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If the function $h(x)=2x-\frac{k}{x}+\frac{k}{3}$ is increasing on $(1,+\infty)$ and decreasing on $(-1,1)$, then the real number $k=\boxed{-2}$. | **Analysis**
This question mainly examines the relationship between the monotonicity of a function and the sign of its derivative, which is a basic problem. First, we differentiate the function $h(x)$, ensuring that the derivative is greater than or equal to $0$ on $(1,+\infty)$ always holds, and the derivative $\leqs... | -2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,866 |
Given sets $A=\{x|x^2-6x+8<0\}$, $B=\{x|(x-a)(x-3a)<0\}$.
(1) If $A\subseteq (A\cap B)$, find the range of $a$;
(2) If $A\cap B=\emptyset$, find the range of $a$. | From the inequality in set $A$, $x^2-6x+80$, $B=\{x|a0$, $B=\{x|a<x<3a\}$, from $A\cap B=\emptyset$, we get $a\geq 4$ or $3a\leq 2$, solving this gives: $0<a\leq \frac{2}{3}$ or $a\geq 4$;
When $a<0$, $B=\{x|3a<x<a\}$, from $A\cap B=\emptyset$, we get $3a\geq 4$ or $a\leq 2$, solving this gives: $a<0$;
When $a=0$, ... | a\leq \frac{2}{3} \text{ or } a\geq 4 | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false | 548,867 |
Given the function $f(x) = \begin{cases} -2x - 3, & x < 2 \\ 2^{-x}, & x \geq 2 \end{cases}$, find the value of $f[f(-3)]$. | First, we find the value of $f(-3)$ using the given function. Since $-3 < 2$, we use the first part of the function definition. Thus,
$$f(-3) = -2 \cdot (-3) - 3 = 6 - 3 = 3.$$
Next, we need to find the value of $f[f(-3)] = f(3)$. Since $3 \geq 2$, we use the second part of the function definition. So,
$$f(3) = 2^{-... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,868 |
Given $S_{k}= \frac{1}{k+1}+ \frac{1}{k+2}+ \frac{1}{k+3}+...+ \frac{1}{2k} (k=1,2,3,…)$, determine the value of $S_{k+1}$
A: $(S_{k}+ \frac{1}{2(k+1)} )$
B: $(S_{k}+ \frac{1}{2k+1}- \frac{1}{2k+2} )$
C: $(S_{k}+ \frac{1}{2k+2}- \frac{1}{k+1} )$
D: $(S_{k}+ \frac{1}{2k+1}+ \frac{1}{2k+2} )$ | **Analysis**
This problem primarily tests the application of mathematical induction. Calculate $a_{k}$ and $a_{k+1}$ respectively, and then observe the relationship between the two.
**Step-by-Step Solution**
1. First, let's write out $S_{k+1}$ based on the given pattern:
$$S_{k+1} = \frac{1}{(k+1)+1} + \frac{1}{(k+1... | (S_{k}+ \frac{1}{2k+1}- \frac{1}{2k+2} ) | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,869 |
The processing speed of a 64-bit quantum computer is about 150,000,000,000 times that of the fastest supercomputer "Sunway TaihuLight" in the world. The number 150,000,000,000 in scientific notation is represented as ( ).
A: 0.15×10^12
B: 1.5×10^11
C: 15×10^10
D: 1.5×10^10 | To convert the number $150,000,000,000$ into scientific notation, we follow the steps below:
1. Identify the significant figures in the number, which are $150$.
2. Count the number of places we move the decimal point to the left to get from $150,000,000,000$ to $1.5$. This count is $11$.
3. Write the number in the for... | \text{B: } 1.5 \times 10^{11} | Logic and Puzzles | MCQ | Yes | Yes | cn_k12 | false | 548,870 |
Given sets $A = \{x \mid y = \sqrt{x^2-5x-14}\}$, $B = \{x \mid y = \lg(-x^2-7x-12)\}$, and $C = \{x \mid m+1 \leq x \leq 2m-1\}$:
1. Find the complement of $A \cup B$ with respect to $\mathbb{R}$;
2. If $A \cup C = A$, determine the range of the real number $m$. | 1. Because the function defined by $A$ involves a square root, the expression under the square root must be non-negative. Thus, we need $x^2 - 5x - 14 \geq 0$. Factoring the quadratic gives $(x-7)(x+2) \geq 0$. This inequality is satisfied when $x \geq 7$ or $x \leq -2$. Therefore, $A = (-\infty, -2] \cup [7, +\infty)$... | m < 2 \text{ or } m \geq 6 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,871 |
Given the sets $M = \{0, 1, 3\}$ and $N = \{x | x \in \{0, 3, 9\}\}$, then $M \cup N =$ ( )
A: $\{0\}$
B: $\{0, 3\}$
C: $\{1, 3, 9\}$
D: $\{0, 1, 3, 9\}$ | The correct answer is $\boxed{\text{D}}$.
(Solution omitted) | \text{D} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,873 |
Given a random variable $\xi$ follows the normal distribution $N(0, \sigma^2)$. If $P(\xi > 2) = 0.023$,
then $P(-2 \leq \xi \leq 2) =$ （ ）
A: 0.477
B: 0.628
C: 0.954
D: 0.977 | Since the random variable $\xi$ follows the normal distribution $N(0, \sigma^2)$,
the normal curve is symmetric about $x=0$.
Given $P(\xi > 2) = 0.023$,
it follows that $P(\xi < -2) = 0.023$.
Therefore, $P(-2 \leq \xi \leq 2) = 1 - 0.023 - 0.023 = 0.954$.
Hence, the answer is $\boxed{0.954}$. | 0.954 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,874 |
If the sequence $\{a_{n}\}$ satisfies: $\exists A$,$B\in R$,$AB\neq 0$, such that for $\forall n\in N^{*}$, $a_{n+2}=Aa_{n+1}+Ba_{n}$, then the sequence $\{a_{n}\}$ is said to have "three-term correlation". Which of the following statements are correct?
A: If the sequence $\{a_{n}\}$ is an arithmetic sequence, then $\... | To analyze the statements given, we approach each option step-by-step:
### Statement A: Arithmetic Sequence
If $\{a_{n}\}$ is an arithmetic sequence, then the common difference $d = a_{n+1} - a_n$ is constant for all $n$. Thus, we can express $a_{n+2}$ in terms of $a_{n+1}$ and $a_n$:
$$
a_{n+2} - a_{n+1} = d = a_{n+... | ABD | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,875 |
Given $$a = \int_{0}^{\frac{\pi}{2}} \left( \cos^2 \frac{x}{2} - \frac{1}{2} \right) dx$$, find the coefficient of the $x^2$ term in the expansion of $$(ax + \frac{1}{2ax})^{10}$$. | Firstly, we evaluate the integral to find the value of $a$.
\begin{align*}
a &= \int_{0}^{\frac{\pi}{2}} \left( \cos^2 \frac{x}{2} - \frac{1}{2} \right) dx \\
&= \int_{0}^{\frac{\pi}{2}} \left( \frac{1 + \cos x}{2} - \frac{1}{2} \right) dx \quad \text{(using the double angle formula $\cos^2 \theta = \frac{1 + \cos 2\th... | \frac{105}{32} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 548,876 |
Given that line $a$ is parallel to plane $\alpha$, $a$ is parallel to plane $\beta$, and $\alpha \cap \beta = b$, then $a$ and $b$ are ( )
A: Intersecting
B: Skew
C: Parallel
D: Coplanar or Skew | Line $a$ is parallel to line $b$, for the following reasons:
Since line $a$ is parallel to plane $\alpha$, and line $a$ is parallel to plane $\beta$,
there must be a line in planes $\alpha$ and $\beta$ that is parallel to $a$, let's denote them as $m$ and $n$ respectively,
which means $a$ is parallel to $m$, and $a$ is... | \text{C} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,877 |
1. If the proposition "$\exists x \in \mathbb{R}, x^2 + 2x + m \leqslant 0$" is a false statement, then the range of real number $m$ is ______.
2. If the line $ax + 4y - l = 0$ is perpendicular to $2x - 5y + 6 = 0$, then the value of $a$ is ______.
3. If the tangent line at point $P_0$ on the curve $y = x^3 + x - 2... | 1. **Analysis:**
This question examines the determination of the truth of a particular proposition and the solution of inequalities that are always true, which is a basic problem.
From the given, we have $x^2 + 2x + m > 0$ always holds,
thus $\Delta = 4 - 4m 1$.
So, the answer is $\boxed{(1, +\infty)}$.
2. ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,878 |
Given the function $f(x)=kx+1$, where the real number $k$ is randomly selected from the interval $[-2,1]$, the probability that $f(x) \geqslant 0$ for $\forall x \in [0,1]$ is __________. | **Analysis**
This problem examines the geometric probability model related to length and the properties of linear functions. We need to find the range of $k$ for which $f(x) \geqslant 0$ for $\forall x \in [0,1]$, and then solve it using the formula for geometric probability.
**Solution**
Since the function is $f(x)... | \dfrac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,879 |
Given that one focus of a hyperbola is $F$, one endpoint of the conjugate axis is $B$, and the distance from focus $F$ to an asymptote is $d$. If $|FB| \geq \sqrt{3}d$, then the range of the eccentricity of the hyperbola is ( ).
A: $(1, \sqrt{2}]$
B: $[\sqrt{2}, +\infty)$
C: $(1, 3]$
D: $[\sqrt{3}, +\infty)$ | [Analysis]
This problem tests the range of the eccentricity of a hyperbola, the formula for the distance from a point to a line, and the student's computational ability. It is relatively basic.
Let $F(c,0)$ and $B(0,b)$. The equation of an asymptote is $bx+ay=0$. Thus, $d=\frac{bc}{\sqrt{b^2+a^2}}=b$ and $|FB|=\sqrt{b... | \text{A: } (1, \sqrt{2}] | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,880 |
Solve the equations:$(1)3x-5=10$;$(2)2x+4\left(2x-3\right)=6-2\left(x+1\right)$. | To solve the given equations step by step:
**For equation (1): $3x - 5 = 10$**
1. Add $5$ to both sides to isolate the term with $x$:
\[3x - 5 + 5 = 10 + 5\]
\[3x = 15\]
2. Divide both sides by $3$ to solve for $x$:
\[x = \frac{15}{3}\]
\[x = 5\]
So, the solution for equation (1) is $\boxed{x = 5}$.
**For equation... | x = \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,881 |
Given lines $l\_1$: $ax+(a+2)y+1=0$, $l\_2$: $ax-y+2=0$. Then "$a=-3$" is the "\_\_\_\_\_\_" condition for "$l\_1 // l\_2$". | When $a=-2$, the two lines become $(-2x+1=0)$, $(-2x-y+2=0)$, respectively. At this time, the two lines are not parallel, so this case is discarded.
When $a \neq -2$, the two lines can be rewritten as:
$y=- \frac {a}{a+2}x- \frac {1}{a+2}$, $y=ax+2$, respectively.
Since $l\_1 // l\_2$,
$\therefore- \frac {a}{a+2}=a... | \text{Sufficient but not necessary} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,882 |
The school starts classes at 8:10 in the morning, and each class lasts for 40 minutes. The class should end at $\boxed{\text{answer}}$ in the morning. | Solution: 8:10 + 40 minutes = 8:50,
Answer: A class lasts for 40 minutes. The first class in the morning starts at 8:10 and ends at 8:50.
Hence, the answer is $\boxed{8:50}$.
Given the starting time of 8:10 and the duration of 40 minutes, the ending time can be found by adding the duration to the starting time. This... | 8:50 | Other | math-word-problem | Yes | Yes | cn_k12 | false | 548,883 |
Given a sequence $\{a_n\}$ with the sum of the first $n$ terms $S_n= \frac {1}{2}n^{2}+ \frac {1}{2}n$,
(1) find the expression for the general term $a_n$;
(2) let $b_n=a_n\cdot2^{n-1}$, find the sum of the first $n$ terms of the sequence $\{b_n\}$, denoted as $T_n$. | Solution:
(1) When $n\geqslant 2$, $a_n=S_n-S_{n-1}= \frac {1}{2}n^{2}+ \frac {1}{2}n- \frac {1}{2}(n-1)^{2}- \frac {1}{2}(n-1)=n$,
When $n=1$, $a_{1}=S_{1}=1$ also fits the formula above,
$\therefore$ The expression for the general term $a_n$ is $a_n=n$,
(2) $b_n=a_n\cdot2^{n-1}=n\cdot2^{n-1}$,
$\therefore T... | T_n=(n-1)\cdot2^{n}+1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,885 |
Calculate: $\sqrt{(-2)^{2}}=\_\_\_\_\_\_$. | To calculate $\sqrt{(-2)^{2}}$, we follow these steps:
1. First, calculate the square of $-2$: $(-2)^{2} = 4$.
2. Then, take the square root of the result: $\sqrt{4} = 2$.
Therefore, $\sqrt{(-2)^{2}} = 2$. So, the answer is $\boxed{2}$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,886 |
The distance from a point $P$ on the hyperbola $4x^{2}-y^{2}+64=0$ to one of its foci equals $1$. What is the distance from point $P$ to the other focus?
A: $17$
B: $16$
C: $15$
D: $13$ | Given the hyperbola $4x^{2}-y^{2}+64=0$,
its standard equation can be rewritten as $\dfrac{y^{2}}{64}-\dfrac{x^{2}}{16}=1$,
thus, $a=8$ and $c=4\sqrt{5}$.
For a point $P$ on the hyperbola, the distance to one of its foci equals $1$.
Let the distance to the other focus be $x$,
then, according to the definition... | A | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,887 |
To implement the State Council's real estate regulation policy and ensure "housing for all," a city accelerated the construction of affordable housing. In 2011, the city government invested 200 million yuan to build 80,000 square meters of affordable housing. It is expected that by the end of 2012, a total investment o... | (1) Let the annual average growth rate of the city government's investment be $x$,
According to the problem, we have: $2+2(1+x)+2(1+x)^2=9.5$,
Simplifying, we get: $x^2+3x-1.75=0$,
Solving, we find $x_1=0.5$, $x_2=-3.5$ (discard),
Answer: The annual growth rate of the city government's investment is $\boxed{50\... | 38 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,888 |
Given that the function $f(x)$ is monotonic on $\mathbb{R}$ and satisfies $f[f(x) - 3^x] = 4$ for any $x \in \mathbb{R}$, find the value of $f(2)$.
A: $4$
B: $8$
C: $10$
D: $12$ | Since $f(x)$ is monotonic on $\mathbb{R}$ and $f[f(x) - 3^x] = 4$ for any $x \in \mathbb{R}$, we can infer that $f(x) - 3^x = k$, where $k$ is a constant.
This implies $f(x) = 3^x + k$.
Substituting $x = k$ into $f(x) = 3^x + k$, we get $f(k) = 3^k + k = 4$. Solving for $k$, we find $k = 1$.
Thus, $f(x) = 3^x + 1$.
... | 10 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,889 |
(1) In the arithmetic sequence $\{a_n\}$, if $a_4=4$ and $a_3+a_5+a_7=15$, then the sum of the first 10 terms $S_{10}=$_______.
(2) In $\triangle ABC$, if $A=60^{\circ}$, $b=1$, and the area of $\triangle ABC=\sqrt{3}$, then $a=$_______.
(3) Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence $\{a_n... | (1) **Analysis**
This problem examines the properties of an arithmetic sequence and the formula for the sum of an arithmetic sequence. First, from $a_3+a_5+a_7=15$, we find $a_5$, then find $d$ and $a_1$, and use the formula for the sum of the first $n$ terms to get the result.
**Solution**
Since $a_3+a_5+a_7=15$,
... | -\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,890 |
Given that $\{a_n\}$ is an arithmetic sequence, if $a_3 + a_4 + a_8 = 9$, then $S_9 =$ ( )
A: 24
B: 27
C: 15
D: 54 | Since $\{a_n\}$ is an arithmetic sequence, we have the property that the sum of the terms equidistant from the beginning and end of a sequence segment is constant. Therefore, $a_3 + a_8 = 2a_5$, and $a_4 = a_5$. From the given condition, $a_3 + a_4 + a_8 = 9$, we can substitute $2a_5 + a_5 = 9$, which simplifies to $3a... | \text{B: 27} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,891 |
In the rectangular coordinate system, the parametric equations of line $l$ are $$\begin{cases} x=4t-1 \\ y=3t- \frac {3}{2}\end{cases}$$ ($t$ is the parameter). Establish a polar coordinate system with the coordinate origin $O$ as the pole and the positive semi-axis of the $x$-axis as the polar axis. The polar equation... | 1. Since the parametric equations of line $l$ are $$\begin{cases} x=4t-1 \\ y=3t- \frac {3}{2}\end{cases}$$ ($t$ is the parameter), eliminating the parameter $t$, we get the general equation of line $l$ as $3x-4y-3=0$.
The polar equation of circle $C$ is ρ2\=2$$\sqrt {2}$$ρsin(θ-$$\frac {π}{4}$$), which can be rewritt... | |PQ|_{\text{min}}=\sqrt{d^2-2}=\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,892 |
Given the system of inequalities \\(\begin{cases}\begin{matrix}3x+4y-10\geqslant 0 \\\\ x\leqslant 4\\\end{matrix} \\\\ y\leqslant 3\\end{cases} \\), which represents the region $D$, a circle with the equation $x^{2}+y^{2}=1$ is drawn passing through any point $P$ in region $D$. The circle intersects at points $A$ and ... | [Analysis]
This problem primarily tests the application of linear programming. By sketching the plane region corresponding to the system of inequalities, we can determine the position of point $P$ when the angle $\alpha$ is at its minimum using the graphical approach. Then, using the double-angle formula for the cosin... | \frac{1}{2} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,893 |
Given that the sequence $\{b_n\}$ is a geometric sequence, and the first term $b_1=1$, common ratio $q=2$, then the sum of the first $10$ terms of the sequence $\{b_{2n-1}\}$ is $(\quad)$
A: $\dfrac{4}{3}(4^{9}-1)$
B: $\dfrac{4}{3}(4^{10}-1)$
C: $\dfrac{1}{3}(4^{9}-1)$
D: $\dfrac{1}{3}(4^{10}-1)$ | **Analysis**
This question tests knowledge related to geometric sequences and is considered a medium-level problem.
It is analyzed that the sum of the first $10$ terms of $\{b_{2n-1}\}$ is the sum of the first $10$ odd terms of the sequence $\{b_n\}$, and the formula for the sum of a geometric sequence can be applied... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,894 |
Given a function $f(x) = ax^3 + bx^2 + cx + d$ ($a, b, c, d \in \mathbb{R}$) defined on $\mathbb{R}$ whose graph is symmetrical with respect to the origin and attains its minimum value of $-2$ at $x = 1$.
(Ⅰ) Determine the interval where $f(x)$ is monotonically increasing;
(Ⅱ) Solve the inequality $f(x) > 5mx^2 - (4m... | (Ⅰ) Since the graph of $f(x)$ is symmetric about the origin, $f(x)$ is an odd function, implying $f(-x) = -f(x)$ for all $x \in \mathbb{R}$. Hence $f(0) = 0$, which gives us $b = 0$ and $d = 0$.
The derivative $f'(x) = 3ax^2 + c$. As $f(x)$ has a minimum value at $x = 1$, we know that $f'(1) = 0$ and $f(1) = -2$. Ther... | \{x \mid x > 4m \text{ or } 0 0 \text{ or } 4m < x < m\} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,895 |
The equation of the asymptotes of the hyperbola $$\frac {x^{2}}{16}- \frac {y^{2}}{9}=-1$$ is ( )
A: $$y=± \frac {3}{4}x$$
B: $$y=± \frac {4}{3}x$$
C: $$y=± \frac {16}{9}x$$
D: $$y=± \frac {9}{16}x$$ | Solution: The equation of the asymptotes of the hyperbola $$\frac {x^{2}}{16}- \frac {y^{2}}{9}=-1$$ is: $$y=± \frac {3}{4}x$$.
Therefore, the answer is: $\boxed{A}$.
By using the equation of the hyperbola, we can directly solve for the equation of the asymptotes.
This question tests the application of the simple... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,896 |
The solution set of the inequality $x^2 - 3x + 2 < 0$ is ($\,\,$):
A: $(-\infty, 1)$
B: $(2, +\infty)$
C: $(-\infty, 1) \cup (2, +\infty)$
D: $(1, 2)$ | The inequality $x^2 - 3x + 2 < 0$ can be factored into $(x-1)(x-2) < 0$, hence $1 < x < 2$.
Therefore, the solution set of the inequality $x^2 - 3x + 2 < 0$ is $\{x \mid 1 < x < 2\}$.
So the answer is $\boxed{\text{D}}$.
By using the method of solving quadratic inequalities, we can obtain the answer. Mastering the m... | \text{D} | Inequalities | MCQ | Yes | Yes | cn_k12 | false | 548,897 |
Given $\sin \left( \alpha+ \frac{\pi}{3} \right)= \frac{12}{13}$, then $\cos \left( \frac{\pi}{6}-\alpha \right)=$ ()
A: $\frac{5}{12}$
B: $\frac{12}{13}$
C: $- \frac{5}{13}$
D: $- \frac{12}{13}$ | **Analysis**
This question mainly tests the simple application of trigonometric function induction formulas and identity transformations, and it is a basic question. According to the problem, we get $\cos \left( \frac{\pi}{6}-\alpha \right)=\sin \left[ \frac{\pi}{2}-\left( \frac{\pi}{6}-\alpha \right) \right]=\sin \le... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,899 |
The equivalent proposition of the converse of the proposition "If p is false, then q is false" is ()
A: If q is false, then p is false
B: If q is false, then p is true
C: If p is true, then q is false
D: If p is true, then q is true | Let's first identify the converse and the contrapositive of the given proposition:
- The original proposition is: "If p is false, then q is false".
- The converse of the original proposition is: "If q is false, then p is false".
- The contrapositive of the original proposition is: "If q is true, then p is true".
By t... | \text{D: If p is true, then q is true} | Logic and Puzzles | MCQ | Yes | Yes | cn_k12 | false | 548,900 |
The vertices of a cube ABCD-A1B1C1D1 with edge length $a$ are all on the surface of a sphere $O$. $E$ and $F$ are the midpoints of edges $AA1$ and $DD1$, respectively. The length of the line segment cut off by sphere $O$ from line $EF$ is __________. | Since the vertices of the cube are on the surface of the sphere, the sphere's center $O$ coincides with the center of the cube. The distance from the center of the cube to any vertex is the radius of the sphere, which is half of the space diagonal of the cube. The space diagonal of the cube is $\sqrt{a^2+a^2+a^2} = a\s... | a\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,901 |
Calculate: $\lg 2 + \lg 5 =$ . | **Answer**: By using the logarithmic operation rules, when adding two logarithms with the same base, the base remains unchanged, and the true numbers are multiplied. Therefore, $\lg 2 + \lg 5 = \lg (2 \times 5) = \lg 10$. Since $\lg 10 = 1$, the final answer is $\boxed{1}$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,902 |
In triangle \\(\triangle ABC\\), the lengths of the sides opposite to angles \\(A\\), \\(B\\), and \\(C\\) are \\(a\\), \\(b\\), and \\(c\\) respectively. Prove that: \\({{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2(bc\cos A+ac\cos B+ab\cos C)\\). | Proof:
By the Law of Cosines, we have \\({{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A\\)
Thus, \\({{b}^{2}}+{{c}^{2}}-{{a}^{2}}=2bc\cos A \quad (1)\\)
Similarly, we get \\({{a}^{2}}+{{c}^{2}}-{{b}^{2}}=2ac\cos B \quad (2)\\)
And \\({{a}^{2}}+{{b}^{2}}-{{c}^{2}}=2ab\cos C \quad (3)\\),
Adding equations (1), (2), and (3)... | {a}^{2} | Geometry | proof | Yes | Yes | cn_k12 | false | 548,903 |
Use the Euclidean algorithm to find the greatest common divisor of 459 and 357. | Since $459 \div 357 = 1$ remainder $102$,
$357 \div 102 = 3$ remainder $51$,
$102 \div 51 = 2$,
Therefore, the greatest common divisor of 459 and 357 is 51.
Hence, the answer is: $\boxed{51}$
By dividing the larger number by the smaller number, we obtain a quotient and a remainder. Then, we divide the previou... | 51 | Number Theory | math-word-problem | Yes | Yes | cn_k12 | false | 548,904 |
Given the parabola $y^{2}=2px (p > 0)$, a line passing through point $T(p,0)$ with a slope of $1$ intersects the parabola at points $A$ and $B$. Find the product of the slopes of lines $OA$ and $OB$, where $O$ is the coordinate origin. | According to the given conditions, the line passing through point $T(p,0)$ with a slope of $1$ intersects the parabola at points $A$ and $B$. The equation of line $AB$ is $y=x-p$.
Solve the system of equations:
$$
\begin{cases}
y = x - p \\
y^{2} = 2px
\end{cases}
$$
Eliminate $x$ to obtain: $y^{2} - 2py - 2p^{2} = ... | -2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,905 |
Select four students from A, B, C, D, and E to participate in mathematics, physics, chemistry, and English competitions, respectively. Given that student A does not participate in the physics and chemistry competitions, there are ______ different competition participation schemes. | **Answer**
To solve this problem, we need to consider the restrictions on student A and then calculate the number of different participation schemes accordingly.
1. Since student A cannot participate in the physics and chemistry competitions, A can only participate in the mathematics or English competitions. This giv... | 72 | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false | 548,906 |
Given that $F_1$ and $F_2$ are the left and right foci of a hyperbola $E$, and point $P$ is on $E$, with $\angle F_1 P F_2 = \frac{\pi}{6}$ and $(\overrightarrow{F_2 F_1} + \overrightarrow{F_2 P}) \cdot \overrightarrow{F_1 P} = 0$, find the eccentricity $e$ of hyperbola $E$.
A: $\sqrt{3} - 1$
B: $\sqrt{3} + 1$
C: $\fr... | Given that $(\overrightarrow{F_2 F_1} + \overrightarrow{F_2 P}) \cdot \overrightarrow{F_1 P} = 0$, we can rewrite it as:
$(\overrightarrow{F_2 P} + \overrightarrow{F_2 F_1}) \cdot (\overrightarrow{F_2 P} - \overrightarrow{F_2 F_1}) = 0$
Expanding the dot product yields:
$\|\overrightarrow{F_2 P}\|^2 - \|\overrightar... | \frac{\sqrt{3} + 1}{2} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,907 |
Given that the center of the ellipse $C$ is at the origin, its foci are on the $x$-axis, the length of the major axis is $4$, and the point $(1, \frac{\sqrt{3}}{2})$ is on the ellipse $C$.
(1) Find the equation of the ellipse $C$;
(2) A line $l$ with a slope of $1$ passes through the right focus of the ellipse and inte... | (1) Since the foci of $C$ are on the $x$-axis and the length of the major axis is $4$, we can set the equation of the ellipse $C$ as: $\frac{x^2}{4} + \frac{y^2}{b^2} = 1 (2 > b > 0)$.
As the point $(1, \frac{\sqrt{3}}{2})$ is on the ellipse $C$, we have $\frac{1}{4} + \frac{3}{4b^2} = 1$.
Solving for $b$, we get $b^2 ... | \frac{8}{5} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,908 |
Given an ellipse $E: \frac{y^2}{a^2} + \frac{x^2}{b^2} = 1$ ($a > b > 0$) with eccentricity $\frac{\sqrt{3}}{2}$, and let $A$ and $F$ be respectively the left vertex and the upper focus point of ellipse $E$. The slope of the line $AF$ is $\sqrt{3}$. Let the line $l: y = kx + m$ intersect the y-axis at a point $P$ diffe... | (Ⅰ) From the given information we have the following system of equations
$$
\begin{cases}
\frac{c}{a} = \frac{\sqrt{3}}{2}, \\
\frac{c}{b} = \sqrt{3}, \\
a^2 = b^2 + c^2.
\end{cases}
$$
Solving this system, we obtain $a = 2$ and $b = 1$.
Thus, the equation of ellipse $E$ is $\frac{y^2}{4} + x^2 = 1$.
(Ⅱ) From the giv... | -2 < m < -1 \text{ or } 1 < m < 2 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,909 |
Given that the diameter of circle $\odot O$ is 10, and line $l$ is a tangent to circle $\odot O$, what is the distance from the center $O$ to line $l$?
A: 2.5
B: 3
C: 5
D: 10 | Given that the diameter of circle $\odot O$ is 10, we can find the radius of the circle by dividing the diameter by 2. This gives us:
\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{10}{2} = 5 \]
Since line $l$ is a tangent to circle $\odot O$, by the properties of a circle and its tangent, the distance from the... | \text{C. 5} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,911 |
Two players, A and B, are playing a table tennis match. It is known that the probability of A winning a game is $0.6$, and the probability of B winning a game is $0.4$. They agree to play a "best-of-3" match, where the first player to win $2$ games wins the match.
$(1)$ Find the probability of A winning the match.
... | ### Solution:
#### Part 1: Probability of A Winning the Match
Given:
- Probability of A winning a game = $0.6$
- Probability of B winning a game = $0.4$
The probability of A winning the match can be calculated by considering all possible ways A can win in a "best-of-3" match:
1. A wins the first two games.
2. A wins... | 0.57 | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false | 548,912 |
Given the line defined by the parametric equations $x=(2-t\sin 30^{\circ})$ and $y=(-1+t\sin 30^{\circ})$, and the circle defined by the equation $x^2+y^2=8$, the line intersects the circle at points $B$ and $C$. Point $O$ is the origin. Calculate the area of triangle $BOC$.
A) $2\sqrt{7}$
B) $\sqrt{30}$
C) $\frac{\sq... | First, let's convert the parametric equations to a Cartesian equation:
$y=1-x$
Next, substitute $y=1-x$ into the circle equation $x^2+y^2=8$:
$x^2+(1-x)^2=8$
Expanding and simplifying the equation, we obtain:
$2x^2-2x-7=0$
Now, let's denote the coordinates of points $B$ and $C$ as $(x_1, y_1)$ and $(x_2, y_2)$, r... | \frac{\sqrt{15}}{2} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,913 |
Given $f(x)=x^{2}+bx+c$ and $f(0)=f(2)$, then $(\ \ \ \ )$
A: $f(-2) < f(0) < f( \frac {3}{2})$
B: $f( \frac {3}{2}) < f(0) < f(-2)$
C: $f( \frac {3}{2}) < f(-2) < f(0)$
D: $f(0) < f( \frac {3}{2}) < f(-2)$ | Since $f(0)=f(2)$,
The axis of symmetry of $f(x)$ is $x=1$, hence $f( \frac {3}{2})=f( \frac {1}{2}).$
Since the graph of $f(x)$ opens upwards,
$f(x)$ is monotonically decreasing on $(-∞,1)$,
Given that $-2 f(0) > f( \frac {1}{2})=f( \frac {3}{2})$,
Therefore, the answer is $\boxed{\text{B}}$.
The solution is ba... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,914 |
Let $m \in \mathbb{R}$, and the complex number $z = (2m^2 + m - 1) + (-m^2 - 2m - 3)i$. If $z$ is a pure imaginary number, then $m = \_\_\_\_\_\_$. | Since the complex number $z = (2m^2 + m - 1) + (-m^2 - 2m - 3)i$ is a pure imaginary number,
it follows that:
$$
\begin{cases}
2m^2 + m - 1 = 0 \\
-m^2 - 2m - 3 \neq 0
\end{cases}
$$
Solving this, we find $m = \frac{1}{2}$.
Therefore, the answer is: $\boxed{\frac{1}{2}}$.
This is directly obtained by setting the rea... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,915 |
Consider a computer system that can only execute one task at a time, and the next task can only start after the completion of the current one. There are three tasks $U$, $V$, $W$ that need to be performed, which take $a$, $b$, $c$ seconds respectively, where $a < b < c$. The "relative waiting time" for a task is define... | The sum of the relative waiting times for the sequence A $U \rightarrow V \rightarrow W$ is:
$$S_{A} = \frac{a}{a} + \frac{a+b}{a+b} + \frac{a+b+c}{a+b+c} = 3,$$
The sum for sequence B $V \rightarrow W \rightarrow U$ is:
$$S_{B} = \frac{b}{b} + \frac{b+c}{b+c} + \frac{b+c+a}{a+b+c} = 2 + \frac{b+c+a}{a+b+c},$$
For se... | A | Logic and Puzzles | MCQ | Yes | Yes | cn_k12 | false | 548,916 |
Given that plane $\alpha$ is parallel to plane $\beta$, and points $A, C$ are on plane $\alpha$, while points $B, D$ are on plane $\beta$, the necessary and sufficient condition for line $AC$ to be parallel to line $BD$ is ( )
A: $AB \parallel CD$
B: $AD \parallel CB$
C: $AB$ intersects $CD$
D: Points $A, B, C, D$ ar... | **Analysis of the Problem:** Since plane $\alpha$ is parallel to plane $\beta$, for line $AC$ to be parallel to line $BD$, lines $AC$ and $BD$ must be coplanar, meaning that points $A, B, C, D$ must be in the same plane. Therefore, the correct answer is $\boxed{D}$.
**Key Point:** Necessary and sufficient condition. | D | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,917 |
The solution to the system of linear equations $\left\{\begin{array}{l}{2x-y=1}\\{x+y=2}\end{array}\right.$ is ______. | To solve the system of linear equations $\left\{\begin{array}{l}{2x-y=1} \text{ (1)}\\{x+y=2} \text{ (2)}\end{array}\right.$, we proceed as follows:
1. **Add Equations (1) and (2):**
By adding equation (1) and equation (2), we aim to eliminate $y$ and solve for $x$.
\begin{align*}
(2x - y) + (x + y) &= 1 + 2... | null | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,918 |
If the line $l_{1}$: $ax+(a+2)y+2=0$ is parallel to the line $l_{2}$: $x+ay-2=0$, then $a=\left(\ \ \right)$
A: $2$
B: $-1$
C: $2$ or $-1$
D: $-2$ or $1$ | To determine the value of $a$ for which the line $l_{1}$: $ax+(a+2)y+2=0$ is parallel to the line $l_{2}$: $x+ay-2=0$, we need to compare their slopes. For two lines to be parallel, their slopes must be equal.
The slope of a line in the form $Ax+By+C=0$ is $-\frac{A}{B}$. Thus, for $l_{1}$, the slope is $-\frac{a}{a+2... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,919 |
Let $z=\frac{i}{{1+i}}$, then the point corresponding to the conjugate of the complex number $z$, $\overline{z}$, in the complex plane is located in which quadrant?
A: First quadrant
B: Second quadrant
C: Third quadrant
D: Fourth quadrant | To solve for $z$, we start with the given expression and simplify it using complex number properties:
\begin{align*}
z &= \frac{i}{{1+i}} \\
&= \frac{i}{{1+i}} \cdot \frac{{1-i}}{{1-i}} \quad \text{(Multiplying by the conjugate of the denominator)} \\
&= \frac{i(1-i)}{(1+i)(1-i)} \\
&= \frac{i - i^2}{1^2 - i^2} \quad ... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,920 |
In the complex plane, the point corresponding to the complex number $\dfrac {2-3i}{i^{3}}$ is in $\text{( }$ $\text{)}$
A: the first quadrant
B: the second quadrant
C: the third quadrant
D: the fourth quadrant | First, we simplify the complex number $\dfrac {2-3i}{i^{3}}$.
We know that $i^2 = -1$, so $i^3 = i^2 \cdot i = -1 \cdot i = -i$.
Now, we can rewrite the given complex number as $\dfrac {(2-3i)}{-i}$.
Next, we multiply both the numerator and the denominator by the conjugate of the denominator, $i$, to eliminate the i... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,921 |
Given the function $f(x) = \frac{1}{2}x - \frac{1}{4}\sin x - \frac{\sqrt{3}}{4}\cos x$, the slope of the tangent line at point $A(x_0, f(x_0))$ is 1. Find the value of $\tan x_0$. | Since $f(x) = \frac{1}{2}x - \frac{1}{4}\sin x - \frac{\sqrt{3}}{4}\cos x$,
we have $f'(x) = \frac{1}{2} - \frac{1}{4}\cos x + \frac{\sqrt{3}}{4}\sin x = \frac{1}{2} + \frac{1}{2}\sin(x - \frac{\pi}{6})$.
Given that the slope of the tangent line at $x_0$ is 1, we have $f'(x_0) = \frac{1}{2} + \frac{1}{2}\sin(x_0 - ... | -\sqrt{3} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 548,922 |
Given that $f(x)$ is an odd function and satisfies $f(2-x)=f(x)$ for any $x \in \mathbb{R}$. When $0 < x \leq 1$, $f(x)=\ln x + 2$. The number of zeros of the function $y=f(x)$ on $[-4, 4]$ is ___. | **Analysis**
This problem tests the application of function properties.
According to the problem, the period of function $f(x)$ is $4$, and it is axially symmetric about the line $x=1+2k$ and centrally symmetric about the point $(2k, 0)$, where $k \in \mathbb{Z}$. We can find the zeros one by one.
**Solution**
Sinc... | 13 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,923 |
On Sunday, taxi driver Xiao Zhang volunteered to shuttle tourists for free on an east-west highway. It is stipulated that east is positive and west is negative. The taxi's itinerary is as follows (in kilometers): +10, -3, +4, -2, +13, -8, -7, -5, -2.
$(1)$ When Xiao Zhang drops off the last tourist at their destinatio... | ### Step-by-Step Solution
#### Part 1: Distance from the Starting Point
To find the distance between Xiao Zhang and the starting point after dropping off the last tourist, we sum up all the distances traveled:
\[
\begin{align*}
& (+10) + (-3) + (+4) + (-2) + (+13) + (-8) + (-7) + (-5) + (-2) \\
&= 10 + 4 + 13 - 3 - ... | 5.4 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,924 |
Among the following sets of numbers, the one that can form a right-angled triangle is:
A: $1$, $2$, $3$
B: $3$, $4$, $5$
C: $7$, $8$, $9$
D: $5$, $10$, $20$ | To determine which set of numbers can form a right-angled triangle, we apply the Pythagorean theorem, which states that in a right-angled 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. We also check if th... | B | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,925 |
Given the following propositions, which one is correct?
A: The direction vector of line $l$ is $\overrightarrow{a}=(1,-1,2)$, and the direction vector of line $m$ is $\overrightarrow{b}=(2,1,-\frac{1}{2})$, then $l$ is perpendicular to $m$.
B: The direction vector of line $l$ is $\overrightarrow{a}=(0,1,-1)$, and the... | To solve the problem, let's analyze each proposition step by step:
**Option A:**
Given $\overrightarrow{a}=(1,-1,2)$ and $\overrightarrow{b}=(2,1,-\frac{1}{2})$, we check if lines $l$ and $m$ are perpendicular by calculating the dot product of $\overrightarrow{a}$ and $\overrightarrow{b}$:
\[
\overrightarrow{a} \cdot... | A \text{ and } D | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,927 |
Let the complex number $z$ satisfy $z(1+i) = |\sqrt{3} - i|$ (where $i$ is the imaginary unit), then the point corresponding to the complex number $z$ in the complex plane is located in ( )
A: The first quadrant
B: The second quadrant
C: The third quadrant
D: The fourth quadrant | Given that the complex number $z$ satisfies $z(1+i) = |\sqrt{3} - i|$ (where $i$ is the imaginary unit),
then $z = \frac{|\sqrt{3} - i|}{1+i} = \frac{2}{1+i} = \frac{2(1-i)}{(1+i)(1-i)} = 1 - i$.
Therefore, the point corresponding to the complex number $z$ in the complex plane, $Z(1, -1)$, is located in the fourth ... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,928 |
Among the following statements:
① Given $a=b$, $b=c$, then $a=c$;
② Dividing both sides of an equation by the same number yields another equation;
③ Multiplying both sides of an equation by $0$, the result is not necessarily an equation;
④ Subtracting the same polynomial from both sides of an equation, the result... | Solution: ① Given $a=b$, $b=c$, according to the substitution property, we can get $a=c$, so ① is correct;
② Dividing both sides of an equation by the same non-zero number, the result is still an equation, so ② is incorrect;
③ Multiplying both sides of an equation by $0$, the result is definitely an equation, so ③ ... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,929 |
A particle moves in a straight line. If the relationship between the distance it travels and time is $s(t) = 4t^2 - 3$ (where the unit of $s(t)$ is meters and the unit of $t$ is seconds), then the instantaneous velocity at $t=5$ is ( )
A: $7m/s$
B: $10m/s$
C: $37m/s$
D: $40m/s$ | To find the instantaneous velocity of a particle moving in a straight line, we differentiate the given distance-time relationship $s(t) = 4t^2 - 3$ with respect to time $t$. This differentiation gives us the velocity function $v(t)$, which represents the instantaneous velocity at any time $t$.
Starting with the given ... | D | Calculus | MCQ | Yes | Yes | cn_k12 | false | 548,930 |
If $\alpha$ and $\beta$ are acute angles, and $\tan(\alpha+\beta)=3$, $\tan\beta=\frac{1}{2}$, find the value of $\alpha$.
A: $\frac{\pi}{3}$
B: $\frac{\pi}{4}$
C: $\frac{\pi}{6}$
D: $\frac{\pi}{12}$ | This problem involves the tangent function of the sum of two angles and the application of the tangent function values of special angles. Pay attention to the relationship between angles, and test your ability to simplify and transform expressions.
Step 1: Given that $\tan(\alpha+\beta)=3$ and $\tan\beta=\frac{1}{2}$,... | \alpha=\frac{\pi}{4} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,931 |
Which of the following survey methods is reasonable?
A: To understand the vision status of middle school students nationwide, choose a comprehensive survey.
B: To determine if a batch of bagged food contains preservatives, choose a comprehensive survey.
C: To test the air quality of a city, choose a sampling survey.... | To evaluate the reasonableness of each survey method, let's consider each option step by step:
- For option $A$: The goal is to understand the vision status of middle school students nationwide. Given the large population involved, conducting a comprehensive survey would be impractical and resource-intensive. Instead,... | C | Logic and Puzzles | MCQ | Yes | Yes | cn_k12 | false | 548,932 |
If angle $\alpha$ satisfies $\alpha= \frac {2k\pi}{3}+ \frac {\pi}{6}$ ($k\in\mathbb{Z}$), then the terminal side of $\alpha$ must be in ( )
A: The first, second, or third quadrant
B: The first, second, or fourth quadrant
C: The first, second quadrant, or on the non-negative half of the x-axis
D: The first, se... | Given $\alpha= \frac {2k\pi}{3}+ \frac {\pi}{6}$ ($k\in\mathbb{Z}$),
When $k=3n$ ($n\in\mathbb{Z}$), $\alpha=2n\pi+ \frac {\pi}{6}$, which is an angle in the first quadrant;
When $k=3n+1$ ($n\in\mathbb{Z}$), $\alpha=2n\pi+ \frac {5\pi}{6}$, which is an angle in the second quadrant;
When $k=3n+2$ ($n\in\mathbb{Z}$... | \text{D} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,933 |
Given the ellipse $$C: \frac {x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}=1(a>b>0)$$, the equation of the directrix corresponding to the focus F(2,0) is x=4.
(Ⅰ) Find the equation of ellipse C;
(Ⅱ) It is known that a line passing through point F1(-2,0) with an inclination angle of $\theta$ intersects the ellipse C at point... | Solution:
(Ⅰ) From the given information: $$\begin{cases} c=2 \\ \frac {a^{2}}{c}=4 \\ c^{2}=a^{2}-b^{2}\end{cases}$$, we find $a^{2}=8$ and $b^{2}=4$.
Therefore, the equation of ellipse C is $$\frac {x^{2}}{8}+ \frac {y^{2}}{4}=1$$.
(Ⅱ) From (Ⅰ), we know that F1(-2,0) is the right focus of the ellipse, and $$e= ... | \frac {16 \sqrt {2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,934 |
Given that the function $f(x)$ satisfies $f(2+x) = f(6-x)$ for any $x$ in its domain $\mathbb{R}$, and when $x \neq 4$, its derivative $f'(x)$ satisfies $xf'(x) > 4f'(x)$. If $9 < a < 27$, then ( )
A: $f(2^{ \sqrt {a}}) < f(6) < f(\log_{3}a)$
B: $f(6) < f(2^{ \sqrt {a}}) < f(\log_{3}a)$
C: $f(\log_{3}a) < f(2^{ \sq... | Since the function $f(x)$ satisfies $f(2+x) = f(6-x)$ for any $x$ in its domain $\mathbb{R}$,
it follows that $f(x)$ is symmetric about the line $x=4$;
Furthermore, when $x \neq 4$, its derivative $f'(x)$ satisfies $xf'(x) > 4f'(x) \Leftrightarrow f'(x)(x-4) > 0$,
therefore, when $x > 4$, $f'(x) > 0$, meaning $f(x)$... | null | Calculus | MCQ | Yes | Yes | cn_k12 | false | 548,935 |
Given two vectors $\overrightarrow {a}$ and $\overrightarrow {b}$ in a plane with an angle of 45° between them, $\overrightarrow {a}$ = (1, -1), and $|\overrightarrow {b}| = 1$, find $|\overrightarrow {a} + 2\overrightarrow {b}| = \_\_\_\_\_\_.$ | Since the angle between $\overrightarrow {a}$ and $\overrightarrow {b}$ is 45° and $|\overrightarrow {a}| = \sqrt{2}$, $|\overrightarrow {b}| = 1$, we have:
$$
\begin{align}
|\overrightarrow {a} + 2\overrightarrow {b}|^2 &= (\overrightarrow {a} + 2\overrightarrow {b}) \cdot (\overrightarrow {a} + 2\overrightarrow {b}) ... | \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,936 |
Factorize the expression: $mx+my=$____. | To factorize the expression $mx + my$, we look for a common factor in both terms. Here, the common factor is $m$. We can then factor out $m$ from both terms, which gives us:
\[
mx + my = m \cdot x + m \cdot y = m(x + y)
\]
Thus, the factorized form of the expression $mx + my$ is $\boxed{m(x + y)}$. | m(x + y) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,937 |
Given a sequence $\{a_n\}$ whose sum of the first $n$ terms is $S_n = 33n - n^2$,
(1) Prove that the sequence $\{a_n\}$ is an arithmetic sequence;
(2) How many terms of $\{a_n\}$ make the sum of its terms maximum;
(3) Let the sequence $\{b_n\}$ be defined by $b_n = |a_n|$, find the sum of the first $n$ terms of $\... | (1) **Proof**: For $n \geq 2$, we have $a_n = S_n - S_{n-1} = 34 - 2n$. When $n = 1$, $a_1 = S_1 = 32 = 34 - 2 \times 1$, which satisfies $a_n = 34 - 2n$,
thus, the general formula for $\{a_n\}$ is $a_n = 34 - 2n$,
so $a_{n+1} - a_n = 34 - 2(n+1) - (34 - 2n) = -2$,
hence, the sequence $\{a_n\}$ is an arithmetic s... | 272 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,938 |
In a bag, there are 6 yellow balls and 4 white balls (all of the same shape and size). Two balls are drawn successively without replacement. Given that the first ball drawn is yellow, the probability that the second ball drawn is also yellow is ( )
A: $$\frac {3}{5}$$
B: $$\frac {1}{3}$$
C: $$\frac {5}{9}$$
D: $$\... | From the problem, after drawing a yellow ball first, since there is no replacement, there are 5 different yellow balls and 4 different white balls left in the bag.
Therefore, given that the first ball drawn is yellow, the probability that the second ball drawn is also yellow is $$\frac {5}{5+4} = \frac {5}{9}$$．
He... | C | Combinatorics | MCQ | Yes | Yes | cn_k12 | false | 548,939 |
Given that vectors $\overrightarrow{p}, \overrightarrow{q}$ satisfy $|\overrightarrow{p}|=8, |\overrightarrow{q}|=6, \overrightarrow{p}\cdot\overrightarrow{q}=24$, find the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$ (denoted as $(\quad)$).
A: $30^{\circ}$
B: $45^{\circ}$
C: $60^{\circ}$
D: $90^{\circ}$ | Since $|\overrightarrow{p}|=8, |\overrightarrow{q}|=6, \overrightarrow{p}\cdot\overrightarrow{q}=24$,
$\cos = \frac{\overrightarrow{p}\cdot\overrightarrow{q}}{|\overrightarrow{p}||\overrightarrow{q}|} = \frac{24}{8 \times 6} = \frac{1}{2}$.
Also, $ \in [0^{\circ}, 180^{\circ}]$.
$\therefore$ The angle between $\ove... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,940 |
If point A (1, 1) lies on the line $mx+ny-3mn=0$, where $mn > 0$, then the minimum value of $m+n$ is \_\_\_\_\_\_. | Given that point A $(1, 1)$ is on the line $mx+ny-3mn=0$, we have:
$$
m(1) + n(1) - 3mn = 0
$$
Therefore:
$$
m + n = 3mn
$$
Since $mn > 0$, this implies:
$$
\frac{1}{m} + \frac{1}{n} = 3
$$
Multiplying both sides of $m + n = 3mn$ by $\frac{1}{3}\left(\frac{1}{m} + \frac{1}{n}\right)$ gives us:
$$
m+n = \frac{1}{3}... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,941 |
Given $$a= \frac {1}{log_{2}\pi }+ \frac {1}{log_{3}\pi }+ \frac {1}{log_{4}\pi }+ \frac {1}{log_{5}\pi }$$ and y=|x-a|, where x is a natural number, find the value of x when y is at its minimum.
A: 2
B: 3
C: 4
D: 5 | First, let's simplify the expression for $a$:
$$a= \frac {1}{log_{2}\pi }+ \frac {1}{log_{3}\pi }+ \frac {1}{log_{4}\pi }+ \frac {1}{log_{5}\pi }$$
Using the change of base formula for logarithms, we can rewrite each term in the sum as a logarithm to the base $\pi$. This gives us:
$$a = log_{\pi}2 + log_{\pi}3 + log_{\... | 4 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,942 |
Given the function $f(x)=\frac{1}{2}x^2+4\ln x$, if there exists a real number $x_0$ such that $1\leqslant x_0\leqslant 3$, which makes the tangent line of the curve $y=f(x)$ at the point $(x_0,f(x_0))$ perpendicular to the line $x+my-10=0$, then the range of values for the real number $m$ is $(\quad\quad)$.
A: $[5,+\... | 1. **Finding the slope of the given line**: The line $x+my-10=0$ can be rewritten in the form $y=-\frac{1}{m}x+\frac{10}{m}$. The slope of this line is $-\frac{1}{m}$.
2. **Finding the derivative of the function**: The derivative of $f(x)=\frac{1}{2}x^2+4\ln x$ is $f'(x)=x+\frac{4}{x}$.
3. **Finding the relationship ... | [4,5] | Calculus | MCQ | Yes | Yes | cn_k12 | false | 548,943 |
Prove using mathematical induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is divisible by 9 for all $n \in \mathbb{N}^*$. To verify for the case when $n = k + 1$ using the inductive hypothesis, it is only necessary to expand:
A: $(k + 3)^3$
B: $(k + 2)^3$
C: $(k + 1)^3$
D: $(k + 1)^3 + (k + 2)^3$ | To solve this problem using mathematical induction, we'll follow the standard two-step process: establish the base case and then prove the inductive step.
**Base Case (n=1):**
We must first prove that the statement holds for $n=1$. We have:
$$1^3 + (1 + 1)^3 + (1 + 2)^3 = 1 + 8 + 27 = 36,$$
which is divisible by 9, si... | \text{A: } (k + 3)^3 | Algebra | proof | Yes | Yes | cn_k12 | false | 548,944 |
Among the following options, which cannot form a set?
A: All positive numbers
B: Numbers equal to 2
C: Numbers close to 0
D: Even numbers not equal to 0 | The elements of a set must satisfy three criteria: definiteness, distinctness, and unorderedness.
The term "numbers close to 0" is indefinite.
Therefore, numbers close to 0 cannot form a set.
Hence, the correct choice is $\boxed{C}$.
This question examines the three criteria that elements of a set must satisfy:... | C | Number Theory | MCQ | Yes | Yes | cn_k12 | false | 548,945 |
In the Cartesian coordinate system $xOy$, the parametric equations of curve $C_1$ are
$$
\begin{cases}
x = 2 + 2\cos \varphi, \\
y = 2\sin \varphi
\end{cases}
(\varphi \text{ as the parameter}).
$$
With the origin $O$ as the pole and the positive x-axis as the polar axis, establish a polar coordinate system. The polar ... | (Ⅰ)
For curve $C_1$ with the given parametric equations
$$
\begin{cases}
x = 2 + 2\cos \varphi, \\
y = 2\sin \varphi
\end{cases},
$$
by eliminating the parameter, we obtain the general equation of curve $C_1$:
$$(x-2)^{2}+y^{2}=4.$$
Since the polar equation of curve $C_2$ is $\rho = 4\sin \theta$,
we have
$$ \rho^2 = ... | \alpha = \dfrac{3\pi}{4} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,946 |
Given points A(4,1,9) and B(10,-1,6), the distance between points A and B is ______. | To find the distance between two points in 3-dimensional space, we use the distance formula:
$$
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
$$
Substituting the given coordinates of points A and B into the formula:
$$
d = \sqrt{(10 - 4)^2 + (-1 - 1)^2 + (6 - 9)^2}
$$
Simplifying the expression:
$$
d = ... | 7 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,947 |
Given a complex number $z$ satisfying $zi=2i+x$ ($x\in \mathbb{R}$), if the imaginary part of $z$ is 2, then $|z|=$ （ ）
A: 2
B: $2\sqrt{2}$
C: $\sqrt{5}$
D: $\sqrt{3}$ | Since the complex number $z$ satisfies $zi=2i+x$ ($x\in \mathbb{R}$),
we can derive $z= \frac{2i+x}{i}=2-xi$.
If the imaginary part of $z$ is 2,
then we get $x=-2$.
Thus, $z=2-2i$.
Therefore, $|z|=2\sqrt{2}$.
Hence, the correct option is: $\boxed{B}$.
This problem involves simplifying a complex number usi... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,948 |
If the function $f(x)=\left\{\begin{array}{l}{-{x}^{2}+2a, x\leqslant -1}\\{ax+4, x>-1}\end{array}\right.$ is a monotonic function on $R$, then the possible values of $a$ are:
A: $0$
B: $1$
C: $2$
D: $3$ | To determine the values of $a$ that make the function $f(x)$
\[f(x)=\left\{\begin{array}{ll}
-{x}^{2}+2a, & x\leqslant -1\\
ax+4, & x>-1
\end{array}\right.\]
monotonic on $\mathbb{R}$, we analyze the function in two parts based on its definition.
**Step 1:** For $x \leqslant -1$, the function is $f(x) = -x^2 + 2a$. Si... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,949 |
Given $\tan \theta = 2$, find the value of $\dfrac{3\sin \theta - 2\cos \theta}{\sin \theta + 3\cos \theta}$. | **Analysis**
This problem examines the relationship between trigonometric functions of the same angle and its application, which is a basic question.
Divide both the numerator and the denominator of the expression by $\cos \theta$ and substitute $\tan \theta = 2$ to get the result.
**Solution**
Since $\tan \theta =... | \dfrac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,950 |
Given the function $f(x)=\log_{\frac{1}{e}}(x^{2}+\frac{1}{e})-\left|\frac{x}{e}\right|$, find the range of $x$ that satisfies $f(x+1) < f(2x-1)$. The options are:
A: $(0,2)$
B: $(-\infty,0)$
C: $(-\infty,0)\cup(2,+\infty)$
D: $(2,+\infty)$ | For $x > 0$, $f(x)=\log_{\frac{1}{e}}(x^{2}+\frac{1}{e})-\frac{x}{e}$ is a decreasing function.
For $x |2x-1|$. Solving this yields: $0 |2x-1|$. This question tests the understanding of function monotonicity, evenness, and transformation thinking, making it a medium-level problem. | null | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,951 |
In the Cartesian coordinate system $xoy$, the parametric equation of curve $C_1$ is $$\begin{cases} x=a\cos t+ \sqrt {3} \\ y=a\sin t\end{cases}$$ (where $t$ is the parameter, $a>0$). In the polar coordinate system with the origin as the pole and the positive $x$-axis as the polar axis, the equation of curve $C_2$ is $... | (1) Since the parametric equation of curve $C_1$ is $$\begin{cases} x=a\cos t+ \sqrt {3} \\ y=a\sin t\end{cases}$$ (where $t$ is the parameter, $a>0$),
the standard equation of $C_1$ is $$(x- \sqrt {3})^{2}+y^{2}=a^{2}$$,
thus $C_1$ is a circle with center at $(\sqrt {3}, 0)$ and radius $a$.
By using $\rho^{2}=x^{2}+y^... | 3 \sqrt {3} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,952 |
A mobile communication company offers two types of services: "Global Call" users pay a monthly rental fee of 50 yuan first, and then 0.4 yuan per minute for calls; "China Mobile" users do not pay a monthly rental fee, and pay 0.6 yuan per minute for calls (referring to local calls). If the call duration in a month is $... | **Analysis:**
(1) Since the mobile communication company offers two types of services: "Global Call" users pay a monthly rental fee of 50 yuan first, and then 0.4 yuan per minute for calls; "China Mobile" users do not pay a monthly rental fee, and pay 0.6 yuan per minute for calls. If the call duration in a month is ... | x > 125 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,953 |
Given points $A(-1,1)$, $B(1,2)$, $C(-2,-1)$, $D(3,4)$, find the projection of vector $\overrightarrow{AB}$ in the direction of vector $\overrightarrow{CD}$. | **Analysis**
This problem mainly examines the scalar product of plane vectors and the projection of vectors, which is a basic question. By using the definition of the scalar product of plane vectors and the projection of vectors, the solution can be obtained.
**Solution**
Given $\overrightarrow{AB}=(2,1)$, $\overrig... | \dfrac{3\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,954 |
In $\triangle ABC$, the sides opposite to angles $A$, $B$, $C$ are respectively $a$, $b$, $c$, and it is given that $a\sin B+ \sqrt{3}a\cos B= \sqrt{3}c.$
(I) Find the magnitude of angle $A$;
(II) Given the function $f(x)=\lambda\cos^2(\omega x+ \frac{A}{2})-3$ $(\lambda > 0,\omega > 0)$ has a maximum value of $2$,... | Solution:
(I) In $\triangle ABC$, since $a\sin B+ \sqrt{3}a\cos B= \sqrt{3}c$,
it follows that $\sin A\sin B+ \sqrt{3}\sin A\cos B= \sqrt{3}\sin C$, and since $C=\pi-(A+B)$,
we have $\sin A\sin B+ \sqrt{3}\sin A\cos B= \sqrt{3}\sin (A+B)= \sqrt{3}(\sin A\cos B+\cos A\sin B)$,
thus $\tan A= \sqrt{3}$, and since ... | \left[-3, \frac{5\sqrt{3}-2}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,955 |
Given an ellipse with foci $F_1$ and $F_2$, and $P$ is a moving point on the ellipse. A perpendicular line is drawn from $F_2$ to the external angle bisector of $\angle F_1PF_2$, and the foot of the perpendicular is $M$. Then, the locus of point $M$ is ( )
A: Circle
B: Ellipse
C: Straight Line
D: One branch of a Hyperb... | The symmetric point $Q$ of $F_2$ with respect to the external angle bisector $PM$ of $\angle F_1PF_2$ lies on the extension of line $F_1Q$.
Therefore, $|F_1Q| = |PF_1| + |PF_2| = 2a$ (the length of the major axis of the ellipse).
Moreover, $OM$ is the median line of $\triangle F_2F_1Q$, so $|OM| = a$.
Hence, the loc... | \text{A: Circle} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,956 |
Simplify: $$(\frac{2}{3})^{0} + 2^{-2} \times (\frac{9}{16})^{-\frac{1}{2}} + (\lg 8 + \lg 125)$$ | Given the expression to be simplified is $$(\frac{2}{3})^{0} + 2^{-2} \times (\frac{9}{16})^{-\frac{1}{2}} + (\lg 8 + \lg 125),$$ let's approach this step-by-step:
1. **Zero Power**:
$$(\frac{2}{3})^{0} = 1$$
Any non-zero number to the power of zero equals 1.
2. **Negative Exponent**:
$$2^{-2} = \frac{1}{2^2... | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,957 |
If on the ellipse $\dfrac {x^{2}}{36}+ \dfrac {y^{2}}{16}=1$, a point $P$ forms perpendicular lines with the two foci $F_{1}$ and $F_{2}$ of the ellipse, then the area of $\triangle PF_{1}F_{2}$ is ( )
A: $36$
B: $24$
C: $20$
D: $16$ | According to the problem, in the ellipse $\dfrac {x^{2}}{36}+ \dfrac {y^{2}}{16}=1$, we have $a= \sqrt {36}=6$, $b= \sqrt {16}=4$,
thus $c= \sqrt {a^{2}-b^{2}}= \sqrt {20}=2 \sqrt {5}$,
By the definition of an ellipse: $|PF_{1}|+|PF_{2}|=2a=12$,
By the Pythagorean theorem, we know: $|PF_{1}|^{2}+|PF_{2}|^{2}=(2c)... | D | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,958 |
The surface area of a cone is three times its base area. What is the central angle of the sector formed by unrolling the cone's lateral surface? | Given that the total surface area of the cone is three times its base area, we can deduce that the ratio of the slant height to the base radius is 2.
Let's denote the base radius of the cone as $r$. Therefore, the slant height of the cone is $2r$. When the cone's lateral surface is unrolled, it forms a sector of a cir... | 180^\circ | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,959 |
Given the function $f(x)$, the equation of the tangent line at point $(x_{0},f(x_{0}))$ is $l$: $y=g(x)$. If the function $f(x)$ satisfies $\forall x \in I$ (where $I$ is the domain of $f(x)$), when $x \neq x_{0}$, the inequality $[f(x)-g(x)](x-x_{0}) > 0$ always holds, then $x_{0}$ is called a "crossing point" of the ... | To solve this, considering the condition that for the function $f(x)$, $\forall x \in I$ (where $I$ is the domain of $f(x)$), when $x \neq x_{0}$, the inequality $[f(x)-g(x)](x-x_{0}) > 0$ always holds, we use the second derivative set to $0$ to solve: $f''(x)=- \frac {1}{x^{2}}-a=0$. It is clear that there is a soluti... | D | Calculus | MCQ | Yes | Yes | cn_k12 | false | 548,960 |
The maximum integer solution of the inequality system $\left\{\begin{array}{l}3x-1 \lt x+1\\2\left(2x-1\right)\leqslant 5x+1\end{array}\right.$ is ( ).
A: $1$
B: $-3$
C: $0$
D: $-1$ | To solve the given system of inequalities, we proceed step by step:
1. **First Inequality: $3x - 1 < x + 1$**
Starting with the first inequality:
\[3x - 1 < x + 1\]
Subtract $x$ from both sides to get the $x$ terms on one side:
\[3x - x - 1 < 1\]
Simplify:
\[2x - 1 < 1\]
Add $1$ to both sides to isolate the $x$ te... | C | Inequalities | MCQ | Yes | Yes | cn_k12 | false | 548,961 |
Given the equation: $\sin^2 5^\circ + \cos^2 35^\circ + \sin 5^\circ \cos 35^\circ = \frac{3}{4}$; $\sin^2 15^\circ + \cos^2 45^\circ + \sin 15^\circ \cos 45^\circ = \frac{3}{4}$; $\sin^2 30^\circ + \cos^2 60^\circ + \sin 30^\circ \cos 60^\circ = \frac{3}{4}$; from this, deduce a general equation that holds for any ang... | Solution: Based on the common characteristics of each equation, where the cosine is the sine degree plus $30^\circ$, and the right side is the constant $\frac{3}{4}$, we can deduce a general equation: $\sin^2 \theta + \cos^2 (\theta + 30^\circ) + \sin \theta \cos (\theta + 30^\circ) = \frac{3}{4}$.
Proof: The left sid... | \text{Proved} | Algebra | proof | Yes | Yes | cn_k12 | false | 548,963 |
If $\int_{0}^{T} x^2 dx = 9$, find the value of the constant $T$. | We begin by evaluating the given definite integral:
$$\int_{0}^{T} x^2 dx = \left[\frac{1}{3}x^3\right]_{0}^{T} = \frac{1}{3}T^3 - \frac{1}{3}(0)^3 = \frac{1}{3}T^3.$$
Equating this to the given value of 9, we have:
$$\frac{1}{3}T^3 = 9.$$
Now, we solve for $T$:
$$T^3 = 27,$$
$$T = \sqrt[3]{27} = 3.$$
So, the value ... | 3 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 548,964 |
If $A$ and $B$ are two independent events, and $P(A) = 0.4$, $P(A+B) = 0.7$, then $P(B) = \, ?$ | Since $A$ and $B$ are two independent events, we have $P(A+B) = P(A) + P(B) - P(A \cap B)$.
Therefore, $0.7 = 0.4 + P(B) - 0.4P(B)$.
This simplifies to $0.6P(B) = 0.3$.
Solving for $P(B)$, we find $P(B) = 0.5$.
Hence, the answer is $\boxed{0.5}$.
**Analysis:** By using the formula $P(A+B) = P(A) + P(B) - P(A \cap ... | 0.5 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,965 |
Given that $f(x)$ is an odd function, and when $x < 0$, $f(x) = \ln (-x) + 2x$, find the equation of the tangent line to the curve $y = f(x)$ at the point $(1, f(1))$. | Since $f(x)$ is an odd function, we have $f(-x) = -f(x)$ for any $x$ in the domain of $f$ (in this case for $x > 0$).
For a negative $x$, we know that $f(x) = \ln(-x) + 2x$. To find $f(x)$ for a positive $x$, we can use the fact that $f$ is odd. Specifically, for $x > 0$, $-x 0$:
\[
f'(x) = \frac{d}{dx}(-\ln(x) + 2x)... | x - y + 1 = 0 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 548,967 |
There are three types of components A, B, and C, with 300 of type B and 200 of type C. A stratified sampling method is used to take a sample of size 45, in which 20 of type A and 10 of type C are sampled. How many total components are there? | Let $N_A$, $N_B$, and $N_C$ be the total number of components A, B, and C, respectively. According to the problem, we have $N_B = 300$ and $N_C = 200$.
Given that $N_A$, along with $N_B$ and $N_C$, are used in a stratified sampling to choose a sample of 45 components, where 20 are A and 10 are C, we can set up the fol... | 900 | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false | 548,968 |
Calculate the following definite integrals:
(1) $\int_{0}^{1} (2x - x^2)dx$;
(2) $\int_{2}^{4} (3 - 2x)dx$;
(3) $\int_{0}^{1} \frac{1}{3}x^2 dx$;
(4) $\int_{0}^{2π} \cos x dx$. | (1) $\int_{0}^{1} (2x - x^2)dx = (x^2 - \frac{1}{3}x^3) \vert_{0}^{1} = (1 - \frac{1}{3}) - (0 - 0) = \boxed{\frac{2}{3}}$;
(2) $\int_{2}^{4} (3 - 2x)dx = (3x - x^2) \vert_{2}^{4} = (12 - 16) - (6 - 4) = \boxed{-6}$;
(3) $\int_{0}^{1} \frac{1}{3}x^2 dx = \frac{1}{9}x^3 \vert_{0}^{1} = (\frac{1}{9} - 0) - (0 - 0) = \b... | 0 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 548,969 |
The coordinates of the midpoint between point P (1, 4, -3) and point Q (3, -2, 5) are ( ).
A: \( (4, 2, 2) \)
B: \( (2, -1, 2) \)
C: \( (2, 1, 1) \)
D: \( (4, -1, 2) \) | To find the midpoint between point P \( (1, 4, -3) \) and point Q \( (3, -2, 5) \), we use the midpoint formula in three dimensions. The midpoint \( M \) can be found using the following calculations for the coordinates \( (x_M, y_M, z_M) \):
\[ x_M = \frac{x_P + x_Q}{2}, \]
\[ y_M = \frac{y_P + y_Q}{2}, \]
\[ z_M = \f... | (2, 1, 1) | Geometry | MCQ | Yes | Yes | cn_k12 | false | 548,970 |
Given the matrix \[ \begin{matrix}5 & 1 \\ 7 & 3\end{matrix} \] has the inverse matrix \[ \begin{matrix}a & b \\ c & d\end{matrix} \], then $a+b+c+d=$ ______. | Since the inverse matrix of \[ \begin{matrix}5 & 1 \\ 7 & 3\end{matrix} \] is \[ \begin{matrix}a & b \\ c & d\end{matrix} \],
it follows that \[ \begin{matrix}5 & 1 \\ 7 & 3\end{matrix} \] \[ \begin{matrix}a & b \\ c & d\end{matrix} \] = \[ \begin{matrix}1 & 0 \\ 0 & 1\end{matrix} \],
thus we have the system of equ... | 0 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,971 |
The complex number $z$ satisfies $|2z-1+i|=4$, and $w=z(1-i)+2+i$.
(1) Find the trajectory $C$ of point $P$ corresponding to $w$ in the complex plane.
(2) In the complex plane, a tangent line is drawn from point $Q(0,4)$ to the trajectory $C$, touching at points $A$ and $B$. Find the equation of line $AB$. | (1) Let $w=x+yi$,
then from $w=z(1-i)+2+i$, we get $z= \frac{w-2-i}{1-i} = \frac{1}{2}[(x-y-1)+(x+y-3)i]$,
Since the complex number $z$ satisfies $|2z-1+i|=4$,
we have $|2z-1+i|^2 = (x-y-2)^2 + (x+y-2)^2 = 2[(x-2)^2+y^2] = 16$,
which simplifies to $(x-2)^2+y^2=8$,
thus, the trajectory $C$ of point $P$ corresp... | x-2y+2=0 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,972 |
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