$, $l \subset \alpha$, then $l \perp \beta$ or $l \parallel \beta$, hence proposition ② is incorrect,
③ If $l \perp m$, $m \perp n$, then $l \parallel n$ or $l$ inter... | \text{D} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,111 |
Given that $m$ and $n$ are two different lines, and $\alpha$, $\beta$, $\gamma$ are three different planes, which of the following statements is correct?
A: If $m\parallel\alpha$ and $n\parallel\alpha$, then $m\parallel n$
B: If $m\parallel\alpha$ and $m\parallel\beta$, then $\alpha\parallel\beta$
C: If $\alpha\perp... | **Analysis**
This question examines the judgment and properties of two planes being parallel, the properties of a plane perpendicular to another plane, and the properties of a line perpendicular to a plane. Special cases should be considered, making it a medium-level question.
By providing counterexamples, it can be ... | D | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,112 |
The equivalent proposition to "For x, y ∈ Z, if x, y are odd, then x + y is even" is ___. | The equivalence of a proposition and its contrapositive relates to their truth values being the same. The contrapositive of the given proposition is: "For x, y ∈ Z, if x + y is not even, then not both x and y are odd."
We can express the original statement formally as: If $x$ and $y$ are odd ($x = 2n + 1$ and $y = 2m ... | \text{For } x, y \in \mathbb{Z}, \text{ if } x + y \text{ is not even, then not both } x \text{ and } y \text{ are odd.} | Number Theory | math-word-problem | Yes | Yes | cn_k12 | false | 549,114 |
Determine the range of values for $b$ such that the line $3x + 4y = b$ intersects the circle $x^2 + y^2 - 2x - 2y + 1 = 0$. | First, rewrite the circle's equation in standard form: $(x-1)^2 + (y-1)^2 = 1$. This gives us the center of the circle $(1, 1)$, and its radius $r = 1$.
For the line $3x + 4y = b$ to intersect the circle $x^2 + y^2 - 2x - 2y + 1 = 0$, the distance $d$ from the circle's center to the line must be less than its radius. ... | (2, 12) | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,115 |
Given $f(\cos x) = \cos 3x$, then $f(\sin x)$ equals ( )
A: $-\sin 3x$
B: $-\cos 3x$
C: $\cos 3x$
D: $\sin 3x$ | **Method 1:** Let $t = \cos x$,
Since $\cos 3x = 4\cos^3 x - 3\cos x$, and $f(\cos x) = \cos 3x = 4\cos^3 x - 3\cos x$,
Therefore, $f(t) = 4t^3 - 3t$,
Thus, $f(\sin x) = 4\sin^3 x - 3\sin x = -\sin 3x$,
Hence, the correct option is $\boxed{\text{A}}$.
**Method 2:** Since $f(\cos x) = \cos 3x$,
Therefore, $f(\sin ... | \text{A} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,116 |
In the rectangular coordinate system $(xOy)$, establish a polar coordinate system with the origin $O$ as the pole, the non-negative semi-axis of $x$ as the polar axis, and the same length unit as the rectangular coordinate system $(xOy)$. Given that the polar coordinate equation of curve $C_{1}$ is $ρ=2\cos θ$ and the ... | 1. The polar coordinate equation of curve $C_{1}$ is $ρ=2\cos θ$, so its Cartesian equation is $(x-1)^2+y^2=1$.
The parametric equation of curve $C_{2}$ is $\begin{cases} x=-\frac{4}{5}t \\ y=-2+\frac{3}{5}t\end{cases}$, so its Cartesian equation is $3x+4y+8=0$.
The distance $d$ between the center $C_{1}(1,0)$ ... | [1-\sqrt{2}, 1+\sqrt{2}] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,117 |
The parabola $y=ax^{2}+bx+c$ passes through $A(-1,3)$ and $B(2,3)$. The solutions of the quadratic equation in $x$, $a(x-2)^{2}-3=2b-bx-c$, are ____. | To solve for the solutions of the quadratic equation $a(x-2)^{2}-3=2b-bx-c$, let's follow the steps closely aligned with the provided solution:
1. Start with the given quadratic equation in $x$: $a(x-2)^{2}-3=2b-bx-c$.
2. Rearrange this equation to match the form of a shifted parabola: $a(x-2)^{2}+b(x-2)+c=0$.
By und... | 1 \text{ or } 4 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,118 |
In $\triangle ABC$, if $\sin A \cos A = \sin B \cos B$, then the shape of $\triangle ABC$ is \_\_\_\_\_\_. | Given that in $\triangle ABC$, $A$, $B$, and $C$ are internal angles, and $\sin A \cos A = \sin B \cos B$,
Using the double angle identity for sine, we can rewrite the equation as $\frac{1}{2} \sin 2A = \frac{1}{2} \sin 2B$, which simplifies to $\sin 2A = \sin 2B$.
This implies that either $2A + 2B = 180^{\circ}$ or ... | \text{Isosceles or right triangle} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,119 |
Given $f(x) = \begin{cases} \cos(\pi x), & \text{if } x<1 \\ f(x-1), & \text{if } x\geq 1 \end{cases}$, find the value of $f\left( \frac {1}{3}\right)+f\left( \frac {4}{3}\right)$.
A: 0
B: $\frac {1}{2}$
C: 1
D: $\frac {3}{2}$ | Since $f(x) = \begin{cases} \cos(\pi x), & \text{if } x<1 \\ f(x-1), & \text{if } x\geq 1 \end{cases}$,
then $f\left( \frac {1}{3}\right) = \cos\left( \frac {\pi}{3}\right) = \frac {1}{2}$,
and $f\left( \frac {4}{3}\right) = f\left( \frac {1}{3}\right) = \cos\left( \frac {\pi}{3}\right) = \frac {1}{2}$,
Therefore... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,120 |
Given that the distance between two parallel lines $l_1: 3x + 4y + 5 = 0$ and $l_2: 6x + by + c = 0$ is 3, find the value of $b+c$. | First, let's rewrite the equation of line $l_1$ ($3x + 4y + 5 = 0$) by multiplying through by 2, to obtain an equivalent form that is more comparable to the second line $l_2$:
$$6x + 8y + 10 = 0$$
Since the lines are parallel, their directional vectors must be proportional, which means that the coefficients of $x$ a... | 48 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,121 |
Given the sets $A = \{x | 2^{x} > \frac{1}{2}\}$ and $B = \{x | x - 1 > 0\}$, find the intersection of $A$ with the complement of $B$ with respect to the real numbers, denoted as $A \cap (\complement_R B)$. | To find the set $A$, we first solve the inequality from the definition of $A$:
\[ 2^{x} > \frac{1}{2} \]
Since $\frac{1}{2} = 2^{-1}$, we can rewrite the inequality as:
\[ 2^{x} > 2^{-1} \]
This implies that $x > -1$, hence:
\[ A = \{x | x > -1\} \]
Next, we solve the inequality given for set $B$:
\[ x - 1 > 0 \]
Solv... | A \cap (\complement_R B) = \{x | -1 < x \leq 1\} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,122 |
Given circles $O_{1}$: $x^{2}+y^{2}-2x-3=0$ and $O_{2}$: $x^{2}+y^{2}-2y-1=0$ intersect at points $A$ and $B$, then ()
A: Circles $O_{1}$ and $O_{2}$ have two common tangents
B: The equation of line $AB$ is $x-y+1=0$
C: There exist two points $P$ and $Q$ on circle $O_{2}$ such that $|PQ| \gt |AB|$
D: The maximum di... | To solve this problem, we first need to rewrite the equations of the given circles in their standard forms and then analyze each option step by step.
**Circle $O_{1}$:**
Given equation: $x^{2}+y^{2}-2x-3=0$
Completing the square for $x$, we get:
\[
x^{2}-2x+1+y^{2}=4 \implies (x-1)^{2}+y^{2}=4
\]
This shows that cir... | \text{A, B, and D} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,123 |
When randomly selecting a student ID from 6 male students and 4 female students, the probability of selecting a male student's ID is ( ).
A: $\frac{2}{5}$
B: $\frac{3}{5}$
C: $\frac{2}{3}$
D: $\frac{3}{4}$ | To solve this problem, we start by identifying the total number of student IDs, which is the sum of male and female students. We have 6 male students and 4 female students, making the total number of students:
\[6 + 4 = 10\]
The probability of selecting a male student's ID is then calculated by dividing the number of... | B | Combinatorics | MCQ | Yes | Yes | cn_k12 | false | 549,124 |
A line segment parallel to the x-axis is 5 units long, with one end point being A (2, 1). What are the coordinates of the other end point B?
A: (−3, 1) or (7, 1)
B: (2, −3) or (2, 7)
C: (−3, 1) or (5, 1)
D: (2, −3) or (2, 5) | Since the line segment is parallel to the x-axis, the y-coordinates of point A and point B must be the same. Therefore, B has the y-coordinate 1.
Now, to find the x-coordinate of B, we consider the length of the line segment. The line segment is 5 units long. Since B could be either to the left or to the right of A, w... | A: (-3, 1) \text{ or } (7, 1) | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,125 |
Which of the following arrays is a Pythagorean triple?
A: $2$, $3$, $4$
B: $0.3$, $0.4$, $0.5$
C: $6$, $8$, $10$
D: $7$, $12$, $15$ | To determine which of the given arrays is a Pythagorean triple, we will check each option by applying the Pythagorean theorem, \(a^2 + b^2 = c^2\), where \(a\) and \(b\) are the legs of a right triangle and \(c\) is the hypotenuse.
**Option A: $2$, $3$, $4$**
We calculate:
\[
2^2 + 3^2 = 4 + 9 = 13 \neq 4^2 = 16
\]
T... | C | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,126 |
Given vector $\overrightarrow{m} =(2,4)$, $|\overrightarrow{n}|= \sqrt{5}$, if the angle between $\overrightarrow{m}$ and $\overrightarrow{n}$ is $\frac{\pi}{3}$, then $|2\overrightarrow{m} -3\overrightarrow{n}|=$ ______. | **Analysis**
This question examines the scalar product of vectors. According to the problem, we have $|\overrightarrow{m}|= \sqrt{2^2+4^2}=2\sqrt{5}$, and $\overrightarrow{m} \cdot \overrightarrow{n}=|\overrightarrow{m}|\cdot|\overrightarrow{n}|\cos \frac{2\pi}{3}=2\sqrt{5}\times\sqrt{5}\times\frac{1}{2}=5$. Therefore... | \sqrt{65} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,127 |
Given the sequence $\{a_n\}$ where $a_1 = 2$ and $a_{n+1} = 4a_n - 3n + 1$ for $n \in \mathbb{N}$:
(1) Prove that the sequence $\{a_n - n\}$ is a geometric sequence;
(2) Find the explicit formula for the $n$-th term of the sequence $\{a_n\}$ and the sum of its first $n$ terms $S_n$. | (1) Proof:
Since $a_1=2$ and $a_{n+1} = 4a_n - 3n + 1$, we can calculate the following term:
$$
a_{n+1} - (n+1) = 4a_n - 3n + 1 - n - 1 = 4(a_n - n).
$$
This implies:
$$
\frac{a_{n+1} - (n+1)}{a_n - n} = 4.
$$
Therefore, the sequence $\{a_n - n\}$ is a geometric sequence with a common ratio of 4.
\[\boxed{\text{The se... | S_n = \frac{n(n+1)}{2} + \frac{4^n - 1}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,128 |
The function $f(x) = x^3 + x + 1$ ($x \in \mathbb{R}$), if $f(a) = 2$, then $f(-a) = \_\_\_\_\_\_$. | Given the function $f(x) = x^3 + x + 1$ ($x \in \mathbb{R}$), if $f(a) = 2$, then $f(-a) = -a^3 - a + 1 = -(a^3 + a + 1) + 2 = -f(a) + 2$
$= -2 + 2 = 0$
Therefore, the answer is $\boxed{0}$.
By utilizing the odd-even properties of the function, we can transform and solve for the function value.
This problem tes... | 0 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,129 |
$|a|=3$, $|b|=\frac{1}{2}$, and $a \gt b$, then $\frac{a}{b}=\_\_\_\_\_\_$. | Given that $|a|=3$ and $|b|=\frac{1}{2}$, we can deduce the possible values of $a$ and $b$ as follows:
1. From $|a|=3$, we have $a=\pm 3$.
2. From $|b|=\frac{1}{2}$, we have $b=\pm\frac{1}{2}$.
Given the condition $a > b$, we need to consider the possible combinations of $a$ and $b$ that satisfy this inequality:
1. ... | 6 \text{ or } -6 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,130 |
Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ that satisfy: $|\overrightarrow{a}| = \sqrt{3}$, $|\overrightarrow{b}| = 1$, and $\overrightarrow{a} \cdot \overrightarrow{b} = -\frac{3}{2}$, find the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ (denoted as $\theta$).
A: $30^{\circ}$
... | 1. We are given the magnitudes of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ as $|\overrightarrow{a}| = \sqrt{3}$ and $|\overrightarrow{b}| = 1$, respectively.
2. The dot product of two vectors is defined as: $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}|\cos\theta$.
... | 150^{\circ} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,131 |
Let the sum of the first $n$ terms of an arithmetic sequence $\{a_n\}$ be $S_n$. Given that $S_3 = 12$, and $2a_1$, $a_2$, $a_3+1$ form a geometric sequence, find $S_n$. | Let the common difference of the arithmetic sequence $\{a_n\}$ be $d$. According to the problem, we have
$$\begin{cases} a_{2}^{2}=2a_{1}(a_{3}+1) \\ 3a_{1}+ \frac {3×2}{2}d=12 \end{cases}$$
Solving these equations, we get
$$\begin{cases} a_{1}=1 \\ d=3 \end{cases}$$
or
$$\begin{cases} a_{1}=8 \\ d=-4 \end{cases}$$
The... | \frac {1}{2}n(3n-1) \text{ or } 2n(5-n) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,132 |
Calculate: $3^{-2}=\_\_\_\_\_\_;\left(\pi -2\right)^{0}=\_\_\_\_\_\_$. | To solve the given problems, we follow the rules of exponents closely.
For the first part, $3^{-2}$, we apply the rule that $a^{-n} = \frac{1}{a^n}$ for any nonzero number $a$ and any integer $n$. Thus, we have:
\[
3^{-2} = \frac{1}{3^2} = \frac{1}{9}.
\]
So, the first part of the solution is $\boxed{\frac{1}{9}}$.... | 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,133 |
Given a moving circle C that passes through a fixed point F(2,0) and is tangent to the line x=-2, the locus of the center C of the circle is E.
(1) Find the equation of the locus E.
(2) If a line l intersects E at points P and Q, and the midpoint of the line segment PQ has coordinates (1,1), find the length |PQ|. | (1) From the given information, the distance from point C to point F is equal to its distance to the line x=-2. Thus, the locus of point C is a parabola with focus F and the directrix x=-2. Hence, the equation of the locus E is $y^2 = 8x$.
(2) From the problem description, it is clear that the slope of line l exists. ... | \frac{\sqrt{119}}{2} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,134 |
To investigate the age demographics of the 1000 athletes participating in a sports event, the ages of 100 athletes were sampled. Which of the following statements is correct regarding this problem?
A: The 1000 athletes constitute the population.
B: Each athlete represents an individual.
C: The 100 athletes sampled repr... | Let's break down the terms used in statistics to analyze the problem:
- **Population** refers to the entire group that you want to draw conclusions about.
- An **Individual** is a single element from the population.
- A **Sample** is a subset of the population that is taken to represent the whole group in your analysi... | A, B, C, D | Other | MCQ | Yes | Yes | cn_k12 | false | 549,135 |
Let the function $f(x)$ be differentiable. Then, $$\lim_{\Delta x \to 0} \frac{f(1+\Delta x)-f(1)}{3\Delta x} =$$ （ ）
A: $f'(1)$
B: $\frac{1}{3}f'(1)$
C: Does not exist
D: None of the above | Solution: $$\lim_{\Delta x \to 0} \frac{f(1+\Delta x)-f(1)}{3\Delta x} = \frac{1}{3} \lim_{\Delta x \to 0} \frac{f(1+\Delta x)-f(1)}{\Delta x} = \frac{1}{3}f'(1),$$
Therefore, the answer is: $\boxed{\text{B}}$.
This can be obtained directly from the definition of the derivative.
This question tests the operation ... | \text{B} | Calculus | MCQ | Yes | Yes | cn_k12 | false | 549,136 |
Determine the equation of the circle that passes through three points A(1, -1), B(1, 4), C(4, 2), and find the center and radius of the circle. | Let's denote the equation of the circle as $x^2 + y^2 + Dx + Ey + F = 0$. Since points A(1, -1), B(1, 4), and C(4, 2) lie on the circle, their coordinates must satisfy the equation. By substituting their coordinates into the equation, we obtain the following system of equations:
$$
\begin{cases}
1 + 1 + D - E + F = 0 \... | \left(x - \frac{3}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{13}{4} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,137 |
Given that the real and imaginary parts of the complex number $\frac{1+ai}{2+i}$ are equal, find the value of the real number $a$.
A: $-1$
B: $-\frac{1}{3}$
C: $\frac{1}{3}$
D: $3$ | First, we simplify the complex number using the division rule for complex numbers:
$$
\frac{1+ai}{2+i} = \frac{(1+ai)(2-i)}{5} = \frac{2+a+(2a-1)i}{5}
$$
Given that the real and imaginary parts of this complex number are equal, we have the equation:
$$
2+a = 2a - 1
$$
Solving for $a$, we find:
$$
a = 3
$$
Therefore, th... | a = 3 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,138 |
If the function $f(x)=2^{|x-1|}$ is monotonically increasing on the interval $[m,+\infty)$, find the minimum value of the real number $m$. | The function $f(x)=2^{|x-1|}$ can be rewritten as a piecewise function:
$$
f(x) =
\begin{cases}
2^{1-x}, & x < 1 \\
2^{x-1}, & x \geq 1
\end{cases}
$$
The function $f(x)$ is monotonically increasing on the interval $[1,+\infty)$.
Given that $f(x)$ is monotonically increasing on the interval $[m,+\infty)$, we have $[... | 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,139 |
Which of the following calculations is correct?
A: $3a+5b=8ab$
B: $a^{6}\div a^{2}=a^{3}$
C: $(-a^{3})a^{3}=a^{6}$
D: $(-2x^{2})^{3}=-8x^{6}$ | To evaluate each option step-by-step:
**Option A: $3a+5b=8ab$**
- This equation attempts to add $3a$ and $5b$, which are unlike terms, and equates them to $8ab$. Since unlike terms cannot be combined in this manner, this option is incorrect.
**Option B: $a^{6}\div a^{2}=a^{3}$**
- Using the rule of exponents for di... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,141 |
If $2^{n+1} \cdot 2^{3}=2^{10}$ ($n$ is a positive integer), then $n=$ \_\_\_\_\_\_\_\_. | **Step 1:**
Recall the rule for multiplying exponents with the same base: $a^m \cdot a^n = a^{m+n}$.
**Step 2:**
Apply this rule to the given equation:
$2^{n+1} \cdot 2^{3}=2^{(n+1)+3}=2^{n+4}$.
**Step 3:**
Set the exponents equal to each other, since the bases are equal:
$n+4=10$.
**Step 4:**
Solve for $n$ by subtr... | n=6 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,142 |
Given vectors $\overrightarrow{m}=(t+1,1)$, $\overrightarrow{n}=(t+2,2)$, if $(\overrightarrow{m}+\overrightarrow{n})\perp(\overrightarrow{m}-\overrightarrow{n})$, find the value of $t$. | This problem can be solved using the necessary and sufficient condition for two vectors to be perpendicular.
Two vectors are perpendicular if and only if their dot product is zero.
We have $\overrightarrow{m}+\overrightarrow{n}=(2t+3,3)$ and $\overrightarrow{m}-\overrightarrow{n}=(-1,-1)$.
The dot product of $(\over... | -3 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,143 |
Given that the equation $3^x - m = 0$ has a real solution, the range of the real number $m$ is ( )
A: $m > 0$
B: $m \leq 1$
C: $0 < m \leq 1$
D: $0 \leq m < 1$ | Since the equation $3^x - m = 0$ has a real solution, it implies that $m = 3^x$ holds. Given that $3^x > 0$, we have $m > 0$.
Therefore, the correct choice is $\boxed{A}$. | A | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,144 |
The medians, altitudes, and angle bisectors corresponding to each side of an equilateral triangle coincide with each other. | **Solution**: In an equilateral triangle, the medians, altitudes, and angle bisectors corresponding to each side coincide with each other.
Therefore, the answer is: medians, altitudes, angle bisectors.
$$ \boxed{\text{medians, altitudes, angle bisectors}} $$ | \text{medians, altitudes, angle bisectors} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,145 |
Given $a > 0$, $b > 0$, and $b = \frac{1-a}{3}$, if $y = 3^a + 27^b$, then the minimum value of $y$ is \_\_\_\_\_\_. | Since $a > 0$, $b > 0$, and $b = \frac{1-a}{3}$,
this implies $a + 3b = 1$,
and $y = 3^a + 27^b \geq 2 \sqrt{3^a \cdot 27^b} = 2 \sqrt{3^{a+3b}} = 2 \sqrt{3}$.
The minimum value, $2 \sqrt{3}$, is achieved if and only if $a = \frac{1}{2}$ and $b = \frac{1}{6}$.
Therefore, the answer is $2 \sqrt{3}$.
From the g... | 2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,146 |
Given the proposition p: For any real number $x$, the inequality $x^2 + ax + a > 0$ always holds true;
and the proposition q: The quadratic equation $x^2 - x + a = 0$ has real roots;
if "p or q" is a true proposition, and "p and q" is a false proposition, find the range of the real number $a$. | For the proposition p to hold true for any real number $x$, the inequality $x^2 + ax + a > 0$ implies that the discriminant $\Delta$ of the corresponding quadratic equation must be less than zero. Hence, we have the condition $\Delta \frac{1}{4},
\end{cases}
\]
which simplifies to $\frac{1}{4} < a < 4$.
If p is false... | (-\infty, 0] \cup \left(\frac{1}{4}, 4\right) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,147 |
Let \\(f(x)\\) be an odd function defined on \\(\mathbb{R}\\). If for \\(x > 0\\), \\(f(x) = 3^{x+1}\\), then \\(f(\log_{3}\frac{1}{2})=\\) \_\_\_\_\_\_\_\_. | **Analysis**
This question examines the odd-even properties of functions. By utilizing the oddness of the function, we get \\(f(\log_{3}\frac{1}{2})=f(-\log_{3}2)=-f(\log_{3}2)\\), and then we can find the result by combining it with the given expression.
**Solution**
Since \\(f(x)\\) is an odd function defined on \... | -6 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,148 |
The solution to the system of equations \\( \begin{cases} x+y=-1 \\\\ x+z=0 \\\\ y+z=1\\end{cases}\\) is \_\_\_\_\_\_. | To solve the system of equations, we have:
\\( \begin{cases} x+y=-1 &① \\\\ x+z=0 &② \\\\ y+z=1 &③\\end{cases}\\)
Adding ①, ②, and ③, we get:
\\(2x+2y+2z=0\\)
Dividing both sides by 2, we have:
\\(x+y+z=0 &④\\)
Now, subtracting ① from ④, we find:
\\(z=1\\)
Subtracting ② from ④, we find:
\\(y=0\\)
Subtracting ③ fro... | \begin{cases} x=-1 \\\\ y=0 \\\\ z=1\\end{cases} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,150 |
Given that $i$ is the imaginary unit, find the modulus of the complex number $z = \frac{4+3i}{1-2i}$. | **Step 1: Understanding the Problem**
The problem requires us to use the rules of complex number operations and the formula for calculating the modulus of a complex number.
**Step 2: Calculating the Modulus**
The modulus of a complex number $z = a + bi$ is given by $|z| = \sqrt{a^2 + b^2}$.
First, we find the modulus ... | |z| = \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,151 |
Every natural number is an integer, and 4 is a natural number, therefore, 4 is an integer. The above syllogism is ( )
A: Correct
B: Incorrect form of reasoning
C: The concept of "natural number" is inconsistent
D: The concept of "integer" is inconsistent | To analyze this problem, let's break down the reasoning presented:
1. Major premise: Every natural number is an integer, which is correct.
2. Minor premise: 4 is a natural number, which is also correct.
3. Conclusion: Therefore, 4 is an integer. This conclusion naturally follows from the premises, making it correct as... | A | Logic and Puzzles | MCQ | Yes | Yes | cn_k12 | false | 549,152 |
A certain Xinhua bookstore has recently acquired a new best-selling book. On the first day, they sold half of the total stock, and on the second day, they sold a quarter more than that, which was 50 books extra. As a result, the bookstore still has 200 books of this kind left. How many of these new best-selling books d... | To solve the problem, let's denote the total number of new best-selling books acquired by the bookstore as $x$. According to the problem:
1. On the first day, half of the total stock was sold, which is $\frac{1}{2}x$.
2. On the second day, a quarter more than the first day's sale was sold, which is $\frac{1}{4}x + 50$... | 1000 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,153 |
If the equation $x^{2}-2x+m=0$ has two equal real roots with respect to $x$, then the value of $m$ is ______. | Given the equation $x^{2}-2x+m=0$, we are told it has two equal real roots. This means the discriminant $\Delta$ must be equal to $0$. The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by $\Delta = b^2 - 4ac$.
For our equation, $a=1$, $b=-2$, and $c=m$. Plugging these values into the formula for t... | 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,154 |
If the line $3x + y + a = 0$ passes through the center of the circle $x^2 + y^2 + 2x - 4y = 0$, then the value of $a$ is ( ).
A: $-1$
B: $1$
C: $3$
D: $-3$ | To solve this problem, we first have to find the center of the circle given by the equation $x^2 + y^2 + 2x - 4y = 0$. We can rewrite the equation of the circle in the standard form to make the center easily identifiable:
$$ x^2 + 2x + y^2 - 4y = 0 $$
To complete the square for $x$ and $y$ separately, we will add and... | B: 1 | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,155 |
The negation of the proposition "$\forall x \in [0, +\infty). x^3 + x \geqslant 0$" is ( )
A: $\forall x \in (-\infty, 0). x^3 + x < 0$
B: $\forall x \in (-\infty, 0). x^3 + x \geqslant 0$
C: $\exists x_0 \in (-\infty, 0). x_0^3 + x_0 < 0$
D: $\exists x_0 \in (0, +\infty). x_0^3 + x_0 \geqslant 0$ | **Analysis**
This question mainly tests the application of universal propositions. Being familiar with the negation of a universal proposition is key to solving this question. It is a common type of question in college entrance exams and is considered a basic question.
**Solution**
Given the problem, the negation of... | C | Inequalities | MCQ | Yes | Yes | cn_k12 | false | 549,156 |
Given the function $f(x) = \sin \omega x - 2 \sqrt{3} \sin^2 \dfrac {\omega x}{2} + \sqrt {3} (\omega > 0)$, where the distance between the two intersections of its graph and the $x$-axis is $\dfrac {\pi}{2}$, find the minimum value of $f(x)$ in the interval $\[0, \dfrac {\pi}{2}\]$.
A: $-2$
B: $2$
C: $- \sqrt {3}$
D:... | Using trigonometric identities, we can rewrite $f(x)$ as:
$$f(x) = \sin \omega x - 2 \sqrt{3} \sin^2 \dfrac {\omega x}{2} + \sqrt {3}$$
Since the distance between the two intersections of the graph and the $x$-axis is $\dfrac {\pi}{2}$, we know that:
$$2x + \dfrac {\pi}{3} \in \[\dfrac {\pi}{3}, \dfrac {4\pi}{3}\]$$
be... | \dfrac {\sqrt{3}}{2} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,157 |
Given a sequence $\{a_n\}$ whose sum of the first $n$ terms is $s_n$, and $s_n + a_n = 2n$ ($n \in \mathbb{N}^*$), which of the following sequences is definitely a geometric sequence?
A: $\{a_n\}$
B: $\{a_n - 1\}$
C: $\{a_n - 2\}$
D: $\{a_n + 2\}$ | Since $s_n + a_n = 2n$ ($n \in \mathbb{N}^*$),
then $s_{n+1} + a_{n+1} = 2n + 2$ ($n \in \mathbb{N}^*$),
subtracting the two equations gives $s_{n+1} - s_n + a_{n+1} - a_n = 2$,
which simplifies to $2a_{n+1} - a_n = 2$,
thus $2(a_{n+1} - 2) - (a_n - 2) = 0$,
which means $2(a_{n+1} - 2) = (a_n - 2)$,
therefo... | \text{C} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,158 |
Let matrix $A= \begin{pmatrix} 2 & 4 \\ 1 & x \end{pmatrix}$, and matrix $B= \begin{pmatrix} 2 & -2 \\ -1 & 1 \end{pmatrix}$. If $BA= \begin{pmatrix} 2 & 4 \\ -1 & -2 \end{pmatrix}$, then $x= \boxed{\text{answer}}$. | Given $A= \begin{pmatrix} 2 & 4 \\ 1 & x \end{pmatrix}$, and $B= \begin{pmatrix} 2 & -2 \\ -1 & 1 \end{pmatrix}$, and $BA= \begin{pmatrix} 2 & 4 \\ -1 & -2 \end{pmatrix}$,
Therefore, $4 \times 2 - 2x = 4$;
Solving this, we get $x=2$;
Hence, the answer is $\boxed{2}$.
This problem involves matrix operations and is c... | 2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,159 |
Given that the sequence $\{a_n\}$ is a decreasing arithmetic sequence, and the sequence $\{b_n\}$ satisfies $b_n=2^{a_n}$, $b_1b_2b_3=64$, $b_1+b_2+b_3=14$, then $a_{2017}=$ .
A: $-2013$
B: $2013$
C: $-2017$
D: $2017$ | **Analysis**
Given the conditions, let's assume the common difference of the arithmetic sequence $\{a_n\}$ is $d(d < 0)$. Since $b_n=2^{a_n}$, for $n\geqslant 2$, $\frac{b_n}{b_{n-1}} = \frac{2^{a_n}}{2^{a_{n-1}}} =2^{a_n-a_{n-1}} =2^d$ is a constant. Therefore, the sequence $\{b_n\}$ is a geometric sequence. By using... | \text{A} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,160 |
Given the function $f(x) = x^3 + 3ax^2 - 9x + 5$, if $f(x)$ has an extreme value at $x = 1$.
(1) Find the value of the real number $a$;
(2) Find the extreme value of the function $f(x)$;
(3) If for any $x \in [-4, 4]$, there is $f(x) < c^2$, find the range of values for the real number $c$. | (1) The derivative of the function is $f'(x) = 3x^2 + 6ax - 9$. Given that $f'(1) = 0$, we have $3 + 6a - 9 = 0$. Solving for $a$, we get $a = 1$.
(2) From part (1), we have $f(x) = x^3 + 3x^2 - 9x + 5$. The derivative is $f'(x) = 3x^2 + 6x - 9$. Let $f'(x) = 0$, we find the critical points $x_1 = -3$ and $x_2 = 1$.
... | 0 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,161 |
Define $\dfrac {n}{p_{1}+p_{2}+\cdots +p_{n}}$ as the "average reciprocal" of $n$ positive numbers $p_{1}$, $p_{2}…p_{n}$. If it is known that the "average reciprocal" of the first $n$ terms of the sequence $\{a_{n}\}$ is $\dfrac {1}{2n+1}$, and $b_{n}= \dfrac {a_{n}+1}{4}$, then $\dfrac {1}{b_{1}b_{2}}+ \dfrac {1}{b_{... | From the problem statement and the definition of "average reciprocal", we have $\dfrac {n}{a_{1}+a_{2}+\cdots +a_{n}}= \dfrac {1}{2n+1}$.
Let the sum of the first $n$ terms of the sequence $\{a_{n}\}$ be $S_{n}$, then $S_{n}=2n^{2}+n$.
When $n=1$, $a_{1}=S_{1}=3$.
For $n\geqslant 2$, $a_{n}=S_{n}-S_{n-1}$
$=(2n^{2}... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,162 |
Given that $\sin(\frac{3\pi}{2} - \alpha) = \frac{3}{5}$, and $\alpha \in (\pi, \frac{3\pi}{2})$, find the value of $\sin 2\alpha$.
A: $-\frac{24}{25}$
B: $\frac{12}{25}$
C: $\frac{24}{25}$
D: $-\frac{12}{25}$ | [Analysis]
This problem involves the application of cofunction identities, relationships among same-angle trigonometric functions, and double-angle formulas. Simplifying the given equation, we get $\cos \alpha = -\frac{3}{5}$, which leads to $\sin \alpha = -\sqrt{1 - \cos^2\alpha} = -\frac{4}{5}$. Then, by using the do... | \sin 2\alpha = \frac{24}{25} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,163 |
Given the function $y=\frac{3\sin x+1}{\sin x+2}$, determine the range of the function. | The function can be rewritten as $y=\frac{3(\sin x+2)-5}{\sin x+2}=3-\frac{5}{\sin x+2}$.
Since $-1 \leq \sin x \leq 1$, it follows that $1 \leq \sin x + 2 \leq 3$, which implies that $\frac{1}{3} \leq \frac{1}{\sin x + 2} \leq 1$.
Therefore, $-2 \leq y \leq \frac{4}{3}$.
Hence, the range of the function is $\boxed{[-2... | [-2, \frac{4}{3}] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,164 |
Given that the central angle of a sector of a circle is $\alpha$ and the radius of the circle is $r$.
(1) If $\alpha = 120^{\circ}$ and $r = 6$, find the length of the arc of the sector.
(2) If the perimeter of the sector is 24, for what value of $\alpha$ (in radians) is the area of the sector ($S$) maximized, and wh... | (1) Given that $\alpha = 120^{\circ} = 120 \times \frac{\pi}{180} = \frac{2\pi}{3}$ and $r = 6$,
the length of the arc is given by $l = \alpha r = \frac{2\pi}{3} \times 6 = \boxed{4\pi}$.
(2) Let the radius of the sector be $r$ and the length of the arc be $l$.
Then, the perimeter of the sector is given by $l + 2r =... | 2 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,165 |
Given that $a$ and $b$ are positive real numbers, and $a + b + \frac{1}{a} + \frac{1}{b} = 5$, determine the range of values for $a + b$.
A: $[1, 4]$
B: $[2, +\infty)$
C: $(2, 4)$
D: $(4, +\infty)$ | Since $a$ and $b$ are positive real numbers, we can apply the AM-GM inequality, which states that the arithmetic mean of two numbers is greater than or equal to their geometric mean. In this case, we have $\frac{a + b}{2} \geq \sqrt{ab}$, which simplifies to $\frac{1}{ab} \geq \frac{4}{(a + b)^2}$.
Given that $a + b +... | [1, 4] | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,166 |
Find the minimum value of the function $f(x) = \sin^2 x + \sqrt{3} \sin x \cos x$ in the interval $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$.
(A) $1$
(B) $\frac{1+\sqrt{3}}{2}$
(C) $\frac{3}{2}$
(D) $1+\sqrt{3}$ | First, let's simplify the expression for $f(x)$. We can use the trigonometric identity for the double angle sine function:
$$
\sin(2x) = 2\sin x \cos x
$$
Therefore, we can write the function $f(x)$ as:
$$
f(x) = \sin^2 x + \frac{\sqrt{3}}{2}\sin(2x)
$$
Next, we will look for critical points of the function in the ... | 1 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,167 |
The opposite of $-\frac{3}{7}$ is
A: $-\frac{7}{3}$
B: $\frac{3}{7}$
C: $\frac{7}{3}$
D: $-\frac{3}{7}$ | The opposite of a number is defined as the number that, when added to the original number, results in zero. Therefore, to find the opposite of $-\frac{3}{7}$, we look for a number that, when added to $-\frac{3}{7}$, equals $0$.
Starting with the given number $-\frac{3}{7}$, the opposite is found by changing the sign ... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,168 |
In the 2016 art exam of a certain high school, there were 6 contestants, including 3 females and 3 males. Now, these six contestants are to perform their talents in sequence. If any two of the three males cannot perform consecutively, and the female contestant A cannot be the first to perform, then the number of possib... | First, insert the 3 male contestants into the 4 gaps formed by the 3 female contestants. Therefore, there are $A_3^3A_4^4=144$ ways. If female contestant A is the first, there are $A_2^2A_3^3=12$ ways.
Thus, if female contestant A cannot be the first, the number of possible sequences for the contestants to perform is... | C | Combinatorics | MCQ | Yes | Yes | cn_k12 | false | 549,169 |
If $\left(x-m\right)\left(x+2\right)=x^{2}+nx-8$, then the value of $m-n$ is:
A: $2$
B: $-2$
C: $-6$
D: $6$ | To solve the given problem, let's start by expanding the left-hand side of the equation:
1. Expand $\left(x-m\right)\left(x+2\right)$:
\begin{align*}
\left(x-m\right)\left(x+2\right) &= x^2 + 2x - mx - 2m \\
&= x^2 + (2-m)x - 2m
\end{align*}
2. Compare the expanded form with the given equation $x^2 + nx -... | 6 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,171 |
Given points $(-4, y_{1})$ and $(2, y_{2})$ lie on the line $y=-x+3$, determine whether $y_{1}$ ____ $y_{2}$. (Fill in $>$, $<$, or $=$) | To solve this problem, we first need to understand how the slope of a line affects the relationship between $y$ values for different $x$ values. The given line equation is $y = -x + 3$.
1. The slope ($k$) of the line is the coefficient of $x$, which is $-1$. This means for every unit increase in $x$, $y$ decreases by... | > | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,172 |
Regarding plane vectors, determine the number of correct propositions from the following:
\\(①\\) The magnitudes of unit vectors are all equal;
\\(②\\) For any two non-zero vectors \\( \overrightarrow{a} \\), \\( \overrightarrow{b} \\), the equation \\(| \overrightarrow{a} + \overrightarrow{b} | < | \overrightarrow{a} ... | \\(①\\) The magnitudes of unit vectors are all equal, which is correct;
\\(②\\) For any two non-zero vectors \\( \overrightarrow{a} \\), \\( \overrightarrow{b} \\), the equation \\(| \overrightarrow{a} + \overrightarrow{b} | < | \overrightarrow{a} | + | \overrightarrow{b} |\\) does not always hold. For example, when \\... | 2 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,173 |
Given that $x > 0$, $n \in \mathbb{N}^*$, based on the following conclusions: $x + \frac{1}{x} \geqslant 2$, $x + \frac{4}{x^2} \geqslant 3$, $x + \frac{27}{x^3} \geqslant 4$, ..., a correct conclusion could be ( )
A: $x + \frac{n^2}{x^n} \geqslant n + 1$
B: $x + \frac{2^n}{x^n} \geqslant n$
C: $x + \frac{n^n}{x^n} ... | Analysis:
This problem primarily tests inductive reasoning skills, with finding the pattern being key. It is a basic-level question.
Answer:
Given that $x + \frac{1}{x} \geqslant 2$, $x + \frac{4}{x^2} \geqslant 3$, $x + \frac{27}{x^3} \geqslant 4$, ..., we observe that:
1. The right side of each inequality is one m... | x + \frac{n^n}{x^n} \geqslant n + 1 | Inequalities | MCQ | Yes | Yes | cn_k12 | false | 549,174 |
Given the universal set $U=\{x\in\mathbb{N}|0<x<8\}$, and $A=\{2,4,5\}$, then the complement of $A$ with respect to $U$, denoted as $\complement_U A$, is ( )
A: $\{1,3,6,7\}$
B: $\{2,4,6\}$
C: $\{1,3,7,8\}$
D: $\{1,3,6,8\}$ | Since the universal set $U=\{x\in\mathbb{N}|0<x<8\}$, and $A=\{2,4,5\}$, then $\complement_U A=\{1,3,6,7\}$.
Therefore, the correct choice is $\boxed{A}$.
**Analysis:** The complement $\complement_U A$ can be determined based on the given information and the operation of the complement. | A | Number Theory | MCQ | Yes | Yes | cn_k12 | false | 549,175 |
Given the function $f(x)=\sin x-2\sqrt{3}\sin^{2}\frac{\pi }{2}$,
(I) Find the smallest positive period of $f(x)$;
(II) Find the minimum value of $f(x)$ in the interval $\left[ 0,\frac{2\pi }{3} \right]$. | (I) Since $f(x)=\sin x+\sqrt{3}\cos x-\sqrt{3}=2\sin (x+\frac{\pi }{3})-\sqrt{3}$,
the smallest positive period of $f(x)$ is $2\pi$;
(II) Since $0\leq x\leq \frac{2\pi }{3}$,
we have $\frac{\pi }{3}\leq x+\frac{\pi }{3}\leq \pi$,
and $f(x)$ attains its minimum value when $x+\frac{\pi }{3}=\pi$, i.e., $x=\frac{2\pi ... | f(\frac{2\pi }{3})=-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,176 |
Given that the point (2, 5) is on the graph of the function f(x) = 1 + a^x (a > 0 and a ≠ 1), find the inverse function of f(x), denoted as f^(-1)(x). | Since the point (2, 5) is on the graph of the function f(x) = 1 + a^x (a > 0 and a ≠ 1),
We have 2 = 1 + a^2. As a > 0, we get a = 2.
Then, y = 1 + 2^x.
Thus, 2^x = y - 1, which gives us x = log_2(y - 1).
So, the inverse function of f(x) is f^(-1)(x) = log_2(x - 1) (x > 1).
Therefore, the answer is $\boxed{f^{-1}(... | f^{-1}(x) = \log_2(x - 1) \text{ (x > 1)} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,177 |
Given $\tan\left(\frac{\pi}{4} + \alpha\right) = 1$, then $\frac{2\sin\alpha + \cos\alpha}{3\cos\alpha - \sin\alpha} =$ \_\_\_\_\_\_. | **Analysis**
This question examines the trigonometric functions of the sum and difference of two angles and the trigonometric identities for the same angle.
From $\tan\left(\frac{\pi}{4} + \alpha\right) = 1$, we get $\frac{1 + \tan\alpha}{1 - \tan\alpha} = 1$, which leads to $\tan\alpha = 0$. Then, $\frac{2\sin\alpha... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,178 |
Points $A(-1, y_{1})$ and $B(1, y_{2})$ lie on the parabola $y=-x^{2}+4x+c$. Then $y_{1}$ ______ $y_{2}$ (fill in "$>$", "$<$", or "$=$"). | Given the parabola equation $y = -x^2 + 4x + c$, we want to compare $y_1$ and $y_2$ for points $A(-1, y_1)$ and $B(1, y_2)$ respectively.
1. The parabola opens downwards because the coefficient of $x^2$ is negative. This is evident from the equation $y = -x^2 + 4x + c$.
2. The axis of symmetry of a parabola given by ... | < | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,179 |
The Shanghai World Expo was held from May 1 to October 31, 2010. That year was a ______ year, with a total of ______ days. If May 1 was a Saturday, then May 31 would be a ______. The total number of days from May 1 to October 31 is ______. | **Analysis:** Divide 2010 by 4 to see if there is a remainder. If there is a remainder, 2010 is a common year; if there is no remainder, it is a leap year. A common year has 365 days, and a leap year has 366 days. Then, based on May 1 being a Saturday, calculate which day of the week May 31 will be. From this, we find ... | 184 | Other | math-word-problem | Yes | Yes | cn_k12 | false | 549,180 |
Given the function $f(x)=\sin(\omega x+\frac{\pi}{4})$ ($\omega>0$) has exactly $3$ axes of symmetry on the interval $\left[0,\pi \right]$. Which of the following statements is correct?
A: The range of $\omega$ is $[{\frac{9}{4},\frac{13}{4}})$
B: $f(x)$ has exactly $3$ different zeros on the interval $(0,\pi)$
C: T... | Given the function $f(x)=\sin(\omega x+\frac{\pi}{4})$ where $\omega>0$, we are to determine which statements among A, B, C, and D are correct, given that $f(x)$ has exactly $3$ axes of symmetry on the interval $\left[0,\pi \right]$.
**Step 1: Analyzing Statement A**
First, we consider the interval for $x$ given as $... | ACD | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,181 |
Given sets $A=\{1,2,3,4\}$ and $B=\{5,6,7,8\}$, if one number is chosen from $A$ and another from $B$, the probability that their product is an even number is ______. | Solution: Given sets $A=\{1,2,3,4\}$ and $B=\{5,6,7,8\}$, if one number is chosen from $A$ and another from $B$,
the total number of basic events is $n=4\times4=16$,
the number of basic events for which the product is an even number is $m= C_{2}^{1} C_{2}^{1} + C_{2}^{1} C_{4}^{1} =12$,
$\therefore$ the probabili... | \dfrac {3}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false | 549,182 |
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