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Given the function $f(x) = \log_2 \frac{2(1+x)}{x-1}$, if $f(a) = 2$, find $f(-a)=$\_\_\_\_\_\_\_. | **Analysis**
This problem involves finding the value of a logarithmic function. It's quite basic.
We first find the value of $a$ using $f(a) = 2$, and then find the value of $f(-a)$.
**Solution**
Given the function $f(x) = \log_2 \frac{2(1+x)}{x-1}$,
We have $f(a) = 2$, which implies $\log_2 \frac{2(1+a)}{a-1} = 2... | 0 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,183 |
If the function $f(x+1) = x^2 - 1$, then $f(2) = \quad$. | **Method 1:** Let $x+1=2$, solving for $x$ gives $x=1$. Substituting into $f(x+1) = x^2 - 1$, we find $f(2) = 0$.
**Method 2:** Let $x+1=t$, solving for $x$ gives $x=t-1$. Substituting into $f(x+1) = x^2 - 1$, we get $f(t) = (t-1)^2 - 1 = t^2 - 2t$.
Thus, the function can be expressed as $f(x) = x^2 - 2x$.
Therefore, ... | 0 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,184 |
Given the inequality $|2x+2|-|x-1|>a$.
(1) When $a=0$, find the solution set of the inequality.
(2) If the inequality has no solution in the interval $[-4, 2]$, find the range of the real number $a$. | (1) When $a=0$, the inequality becomes $|2x+2|-|x-1|>0$,
we can get $$\begin{cases} x0\end{cases}$$①, or $$\begin{cases} -1\leq x0\end{cases}$$②, or $$\begin{cases} x\geq 1 \\ 2x+2-(x-1)>0\end{cases}$$③.
Solving ① gives $x-\frac{1}{3}\}}$.
(2) When $x\in[-4, 2]$, $f(x)=|2x+2|-|x-1|=$ $$\begin{cases} -x-3, & xa$ h... | a\leq 3 | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false | 549,185 |
The graph of the function $f(x)$ is translated 1 unit to the right, and the resulting graph is symmetric about the y-axis with the curve $y = e^x$. Then, $f(x) =$ ( )
A: $e^{x+1}$
B: $e^{x-1}$
C: $e^{-x+1}$
D: $e^{-x-1}$ | The curve symmetric about the y-axis to $y = e^x$ is $y = e^{-x}$. Translating $y = e^{-x}$ 1 unit to the left gives $y = e^{-(x+1)}$, which means $f(x) = e^{-x-1}$. Therefore, the correct answer is $\boxed{\text{D}}$. | \text{D} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,186 |
Let ellipse $C$: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b > 0)$ have its left and right foci as $F_{1}$ and $F_{2}$, respectively. If point $P$ on $C$ satisfies $PF_{2} \perp F_{1}F_{2}$ and $\angle PF_{1}F_{2} = 30^{\circ}$, then the eccentricity of $C$ is ( )
A: $\frac{\sqrt{6}}{6}$
B: $\frac{1}{3}$
C... | **Analysis**
This question examines the basic properties of an ellipse. By finding the lengths of $|PF_{1}|$, $|PF_{2}|$, and $|F_{1}F_{2}|$, and using the definition of an ellipse along with the Pythagorean theorem, we can solve the problem.
**Solution**
Let $|PF_{1}| = x$,
Since $PF_{2} \perp F_{1}F_{2}$ and $\an... | D | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,187 |
Given the universal set $U=\{x|x\leq 4\}$, set $A=\{x|-2<x<3\}$, and set $B=\{x|-3\leq x\leq 2\}$, find $A\cap B$, $\complement_U (A\cup B)$, $(\complement_U A)\cup B$, $A\cap (\complement_U B)$, and $(\complement_U A)\cup (\complement_U B)$. | Since the universal set $U=\{x|x\leq 4\}$, set $A=\{x|-2<x<3\}$, and set $B=\{x|-3\leq x\leq 2\}$,
- The intersection of $A$ and $B$ is $A\cap B=\{x|-2<x\leq 2\}$.
- The complement of $A\cup B$ in $U$ is $\complement_U (A\cup B)=(-\infty, -3)\cup [3, 4]$.
- The union of the complement of $A$ in $U$ with $B$ is $(\comp... | (-\infty, -2]\cup (2, 4] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,188 |
$$3^{\log_{3}4}-27^{\frac{2}{3}}-\lg0.01+\ln e^{3} = \boxed{?}$$
A: 14
B: 0
C: 1
D: 6 | We have $$3^{\log_{3}4}-27^{\frac{2}{3}}-\lg0.01+\ln e^{3} = 4-3^{3\times\frac{2}{3}}-\lg10^{-2}+3\ln e = 4-9+2+3 = 0,$$
Therefore, the correct choice is $\boxed{\text{B}}$.
This problem mainly tests the calculation of exponential and logarithmic functions. It can be solved directly using the formulas for exponent... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,189 |
Place 6 cards labeled 1, 2, 3, 4, 5, 6 into 3 different envelopes, with 2 cards in each envelope. If the cards labeled 1 and 2 are placed in the same envelope, how many different arrangements are there? | This problem is a step-by-step counting problem.
First, choose one of the 3 envelopes to place cards 1 and 2, which can be done in $3$ different ways.
Next, choose 2 out of the remaining 4 cards to place in another envelope, which can be done in $C_{4}^{2}=6$ ways.
The remaining 2 cards are then placed in the las... | 18 | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false | 549,190 |
Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ satisfy $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{a} + \overrightarrow{b}| = 1$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is ______. | From the given conditions, we have $|\overrightarrow{a} + \overrightarrow{b}| = 1$,
which implies $\overrightarrow{a}^2 + 2\overrightarrow{a} \cdot \overrightarrow{b} + \overrightarrow{b}^2 = 1$,
Since $|\overrightarrow{a}| = |\overrightarrow{b}|$,
we get $\overrightarrow{a} \cdot \overrightarrow{b} = -\frac{1}{2}$,... | \frac{2\pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,191 |
When $x=m$ and $x=n$ (where $m \neq n$), the values of the quadratic function $y=x^2-2x+3$ are equal. What is the value of this function when $x=m+n$? | Given that the function values are identical for $x=m$ and $x=n$, and observing that the function can be expressed as
$$y=(x-1)^2+2,$$
we can deduce that the points with coordinates $(m, y)$ and $(n, y)$ are symmetric about the line $x=1$. This symmetry implies that the midpoint of the line segment connecting the two... | y=3 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,192 |
Given vector $\overrightarrow{b}=(3,4)$, and $\overrightarrow{a}•\overrightarrow{b}=10$, then the projection vector of vector $\overrightarrow{a}$ on vector $\overrightarrow{b}$ is ()
A: $(\frac{3}{5},\frac{4}{5})$
B: $(\frac{4}{5},\frac{3}{5})$
C: $(\frac{6}{5},\frac{8}{5})$
D: $(\frac{8}{5},\frac{6}{5})$ | To solve for the projection vector of vector $\overrightarrow{a}$ on vector $\overrightarrow{b}$, we start with the given vectors and the dot product between them:
Given:
- $\overrightarrow{b}=(3,4)$,
- $\overrightarrow{a}•\overrightarrow{b}=10$,
The formula for the projection of $\overrightarrow{a}$ on $\overrightar... | \text{C} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,193 |
Given the lengths of the sides of a triangle are 8cm, 10cm, and 12cm, find the perimeter of the triangle formed by connecting the midpoints of the sides of the original triangle. | **Analysis of the problem**: According to the Midsegment Theorem, the lengths of the sides of the new triangle are half of the original triangle's sides. Therefore, the perimeter of the new triangle is half of the original triangle's perimeter, which is $15cm$.
**Key point**: Midsegment Theorem.
Hence, the perimeter ... | 15cm | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,194 |
Given that $\cos \left(x - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$, find the value of $\cos x + \cos\left(x - \frac{\pi}{3}\right)$.
(A) $-1$
(B) $1$
(C) $\frac{2\sqrt{3}}{3}$
(D) $\sqrt{3}$ | Since $\cos \left(x - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$, we aim to express $\cos x + \cos\left(x - \frac{\pi}{3}\right)$ in a usable form.
Firstly, recall the cosine addition formula: $\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$.
Applying this formula, we can express $\cos\left(x - \fra... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,195 |
Cancer is one of the diseases that seriously threaten human health. Among the following descriptions, the incorrect one is ( ).
A: Due to the influence of the virus, AIDS patients have a higher probability of developing cancer than normal people.
B: Cancer cells are the result of abnormal cell differentiation, can pr... | The knowledge points required include the characteristics of cancer cells and related knowledge about carcinogens. The virus in AIDS patients attacks T cells, making the immune system unable to timely discover and clear cells that have undergone malignant transformation, so the probability of developing cancer is highe... | \text{D} | Other | MCQ | Yes | Yes | cn_k12 | false | 549,198 |
Given a hyperbola with the equation $\frac {x^{2}}{a^{2}} - \frac {y^{2}}{b^{2}} = 1$ (where $a > 0$, $b > 0$), a point $P(x_0, y_0)$ on the hyperbola has a tangent line with the equation $\frac {x_{0}x}{a^{2}} - \frac {y_{0}y}{b^{2}} = 1$. If a point $P(x_0, y_0)$ on the hyperbola ($a \leq x_0 \leq 2a$) has a tangent ... | A tangent line is drawn to the hyperbola $\frac {x^{2}}{a^{2}} - \frac {y^{2}}{b^{2}} = 1$ (where $a > 0$, $b > 0$) at a point $P(x_0, y_0)$ ($a \leq x_0 \leq 2a$). The equation of the tangent line is $\frac {x_{0}x}{a^{2}} - \frac {y_{0}y}{b^{2}} = 1$. Substituting the point $N(0, b)$ into the tangent line equation, w... | B | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,199 |
To alleviate traffic pressure, a special railway line is constructed between two cities in a certain province, with a train serving as public transportation. It is known that the daily round trips ($y$) is a first-order function of the number of carriages dragged each time ($x$). If this train drags 4 carriages each ti... | (1) Let $y = kx + m (k ≠ 0)$,
According to the problem, we can derive the following system of equations:
$$\begin{cases} 4k + m = 16 \\ 6k + m = 10 \end{cases} \Rightarrow \begin{cases} k = -3 \\ m = 28 \end{cases}$$
Therefore, the function of $y$ with respect to $x$ is: $y = -3x + 28$.
(2) Let $g(x) = 110xy = 110x(-3... | 14300 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,201 |
Given the function f(x) = ax^2 - 2lnx (a ∈ R).
1. Discuss the monotonicity of the function f(x);
2. When $a= \frac {1}{e^{2}}$, if the two zeros of the function y=f(x) are x₁ and x₂ (x₁ ln2 + 1. | 1. It is easy to know that the domain of f(x) is (0, +∞),
f'(x) = 2ax - $\frac {2}{x}$ = $\frac {2(ax^{2}-1)}{x}$,
When a ≤ 0, f'(x) 0, f'(x) > 0. Let f'(x) = 0, we get x = $\sqrt { \frac {1}{a}}$,
Therefore, f(x) is decreasing on (0, $\sqrt { \frac {1}{a}}$) and increasing on ($\sqrt { \frac {1}{a}}$, +∞)... | \text{ln(x₁ + x₂) > ln2 + 1} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,202 |
The relationship among three numbers $a=0.4^{2}$, $b=\log_{2}0.4$, $c=2^{0.4}$ is ( ).
A: $a < c < b$
B: $b < a < c$
C: $a < b < c$
D: $b < c < a$ | Since $a=0.4^{2}$ belongs to the interval $(0,1)$, $b=\log_{2}0.4 1$,
it follows that $b < a < c$.
Therefore, the correct choice is $\boxed{B}$.
This conclusion can be reached by utilizing the monotonicity of the exponential and logarithmic functions.
This question tests the understanding of the monotonicity o... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,203 |
In the polar coordinate system, the curve $$ρ=4\sin\left(θ- \frac {π}{3}\right)$$ is symmetric about ( )
A: The line $$θ= \frac {π}{3}$$
B: The line $$θ= \frac {5}{6}π$$
C: The point $$(2， \frac {π}{3})$$
D: The pole | Solution: Convert the original polar equation $$ρ=4\sin\left(θ- \frac {π}{3}\right)$$ to:
$$ρ^2=2ρ\sinθ-2\sqrt{3}ρ\cosθ,$$
which, when converted to Cartesian coordinates, becomes: $$x^2+y^2+2\sqrt{3}x-2y=0.$$
This represents a circle with its center at $(-\sqrt{3}, 1)$. The polar equation of a line passing throug... | \text{B} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,204 |
Given that the period of the function $f(x)=\sin^{2}(ωx)-\frac{1}{2}\ (ω > 0)$ is $π$, if its graph is shifted right by $a$ units $(a > 0)$, and the resulting graph is symmetric about the origin, then the minimum value of the real number $a$ is $(\ \ \ \ \ \ )$
A: $\frac{π}{4}$
B: $\frac{π}{2}$
C: $\frac{3π}{4}$
D: $π$ | From the function $f(x)=\sin^{2}(ωx)-\frac{1}{2}=-\frac{1}{2}\cos(2ωx)\ (ω > 0)$ with a period of $\frac{2π}{2ω}=π$, we can deduce that $ω=1$.
Thus, $f(x)=-\frac{1}{2}\cos(2x)$.
If the graph of this function is shifted right by $a$ units $(a > 0)$, we get $y=-\frac{1}{2}\cos(2(x-a))=-\frac{1}{2}\cos(2x-2a)$.
Given t... | \frac{π}{4} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,205 |
For two non-intersecting space lines $a$ and $b$, there must exist a plane $\alpha$ such that ( )
A: $a \subset \alpha$, $b \subset \alpha$
B: $a \perp \alpha$, $b \perp \alpha$
C: $a \subset \alpha$, $b \perp \alpha$
D: $a \subset \alpha$, $b \parallel \alpha$ | Since the space lines $a$ and $b$ do not intersect,
the positional relationship between $a$ and $b$ could either be parallel or skew.
Now, let's evaluate each option:
For option A, when $a$ and $b$ are skew lines, there does not exist a plane $\alpha$
such that $a \subset \alpha$ and $b \subset \alpha$ simultan... | \text{D} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,206 |
Determine the value of $\sin 240^{\circ}=$ \_\_\_\_\_\_. | According to the trigonometric identity $\sin (180^{\circ}+\alpha)=-\sin \alpha$, we have:
\begin{align*}
\sin 240^{\circ} &= \sin (180^{\circ}+60^{\circ}) \\
&= -\sin 60^{\circ} \\
&= -\frac{\sqrt{3}}{2}.
\end{align*}
Therefore, the answer is: $\boxed{-\frac{\sqrt{3}}{2}}$.
To solve this problem, we applied the trigo... | -\frac{\sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,207 |
To make the fraction $\frac{1}{x+1}$ meaningful, the condition that $x$ should satisfy is:
A: $x\neq -1$
B: $x\neq 0$
C: $x\neq 1$
D: $x \gt 1$ | To ensure the fraction $\frac{1}{x+1}$ is meaningful, the denominator, $x+1$, must not be equal to zero. This is because division by zero is undefined in mathematics. Therefore, we set up the inequality:
\[x + 1 \neq 0\]
Subtracting $1$ from both sides to solve for $x$, we get:
\[x \neq -1\]
This means that for the... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,208 |
Given the function $f(x)=(2-a)\ln x+\frac{1}{x}+2ax (a\in R)$,
(I) Discuss the monotonicity of the function $f(x)$;
(II) If for any $a\in (-3,-2)$, ${{x}_{1}},{{x}_{2}}\in [1,3]$, the inequality $(m+\ln 3)a-2\ln 3 > \left| f({{x}_{1}})-f({{x}_{2}}) \right|$ always holds, find the range of values for the real number $... | (I) The derivative of the function is ${{f}^{{'}}}(x)=\frac{(2-a)x-1+2a{{x}^{2}}}{{{x}^{2}}}=\frac{(2x-1)(ax+1)}{{{x}^{2}}}$, $x > 0$
① When $a=0$, ${{f}^{{'}}}(x)=\frac{(2x-1)}{{{x}^{2}}}$,
$f(x)$ is increasing on $x\in (\frac{1}{2},+\infty )$ and decreasing on $x\in (0,\frac{1}{2})$;
② When $a > 0$, ${{f}^{{'}}}(x... | m\leqslant -\frac{13}{3} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,209 |
Given $a^2b^2 + a^2 + b^2 + 16 = 10ab$, find the value of $a^2 + b^2$. | Since $a^2b^2 + a^2 + b^2 + 16 = 10ab$,
we have $a^2b^2 + a^2 + b^2 + 16 - 10ab = 0$,
which can be rewritten as $a^2b^2 - 8ab + 16 + a^2 + b^2 - 2ab = 0$,
thus, $(ab - 4)^2 + (a - b)^2 = 0$,
which implies $ab = 4$ and $a - b = 0$,
therefore, $a = b = \pm 2$;
hence, $a^2 + b^2 = 8$.
So, the answer is $\box... | 8 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,210 |
$(1)$ Calculate: $\sqrt{16}+{(1-\sqrt{3})^0}-{2^{-1}}$.
$(2)$ Solve the system of inequalities: $\left\{{\begin{array}{l}{-2x+6≥4}\\{\frac{{4x+1}}{3}＞x-1}\end{array}}\right.$, and write down the non-negative integer solutions of this system. | ### Solution:
#### Part 1: Calculate $\sqrt{16}+{(1-\sqrt{3})^0}-{2^{-1}}$
- We start by calculating each term individually:
- $\sqrt{16} = 4$ because $4^2 = 16$.
- $(1-\sqrt{3})^0 = 1$ because any non-zero number raised to the power of $0$ equals $1$.
- $2^{-1} = \frac{1}{2}$ because $2^{-1}$ means $1$ divided... | 0, 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,211 |
(1) Given that $a$ and $b$ are positive real numbers, prove that: $\frac{1}{a^{2}} + \frac{1}{b^{2}} + ab \geqslant 2 \sqrt{2}$.
(2) Given that $a > 0$, $b > 0$, $c > 0$, and $a^{2} + b^{2} + c^{2} = 4$, find the maximum value of $ab + bc + ac$. | (1) Proof:
Since $a$ and $b$ are positive real numbers, we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality twice as follows:
- First, $\frac{1}{a^{2}} + \frac{1}{b^{2}} \geqslant \frac{2}{ab}$
- Second, $\frac{2}{ab} + ab \geqslant 2\sqrt{2}$
Combining these inequalities yields: $\frac{1}{a^{2}} + \fra... | 4 | Inequalities | proof | Yes | Yes | cn_k12 | false | 549,212 |
In the Cartesian coordinate system, point A has coordinates (1, 1, 2) and point B has coordinates (2, 3, 4). Find the length of segment AB, denoted as |AB|. | Given that point A has coordinates (1, 1, 2) and point B has coordinates (2, 3, 4) in the Cartesian coordinate system, we can find the distance |AB| by applying the distance formula for two points in space. Thus:
\[
|AB| = \sqrt{(2-1)^2 + (3-1)^2 + (4-2)^2}.
\]
Calculate the individual squared differences:
\[
(2-1)^... | |AB| = 3 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,213 |
Solve the equations:$(1)x^{2}+3x-1=0$;$(2)\left(x+2\right)^{2}=\left(x+2\right)$. | To solve the given equations, we proceed as follows:
**For equation (1): $x^2 + 3x - 1 = 0$**
Given $a=1$, $b=3$, and $c=-1$, we calculate the discriminant ($\Delta$) as follows:
\[
\Delta = b^2 - 4ac = 3^2 - 4(1)(-1) = 9 + 4 = 13
\]
Since $\Delta = 13 > 0$, there are two distinct real roots for the equation. These r... | -1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,214 |
Solve the problem: The weight of Mother Hen is 2.3 kilograms, and the weight of Baby Chick is 400 grams. How much heavier is Mother Hen than Baby Chick in kilograms? | **Answer:** Upon careful observation of the conditions given in the problem, it is evident that the units are not uniform. Therefore, it is necessary to first unify the units. Since the question asks for the answer in kilograms, we should convert grams into kilograms. 400 grams is equal to 0.4 kilograms. Subtracting 0.... | 1.9 \text{ kilograms} | Other | math-word-problem | Yes | Yes | cn_k12 | false | 549,215 |
Prove the proposition using the method of contradiction: "If a quadratic equation with real coefficients $ax^2+bx+c=0$ ($a \neq 0$) has real roots, then $b^2-4ac \geq 0$." Which of the following assumptions is correct?
A: Assume $b^2-4ac \leq 0$
B: Assume $b^2-4ac 0$ | To prove a mathematical proposition using the method of contradiction, we should first negate the conclusion to be proved, obtaining the opposite of the conclusion.
The negation of the proposition: "If a quadratic equation with real coefficients $ax^2+bx+c=0$ ($a \neq 0$) has real roots, then $b^2-4ac \geq 0$" is "$b... | \text{B} | Algebra | proof | Yes | Yes | cn_k12 | false | 549,216 |
Given the proposition: "$\exists x\in R$, $x^{2}+ax-4a=0$" is a false proposition, then the range of real number $a$ is ()
A: $\{a\left|\right.-16\leqslant a\leqslant 0\}$
B: $\{a\left|\right.-16 \lt a \lt 0\}$
C: $\{a\left|\right.-4\leqslant a\leqslant 0\}$
D: $\{a\left|\right.-4 \lt a \lt 0\}$ | To solve the problem, we need to analyze the given proposition and its implications on the value of $a$. The proposition states that there does not exist a real number $x$ such that $x^{2}+ax-4a=0$ is true. This means the quadratic equation $x^{2}+ax-4a=0$ has no real roots.
Step 1: Recall the condition for a quadrat... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,217 |
A maintenance team departs from point $A$ to inspect a road in an east-west direction. If traveling east is considered positive and traveling west is considered negative, the team's travel records for the day are as follows: (unit: $km$)
| First Time | Second Time | Third Time | Fourth Time | Fifth Time | Sixth Time |... | To solve the problem step by step, we follow the given information and calculations closely:
### Part 1: Distance from Point $A$
We need to calculate the final position of the maintenance team relative to point $A$ after their day's travel. The travel records are given as movements in kilometers, with eastward moveme... | 12.3 \text{ liters} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,218 |
Given that the set $M$ is the domain of the function $y= \frac {1}{ \sqrt {1-2x}}$, and the set $N$ is the range of the function $y=x^{2}-4$, then $M\cap N=$ ()
A: $\{x|x\leqslant \frac {1}{2}\}$
B: $\{x|-4\leqslant x < \frac {1}{2}\}$
C: $\{(x,y)|x < \frac {1}{2}\}$ and $y\geqslant -4$
D: $\varnothing $ | Solve $1-2x > 0$ to get $x < \frac {1}{2}$;
Therefore, $M=\{x|x < \frac {1}{2}\}$;
$y=x^{2}-4\geqslant -4$;
Therefore, $N=\{y|y\geqslant -4\}$;
Therefore, $M\cap N=\{x|-4\leqslant x < \frac {1}{2}\}$.
Hence, the correct option is $\boxed{B}$.
To find the set $M$, we need to determine the domain of the funct... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,219 |
Given the set $A=\{x|ax^{2}+2x+a=0\}$, if the set $A$ has exactly two subsets, then the value of $a$ is ______. | Let's analyze the given problem step by step:
1. The set $A$ has exactly two subsets. This implies that $A$ must have exactly one element. If $A$ had more than one element, it would have more than two subsets.
2. Considering the equation $ax^2 + 2x + a = 0$, it must have only one root for $A$ to have exactly one elem... | 0 \text{ or } 1 \text{ or } -1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,220 |
Given the function f(x) = lnx - (x+1)/(x-1),
1. Discuss the monotonicity of f(x) and prove that f(x) has exactly two zero points.
2. Let x0 be a zero point of f(x). Prove that the tangent line of the curve y=lnx at point A(x0, lnx0) is also the tangent line of the curve y=e^x. | 1. The function f(x) = lnx - (x+1)/(x-1) is defined on the domain (0, 1) ∪ (1, +∞).
The derivative f'(x) = 1/x + 2/(x-1)² > 0 for x > 0 and x ≠ 1.
Therefore, f(x) is strictly increasing on both intervals (0, 1) and (1, +∞).
a. On the interval (0, 1), the function takes the values 1/e² and 1/e. By the defin... | \text{The function f(x) has exactly two zero points, and the tangent line of y = lnx at a zero point is also the tangent line of y = e^x.} | Calculus | proof | Yes | Yes | cn_k12 | false | 549,221 |
In triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. It is known that $b\sin(C+\frac{π}{3})-c\sin B=0$.
$(1)$ Find the value of angle $C$.
$(2)$ If the area of $\triangle ABC$ is $10\sqrt{3}$ and $D$ is the midpoint of $AC$, find the minimum value of $BD$. | ### Solution:
#### Part (1): Finding the value of angle $C$
Given that $b\sin(C+\frac{π}{3})-c\sin B=0$, we can expand and simplify using trigonometric identities:
1. Start with the given equation: $b\sin(C+\frac{π}{3})-c\sin B=0$.
2. Apply the sine addition formula: $b\left(\sin C\cos\frac{π}{3} + \cos C\sin\frac{π... | 2\sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,222 |
A and B each take one shot. If the probability of both hitting the target is 0.6, find:
(1) The probability that both hit the target.
(2) The probability that exactly one of them hits the target.
(3) The probability that at least one of them hits the target. | (1) Since A and B each take one shot and the probability of hitting the target is 0.6 for both,
the events of hitting the target are independent of each other.
Therefore, the probability that both hit the target is $0.6 \times 0.6 = 0.36$.
So, the answer is $\boxed{0.36}$.
(2) Exactly one person hitting the targ... | 0.84 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,223 |
If the variance of a set of data \\({x_1},{x_2},\cdots,{x_n}\\) is \\(1\\), then the variance of \\(2{x_1}+4,2{x_2}+4,\cdots,2{x_n}+4\\) is \\((\ )\ )
A: \\(1\\)
B: \\(2\\)
C: \\(4\\)
D: \\(8\\) | **Analysis**
This question examines the calculation and application of variance. The key to solving the problem is to find the relationship between the variance of the new data and the original data, which is a basic question.
According to the data \\(2x_1+4\\), \\(2x_2+4\\), \\(\ldots\\), \\(2x_n+4\\), the variance ... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,224 |
The graph of the function $y=f(x)$ is symmetric to the graph of the function $y=g(x)$ with respect to the line $x+y=0$. The inverse function of $y=f(x)$ is ( )
A: $y=g(x)$
B: $y=g(-x)$
C: $y=-g(x)$
D: $y=-g(-x)$ | Let $P(x, y)$ be any point on the graph of the inverse function of $y=f(x)$,
then the point $P$ is symmetric to point $P'(y, x)$ with respect to the line $y=x$, and the point $P'(y, x)$ lies on the graph of $y=f(x)$.
Since the graph of the function $y=f(x)$ is symmetric to the graph of the function $y=g(x)$ with re... | \text{D} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,225 |
Given the function $$f(x)= \frac {2^{x}-1}{2^{x}+1}+x+\sin x$$, if the positive real numbers $a$ and $b$ satisfy $f(4a)+f(b-9)=0$, then the minimum value of $$\frac {1}{a}+ \frac {1}{b}$$ is \_\_\_\_\_\_. | According to the problem, for the function $$f(x)= \frac {2^{x}-1}{2^{x}+1}+x+\sin x$$,
we have $$f(-x)= \frac {2^{-x}-1}{2^{-x}+1}+(-x)+\sin(-x)$$ = $$-\frac {2^{x}-1}{2^{x}+1}-x-\sin x$$ = $$-f(x)$$,
thus, the function $f(x)$ is an odd function.
The derivative of $y=x+\sin x$ is $y'=1+\cos x \geq 0$, indicating... | 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,226 |
If the isometric drawing of a horizontally placed figure is an isosceles trapezoid with a bottom angle of 60°, and both the legs and the top base are 1, then the area of the original plane figure is \_\_\_\_\_\_. | To solve this, consider the isometric drawing of a horizontally placed figure. The top base is 1, the height is $\frac{\sqrt{3}}{2}$, and the bottom base is 2.
The area, $S$, can be calculated as $S= \frac{1}{2} \times (1+2) \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}$.
Therefore, the area of the original plane... | \frac{3\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,227 |
Given that line $a$ is parallel to plane $\alpha$, and line $b$ is contained within plane $\alpha$, the positional relationship between line $a$ and line $b$ is ( )
A: Parallel
B: Intersecting or skew
C: Skew
D: Parallel or skew | Since line $a$ is parallel to plane $\alpha$, and line $b$ is within plane $\alpha$, it follows that $a$ is parallel to $b$, or $a$ and $b$ are skew.
Therefore, the answer is: parallel or skew.
**Analysis:** From line $a$ being parallel to plane $\alpha$, and line $b$ being within plane $\alpha$, we know $a$ is para... | \text{D: Parallel or skew} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,228 |
The maximum value of the function $y=2x^3-3x^2-12x+5$ on the interval $[0,3]$ is _____. | To find the maximum value of the function on the given interval, we first need to determine the critical points by finding the derivative of the function and setting it equal to zero.
The derivative of the function is:
$$y' = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 5) = 6x^2 - 6x - 12.$$
To find the critical points, we set ... | 5 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,229 |
Master Wang went to the fruit wholesale market to purchase apples. He was interested in the apples from two stores, $A$ and $B$, which have the same quality of apples. The retail price of the apples in both stores is $6$ yuan per kilogram, but the wholesale prices are different. Store $A$ has the following rules:
- For... | ### Solution:
#### Part 1: Comparing Costs for 800 Kilograms
**Store A:**
The cost calculation for 800 kilograms of apples at Store A is as follows:
\[
\text{Cost at Store A} = 800 \times 6 \times 92\% = 800 \times 6 \times 0.92 = 4416 \text{ yuan}
\]
**Store B:**
The cost calculation for 800 kilograms of apples a... | 4.5x + 1200 \text{ yuan} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,230 |
Given $f(x) = x^2 - ax + \ln x$, where $a \in \mathbb{R}$.
1. When $a=3$, find the minimum value of the function $f(x)$.
2. Let $g(x) = x^2 - f(x)$, determine if there exists a real number $a$ such that when $x \in [1, e]$ (where $e$ is the base of the natural logarithm), the function $g(x)$ attains its minimum value o... | 1. Given the problem, we have $f(x) = x^2 - 3x + \ln x$, so $$f'(x) = 2x - 3 + \frac{1}{x}$$.
Setting $f'(x) = 0$, we get $$x = \frac{1}{2}$$ or $x=1$.
Solving $f'(x) > 0$, we find: $0 1$,
and solving $f'(x) < 0$, we find: $\frac{1}{2} < x < 1$.
Thus, $f(x)$ is increasing on $(0, \frac{1}{2})$ and $(1, +\infty)$, and ... | 1 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,231 |
A cube has one face labeled with $1$, two faces labeled with $2$, and three faces labeled with $3$. Find the following probabilities after rolling this cube:$(1)$ The probability of getting a $2$ facing up;$(2)$ The number with the highest probability of facing up;$(3)$ If it is determined that player A wins when the n... | ### Step-by-Step Solution
**(1) Probability of Getting a $2$ Facing Up**
Given that a cube has $6$ faces in total and two of these faces are labeled with $2$, the probability of rolling a $2$ can be calculated as follows:
- Total number of outcomes (faces) = $6$
- Favorable outcomes (faces labeled with $2$) = $2$
T... | \text{equal} | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false | 549,232 |
To promote traditional culture, a certain junior high school organized a traditional culture recitation competition for the second grade. Xiaoming made a table based on the scores given by nine judges to a certain contestant in the competition. If the highest and lowest scores are removed, the data in the table will de... | To solve this problem, let's analyze the impact of removing the highest and lowest scores on each of the given statistical measures: Median, Mode, Mean, and Variance.
1. **Median**: The median is the middle value of a data set when it is ordered from least to greatest. Removing the highest and lowest scores from the d... | \text{A: Median} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,233 |
Given that $\overrightarrow{a}=(1,2)$, $\overrightarrow{b}=(-3,4)$, and $\overrightarrow{c}= \overrightarrow{a}+λ \overrightarrow{b}(λ∈R)$.
(1) If $\overrightarrow{b}⊥ \overrightarrow{c}$, find the value of $|\overrightarrow{c}|$;
(2) What value of $λ$ makes the angle between $\overrightarrow{c}$ and $\overrightarrow{a... | (1) Since $\overrightarrow{a}=(1,2)$ and $\overrightarrow{b}=(-3,4)$,
we have $\overrightarrow{c}= \overrightarrow{a}+λ \overrightarrow{b}=(1-3λ,2+4λ)$.
Given that $\overrightarrow{b}⊥ \overrightarrow{c}$,
we have $\overrightarrow{b}⋅ \overrightarrow{c}=-3(1-3λ)+4(2+4λ)=5+25λ=0$,
which gives $λ=- \dfrac {1}{5}$.
Thus, ... | \overrightarrow{c}//\overrightarrow{a} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,234 |
Transform the sine curve $y = \sin x$ through the scaling transformation $\begin{cases} x'= \frac {1}{2}x \\ y'=3y \end{cases}$ to obtain the equation of the curve. The period of the resulting curve is ( )
A: $\frac {\pi}{2}$
B: $\pi$
C: $2\pi$
D: $3\pi$ | Given $\begin{cases} x'= \frac {1}{2}x \\ y'=3y \end{cases}$,
We have $\begin{cases} x=2x' \\ y= \frac {1}{3}y' \end{cases}$,
Thus, $\frac {1}{3}y' = \sin 2x'$, i.e., $y' = 3\sin 2x'$,
Therefore, the period of the transformed curve is $\frac {2\pi}{2} = \pi$.
Hence, the correct option is $\boxed{B}$.
This solution... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,235 |
If function $f(x)$ is increasing on some interval $I$ within its domain $D$, and $y=\frac{f(x)}{x}$ is decreasing on $I$, then $y=f(x)$ is called a "weakly increasing function" on $I$.
(1) Please judge whether $f(x)=x+4$ and $g(x)=x^2+4x+2$ are "weakly increasing functions" on the interval $x \in (1, 2)$ and briefly ex... | (1) Since $f(x)=x+4$ is increasing on (1, 2), and $$F(x)=\frac{f(x)}{x}=\frac{x+4}{x}=1+ \frac{4}{x}$$ is decreasing on (1, 2) (since $\frac{4}{x}$ is decreasing), $f(x)=x+4$ is a "weakly increasing function" on (1, 2);
$g(x)=x^2+4x+2$ is increasing on (1, 2), but $$\frac{g(x)}{x}=\frac{x^2+4x+2}{x}=x+4+\frac{2}{x}$$ i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,236 |
(1) Calculate: $16\;^{ \frac {1}{2}}+(\frac{1}{81})^{-0.25}-(-\frac{1}{2})^{0}$
(2) Simplify: $((2a\;^{ \frac {1}{4}}b\;^{- \frac {1}{3}})(-3a\;^{- \frac {1}{2}}b\;^{ \frac {2}{3}})\div(-\frac{1}{4}a\;^{- \frac {1}{4}}b\;^{- \frac {2}{3}}))$ | (1)
First, we simplify each term separately:
$16^{ \frac {1}{2}} = \boxed{4}$, because 16 is the square of 4.
$(\frac{1}{81})^{-0.25} = (\frac{1}{81})^{ \frac {1}{4}} = \boxed{\frac{1}{3}}$, because 81 raised to the power of $\frac{1}{4}$ equals 3, and the reciprocal of 3 is $\frac{1}{3}$.
$(-\frac{1}{2})^{0} = \box... | 24b | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,237 |
Given that \\(\{i,j,k\}\\) is a unit orthogonal basis in space, and \\(a=-2i+2j-2k\\), \\(b=i+4j-6k\\), \\(c=xi-8j+8k\\), if vectors \\(a\\), \\(b\\), and \\(c\\) are coplanar, then the coordinates of vector \\(c\\) are \_\_\_\_\_\_\_\_. | **Analysis**
This problem examines the theorem of coplanar vectors in space and their coordinate representation. Since \\(\vec{a}, \vec{b}, \vec{c}\\) are coplanar, we can set \\(\vec{c}=\lambda \vec{a}+\mu \vec{b}\\), and then solve it using the operations of addition and subtraction of spatial vectors.
**Solution**... | (8,-8,8) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,238 |
Consider a hyperbola given by the equation $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\,(a > 0, b > 0)$ with two asymptotes $l\_1$ and $l\_2$, and its right focus $F$. If the point $M$, which is the symmetric point of $F$ with respect to the line $l\_1$, lies on $l\_2$, then the eccentricity of the hyperbola is $(\ ... | Let $l\_1$ and $l\_2$ be the two asymptotes of the hyperbola given by the equation $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\,(a > 0, b > 0)$. Without loss of generality, assume $l\_1$ is given by $y = \frac{b}{a}x$ and $l\_2$ is given by $y = -\frac{b}{a}x$.
Given that the symmetric point $M$ of the right focus ... | B | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,239 |
Which of the following functions is both an odd function and monotonically increasing on the interval $\left(0,1\right)$?
A: $y=\lg x$
B: $y=\frac{3}{x}$
C: $y=2|x|$
D: $y=\sin x$ | To determine which of the given functions is both an odd function and monotonically increasing on the interval $\left(0,1\right)$, we analyze each option step by step:
- **Option A: $y=\lg x$**
- **Odd/Even Function Check:** A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain. Since $\lg(-x)$ is u... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,240 |
Given that $\left(m-2\right)x^{|m|}+x=1$ is a quadratic equation in $x$, the possible values of $m$ are ______. | Given the equation $\left(m-2\right)x^{|m|}+x=1$ is a quadratic equation in $x$, we know that the exponent of $x$ must be 2 for it to be quadratic. Therefore, we have:
1. The condition for the equation to be quadratic is $|m| = 2$.
2. However, we also need to ensure that the coefficient of $x^2$ is not zero, which mea... | -2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,241 |
Given the complex numbers $z_{1}$ and $z_{2}$ are symmetrical with respect to the imaginary axis in the complex plane. If $z_{1}=1-2i$, where $i$ is the imaginary unit, then the imaginary part of $\frac{z_{2}}{z_{1}}$ is ( ).
A: $- \frac {4}{5}$
B: $\frac {4}{5}$
C: $- \frac {3}{5}$
D: $\frac {3}{5}$ | Since $z_{1}=1-2i$,
According to the problem statement, since $z_{2}$ is symmetrical with $z_{1}$ with respect to the imaginary axis, $z_{2}$ is the reflection of $z_{1}$ in the imaginary axis. This means that $z_{2}$ has the same imaginary part as $z_{1}$, but the real part will have the opposite sign. Therefore, $z_... | -\frac {4}{5} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,242 |
"The statement 'All multiples of 9 are also multiples of 3, and a certain odd number is a multiple of 9, therefore that odd number is a multiple of 3.' The reasoning above is ( )"
A: The minor premise is wrong
B: The conclusion is wrong
C: Correct
D: The major premise is wrong | Let's examine the given statements and the reasoning to arrive at the correct answer.
- Major premise: All multiples of 9 are also multiples of 3.
- Minor premise: A certain odd number is a multiple of 9.
- Conclusion: Therefore, that odd number is a multiple of 3.
Firstly, let's address the major premise. Every mult... | \text{C: Correct} | Number Theory | MCQ | Yes | Yes | cn_k12 | false | 549,243 |
A rectangular sheet of iron with a length of 90cm and a width of 48cm is used to make an open-top rectangular container. By cutting out a small square at each of the four corners, then folding up the sides at a 90-degree angle and welding them together, the volume of the resulting rectangular container will be maximize... | When the side length of the cut-out squares is $\boxed{10\text{cm}}$, the volume of the rectangular container is maximized. | 10\text{cm} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,244 |
Given that $\overrightarrow{a}, \overrightarrow{b}$ are plane vectors, and $\overrightarrow{a}=(1,2)$.
$(1)$ If $\overrightarrow{b}=(1,1)$, and $k\overrightarrow{a}-\overrightarrow{b}$ is perpendicular to $\overrightarrow{a}$, find the value of the real number $k$;
$(2)$ If $\overrightarrow{a} \parallel \overrighta... | ### Step-by-Step Solution
#### Part (1)
Given $\overrightarrow{a}=(1,2)$ and $\overrightarrow{b}=(1,1)$, we can express $k\overrightarrow{a}-\overrightarrow{b}$ as follows:
\[
k\overrightarrow{a}-\overrightarrow{b} = (k \cdot 1 - 1, k \cdot 2 - 1) = (k-1, 2k-1)
\]
Since $k\overrightarrow{a}-\overrightarrow{b}$ is pe... | \overrightarrow{b}=(-2,-4) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,245 |
A sector of a circle has a central angle of ${{120}^{\circ }}$ and an area of ${\pi}$. Find the length of the arc of this sector. | **Analysis**: This problem requires the use of the formulas for the area of a sector and the length of an arc. First, we need to find the radius of the sector using the area formula, and then substitute it into the arc length formula to solve for the length of the arc.
**Step 1**: Let $R$ be the radius of the sector. ... | \frac{2\sqrt{3}\pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,246 |
Let the set $A=\{1,2,3\}$, $B=\{2,3,4\}$, and $M=\{x|x=ab,a\in A,b\in B\}$. Then, the number of elements in $M$ is ( ).
A: $5$
B: $6$
C: $7$
D: $8$ | **Analysis**
This question examines the relationship between elements and sets, which is a basic topic.
**Answer**
Given $A=\{1,2,3\}$ and $B=\{2,3,4\}$, $M=\{x|x=ab,a\in A,b\in B\}$,
Thus, $M=\{2,3,4,6,8,9,12\}$.
Therefore, there are $7$ elements in $M$.
Hence, the correct choice is $\boxed{C}$. | C | Number Theory | MCQ | Yes | Yes | cn_k12 | false | 549,247 |
In the first quadrant, the area enclosed by the lines $y=2x$, $y=\frac{1}{2}x$, and the curve $y=\frac{1}{x}$ is $\_\_\_\_\_\_\_\_\_.$ | This problem tests the rules of differentiation, the application of the fundamental theorem of calculus, and the integration of geometric and algebraic concepts. It is of moderate difficulty.
First, determine the coordinates of the intersection points. Then, represent the area of the curvilinear trapezoid using defini... | \ln{2} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,248 |
In triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. If $\sqrt{3}\sin B + 2\cos^2\frac{B}{2} = 3$ and $\frac{\cos B}{b} + \frac{\cos C}{c} = \frac{\sin A \sin B}{6\sin C}$, then the area of the circumcircle of $\triangle ABC$ is ( ).
A: $12\pi$
B: $16\pi$
C:... | To solve this problem, we start by analyzing the given equation $\sqrt{3}\sin B + 2\cos^2\frac{B}{2} = 3$. We can rewrite the cosine term using the double angle formula for cosine, which gives us:
\[
\sqrt{3}\sin B + 2\left(\frac{1+\cos B}{2}\right) = 3
\]
Simplifying this, we obtain:
\[
\sqrt{3}\sin B + \cos B = 2
\]
... | \text{B: } 16\pi | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,249 |
Given a function $f(x)$ defined on $\mathbb{R}$ that is both odd and periodic with a period of $3$, and for $x \in \left(0, \frac{3}{2}\right)$, $f(x) = \sin(\pi x)$, $f\left(\frac{3}{2}\right) = 0$, find the number of zeros of the function $f(x)$ in the interval $[0,6]$.
A: 9
B: 7
C: 5
D: 3 | Since the function $f(x)$ is given to be an odd function with the period of $3$ defined on $\mathbb{R}$, and for $x \in \left(0, \frac{3}{2}\right)$, we have $f(x) = \sin(\pi x)$, then to find the roots of $f(x) = 0$ within this interval, we solve the equation $$ \sin(\pi x) = 0 $$ to get $$ x = k, \quad k \in \mathbb{... | 9 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,250 |
Which of the following derivative operations is correct?
A: $(\frac{cosx}{x})'=-sinx$
B: $(log_2x)'=\frac{1}{xln2}$
C: $(2^{x})'=2^{x}$
D: $(x^{3}e^{x})'=3x^{2}e^{x}$ | To determine which derivative operation is correct, let's evaluate each option step by step:
**Option A:**
Given function: $f(x) = \frac{\cos x}{x}$
Using the quotient rule for derivatives, $[f(x)/g(x)]' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$, we find:
$f'(x) = \left(\frac{\cos x}{x}\right)' = \frac{-\sin x \cdo... | \text{B} | Calculus | MCQ | Yes | Yes | cn_k12 | false | 549,251 |
Given the sequences ${a_n}$ and ${b_n}$ with their first $n$ terms sums $S_n$ and $T_n$ respectively, such that $a_n > 0$, $6S_n = a_n^2 + 3a_n - 4$ ($n \in \mathbb{N}^*$), and $b_n = \frac{1}{(a_n - 1)(a_{n+1} - 1)}$. If for any $n \in \mathbb{N}^*$, $k > T_n$ always holds true, find the minimum value of $k$ ( ).
A: ... | Since $a_n > 0$ and $6S_n = a_n^2 + 3a_n - 4$ ($n \in \mathbb{N}^*$),
We can deduce that $6a_1 = 6S_1 = a_1^2 + 3a_1 - 4$, which gives $a_1 = 4$.
For $n \geq 2$, we have $6S_{n-1} = a_{n-1}^2 + 3a_{n-1} - 4$ and $6S_n = a_n^2 + 3a_n - 4$.
Subtracting these two equations yields $6a_n = a_n^2 + 3a_n - 4 - a_{n-1}^2 - ... | \frac{1}{9} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,252 |
Given a point $P$ on the hyperbola $y^{2}-4x^{2}=16$ is at a distance of $2$ from one of its foci, then the distance from point $P$ to the other focus is ______. | **Analysis**
This question examines the equation and definition of a hyperbola, the method of thinking through classification discussion, and computational ability. It is a basic and commonly mistaken question. By finding the values of $a$ and $c$ for the hyperbola, and setting $|PF_{1}|=2$, we can use the definition ... | 10 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,253 |
Given the inequality $|x+3|+|x+m| \geq 2m$ with respect to $x$ and its solution set is $\mathbb{R}$.
1. Find the range of values for the real number $m$.
2. Given $a > 0$, $b > 0$, $c > 0$ and $a + b + c = m$, find the minimum value of $a^2 + 2b^2 + 3c^2$ and the corresponding values of $a$, $b$, and $c$ when $m$ is a... | 1. Since $|x+3|+|x+m| \geq |(x+3) - (x+m)| = |m-3|$,
The equality holds when $-3 \leq x \leq -m$ or $-m \leq x \leq -3$. We have $|m-3| \geq 2m$,
Solving this inequality, we get $m-3 \geq 2m$ or $m-3 \leq -2m$, which gives $m \leq -3$ or $m \leq 1$.
Therefore, the range of $m$ is $m \leq 1$.
2. From part 1... | \frac{6}{11} | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false | 549,255 |
Determine the value of $\log_{2}\left( \cos \frac{7\pi }{4} \right)$.
A: $-1$
B: $-\frac{1}{2}$
C: $\frac{1}{2}$
D: $\frac{\sqrt{2}}{2}$ | 1. **Analyze the problem**: This question involves the use of trigonometric identities and logarithm operations. By applying these concepts, we can find the solution.
2. **Solve the problem**:
First, recall that $\cos(\theta) = \cos(-\theta)$. Using this identity, we can rewrite the given expression as:
$$\log_{... | \frac{1}{2} \cdot 1 - 1 = -\frac{1}{2} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,256 |
Given that the inequality $x^2 + x < \dfrac{a}{b} + \dfrac{b}{a}$ holds for any positive real numbers $a$ and $b$, determine the range of the real number $x$.
A: $(-2,0)$
B: $(-\infty,-2)\cup(1,+\infty)$
C: $(-2,1)$
D: $(-\infty,-4)\cup(2,+\infty)$ | We have the inequality $x^2 + x 1$, both factors are positive, leading to a positive product, which again does not satisfy the inequality.
Therefore, the range of $x$ for which the inequality $x^2 + x < \dfrac{a}{b} + \dfrac{b}{a}$ holds true is $\boxed{(-2, 1)}$. | (-2, 1) | Inequalities | MCQ | Yes | Yes | cn_k12 | false | 549,257 |
The vertex coordinates of the parabola $y=x^2+2x-3$ are. | Since the parabola $y=x^2+2x-3$ can be rewritten as: $y=(x+1)^2-4$,
Therefore, the coordinates of its vertex are $(-1, -4)$.
Hence, the answer is $\boxed{(-1, -4)}$. | (-1, -4) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,258 |
Given that $S_n$ is the sum of the first $n$ terms of a geometric sequence $\{a_n\}$, and $S_n = 2a_n - 1$, find the common ratio $q$ of the sequence $\{a_n\}$. | Since $S_n$ represents the sum of the first $n$ terms of a geometric sequence $\{a_n\}$, we can use the relationship $S_n = 2a_n - 1$ to obtain the values of $a_1$ and $a_2$, which will then allow us to find the common ratio of the sequence.
For $n=1$, we have:
$S_1 = a_1$,
so substituting into the given equation, we ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,259 |
If the solution set of the inequality $x^{2} - 2x + 3 \leq a^{2} - 2a - 1$ on $\mathbb{R}$ is empty, then the range of the real number $a$ is ______. | To find the range of $a$ for which the solution set of the given inequality is empty, we rearrange the inequality:
$$x^{2} - 2x + 3 - a^{2} + 2a + 1 \leq 0$$
This can be rewritten as the quadratic inequality:
$$(x - 1)^{2} + 2 - a^{2} + 2a + 1 \leq 0$$
Since $(x - 1)^{2} \geq 0$ for all real numbers $x$, the inequa... | \{a \mid -1 < a < 3\} | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false | 549,260 |
The absolute value of $-11$ is ( ).
A: $1$
B: $11$
C: $-\frac{1}{11}$
D: $\frac{1}{11}$ | To solve the problem, we start by understanding what the absolute value function does. The absolute value of a number is the distance of that number from $0$ on the number line, without considering direction. Therefore, the absolute value of any number is always non-negative.
Given the number $-11$, we apply the absol... | B | Number Theory | MCQ | Yes | Yes | cn_k12 | false | 549,261 |
In the arithmetic sequence ${a_n}$, ${a_1=8,a_3=4}$.
(1) Let the sum of the first $n$ terms of the sequence ${a_n}$ be ${S_n}$. Find the maximum value of ${S_n}$ and the value of $n$ that makes ${S_n}$ maximum.
(2) Let ${b_n=\frac{1}{n(12-{a_n})}\ (n\in\mathbb{N}^*)}$. Find ${T_n=b_1+b_2+...+b_n\ (n\in\mathbb{N}^*)}$... | (1) The arithmetic sequence ${a_n}$ has a common difference of ${d=\frac{{a_3-a_1}}{3-1}=-2}$.
Thus, ${a_n=10-2n}$.
Hence, ${S_n=a_1+a_2+...+a_n=\frac{n(a_1+a_n)}{2}=\frac{n(8+10-2n)}{2}=-n^2+9n=-\left(n-\frac{9}{2}\right)^2+\frac{81}{4}}$,
Therefore, when ${n}$ takes the value of ${4}$ or ${5}$, ${S_n}$ is maximum,... | \frac{n}{2(n+1)} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,262 |
The number of roots of the function $f(x) = x^2 - 2x + 3$ is ( )
A: 0
B: 1
C: 2
D: 3 | Since $f(x) = x^2 - 2x + 3$, the discriminant $\Delta = b^2 - 4ac = (-2)^2 - 4 \times 3 < 0$,
Therefore, the corresponding equation $x^2 - 2x + 3 = 0$ has no real roots. Hence, the number of roots of the function $f(x) = x^2 - 2x + 3$ is 0.
Thus, the correct choice is $\boxed{A}$. | A | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,263 |
Given the function $f(x)=(1-k)x+ \frac{1}{e^{x}}$.
(I) Find the monotonic intervals of the function $f(x)$;
(II) When $k=0$, there exists a tangent line to the function curve $f(x)$ passing through point $A(0,t)$. Find the range of $t$ values. | (I) The domain of the function is $\mathbb{R}$,
so $f'(x)= \frac{(1-k)e^{x}-1}{e^{x}}$,
$\quad\quad$ (1) When $k\geqslant 1$, $f'(x) 0$, $f(x)$ is increasing on $(-\ln (1-k),+\infty)$.
(II) Let the coordinates of the tangent point be $(x_{0},y_{0})$,
then the tangent line equation is $y-y_{0}=f'(x_{0})(x-x_{0})$
i.e.... | 1 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,264 |
A plane that is 1 unit away from the center of a sphere cuts the sphere, resulting in a cross-sectional area of $\pi$. The volume of the sphere is \_\_\_\_\_\_. | Solution: Cutting the sphere with a plane results in a cross-sectional area of $\pi$, which means the radius of the small circle is 1.
Given the distance from the center of the sphere to this cross-section is 1, the radius of the sphere is $r= \sqrt{1+1} = \sqrt{2}$.
Therefore, the volume of the sphere is: $\frac{4... | \frac{8\sqrt{2}}{3}\pi | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,265 |
In an acute triangle \\(ABC\\), prove that: \\(\sin A + \sin B + \sin C > \cos A + \cos B + \cos C\\). | Proof: Since \\(\triangle ABC\\) is an acute triangle,
it follows that \\(A + B > \frac{\pi}{2}\\),
which implies \\(A > \frac{\pi}{2} - B\\),
Since \\(y = \sin x\\) is an increasing function on \\((0, \frac{\pi}{2})\\),
it follows that \\(\sin A > \sin ( \frac{\pi}{2} - B) = \cos B\\),
Similarly, we can obtain \\... | \sin A + \sin B + \sin C > \cos A + \cos B + \cos C | Inequalities | proof | Yes | Yes | cn_k12 | false | 549,266 |
Given $x+y=1$, where $x$ and $y$ are positive numbers, find the minimum value of $\frac{1}{x}+\frac{4}{y}$. | To find the minimum value of $\frac{1}{x}+\frac{4}{y}$ given $x+y=1$ where $x, y > 0$, we begin by expressing the given condition and applying algebraic manipulations.
Starting with the given condition $x+y=1$, we express $\frac{1}{x}+\frac{4}{y}$ in terms of $x$ and $y$:
\begin{align*}
\frac{1}{x}+\frac{4}{y} &= \fra... | 9 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,267 |
If an integer can be expressed in the form of $a^{2}+b^{2}$ (where $a$ and $b$ are positive integers), then this number is called a "Fengli number." For example, $2$ is a "Fengli number" because $2=1^{2}+1^{2}$, and for example, $M=x^{2}+2xy+2y^{2}=\left(x+y\right)^{2}+y^{2}$ (where $x$ and $y$ are positive integers), ... | ### Step-by-Step Solution
#### Part 1: Is $11$ a "Fengli number"?
To determine if $11$ is a "Fengli number," we need to check if it can be expressed in the form of $a^{2}+b^{2}$ or $\left(x+y\right)^{2}+y^{2}$, where $a$, $b$, $x$, and $y$ are positive integers.
- For small values of $a$ and $b$, the possible sums $... | m = \pm 4 | Number Theory | math-word-problem | Yes | Yes | cn_k12 | false | 549,268 |
If set $A=\{x|x^{2}+x-6=0\}$, $B=\{x|x\cdot m+1=0\}$, and $B\subseteq A$, then the set of possible values for $m$ is ____. | To solve for the set $A$, we start with the equation given for the elements of $A$:
$$x^{2}+x-6=0$$
Factoring this quadratic equation, we find:
$$(x+3)(x-2)=0$$
Thus, the solutions for $x$ are:
$$x=-3 \quad \text{or} \quad x=2$$
Therefore, the set $A$ is:
$$A=\{-3,2\}$$
For set $B$, we analyze it based on the g... | \left\{-\frac{1}{2}, 0, \frac{1}{3}\right\} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,269 |
The line $l$ passes through point $P(1,4)$ and intersects the positive half of the $x$-axis and the positive half of the $y$-axis at points $A$ and $B$, respectively, with $O$ being the origin.
$(1)$ When $|OA|+|OB|$ is minimized, find the equation of $l$;
$(2)$ If the area of $\triangle AOB$ is minimized, find the equ... | Solution:
$(1)$ According to the problem, let the coordinates of $A$ be $(a,0)$, and the coordinates of $B$ be $(0,b)$, where $(a, b > 0)$,
then the equation of line $l$ is: $\dfrac{x}{a}+ \dfrac{y}{b}=1$,
Since line $l$ passes through point $P(1,4)$, we have $\dfrac{1}{a}+ \dfrac{4}{b}=1$,
Also, since $|OA|=a$, $|OB|=... | 4x+y-8=0 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,270 |
The straight line passes through the origin and the point $(-1, -1)$. Calculate the inclination angle of the line. | The inclination angle of a line is the angle that the line makes with the positive direction of the x-axis. The slope of the line passing through the origin $(0,0)$ and the point $(-1, -1)$ can be found using the formula:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
Substituting the coordinates of the two points we have:
$... | 45^\circ | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,271 |
In $\triangle ABC$, $a=3$, $b=2\sqrt{6}$, $\angle B = 2\angle A$.
(I) Find the value of $\cos A$;
(II) Find the value of $c$. | (I) Given that $a=3$, $b=2\sqrt{6}$, and $\angle B = 2\angle A$,
In $\triangle ABC$, by the sine law, we have $\frac{3}{\sin A} = \frac{2\sqrt{6}}{\sin 2A}$.
So, $\frac{2\sin A \cos A}{\sin A} = \frac{2\sqrt{6}}{3}$.
Hence, $\cos A = \frac{\sqrt{6}}{3}$.
(II) From (I), we know $\cos A = \frac{\sqrt{6}}{3}$,
So, $\... | 5 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,273 |
A certain shopping mall purchased a batch of daily necessities at a unit price of $20$ yuan. If they are sold at a unit price of $30$ yuan, then 400 items can be sold in half a month. According to sales experience, increasing the selling price will lead to a decrease in sales volume. Specifically, for every $1$ yuan in... | To solve this problem, we first establish the relationship between the selling price, sales volume, and profit. Let's denote the selling price by $x$ yuan and the sales profit by $y$ yuan. From the information given:
1. The base case is when the selling price is $30$ yuan, and 400 items are sold.
2. For every $1$ yua... | 35 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,274 |
If the one-variable quadratic equation $x^{2}-2x+m=0$ has two distinct real roots with respect to $x$, then the range of $m$ is ______. | To determine the range of $m$ for which the quadratic equation $x^{2}-2x+m=0$ has two distinct real roots, we need to analyze the discriminant $\Delta$ of the equation. The discriminant $\Delta$ is given by the formula $\Delta = b^{2} - 4ac$, where $a$, $b$, and $c$ are the coefficients of $x^{2}$, $x$, and the constan... | m < 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,275 |
Given a function $f(x)$ defined on $(0, +\infty)$, its derivative $f′(x)$ satisfies $$x^{2}f′(x) + 2xf(x) = \frac{\ln x}{x},$$ and $f(e) = \frac{1}{4e}$, where $e$ is the base of the natural logarithm. Find the solution set of the inequality $$f(x) + \frac{1}{x} > \frac{5}{4e}.$$ | Starting with the given differential equation $$x^2f'(x) + 2xf(x) = \frac{\ln x}{x},$$ we rewrite it to represent the derivative of a product:
$$\Longrightarrow (x^2f(x))' = \frac{\ln x}{x}.$$
Integrating both sides with respect to $x$ gives us:
$$x^2f(x) = \frac{1}{2}(\ln x)^2 + C,$$
where $C$ is the constant of integ... | (0,e) | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,277 |
Given that $a$ and $b$ are real numbers, and $b^{2}+(4+i)b+4+ai=0$ (where $i$ is the imaginary unit), find the value of $|a+bi|$. | [Analysis]
This problem involves the operations of complex numbers and the calculation of the modulus of a complex number, which is a basic question.
[Answer]
Since $b^{2}+(4+i)b+4+ai=0$,
we can separate the real and imaginary parts to get the system of equations:
$$\begin{cases}b^{2}+4b+4=0 \\ b+a=0\end{cases}$$
Solv... | 2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,278 |
Given a sector with a radius of $10cm$ and a perimeter of $45cm$, find the central angle of the sector in radians. | **Analysis**
From the given information, we can calculate the arc length of the sector. By substituting into the formula $\alpha= \frac{l}{r}$, we can find the solution. This question tests the formula for arc length and is considered a basic problem.
**Solution**
Given that the radius of the sector $r=10$ and the p... | 2.5 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,279 |
In recent years, running has become a lifestyle for more and more people. According to official data, the number of participants in the half marathon in Shanghai reached 78,922 in the year 2023. Express the number 78,922 in scientific notation as ______. | To express the number 78,922 in scientific notation, we follow these steps:
1. Identify the decimal point's original position. In 78,922, it's implicitly after the last digit (2).
2. Move the decimal point to the right of the first non-zero digit. Moving it between 7 and 8 gives us 7.8922.
3. Count the number of place... | 7.8922 \times 10^{4} | Number Theory | math-word-problem | Yes | Yes | cn_k12 | false | 549,280 |
Which of the following statements about a parallelepiped is correct?
A: A parallelepiped has 12 vertices.
B: A cube is not a parallelepiped.
C: A parallelepiped has 12 edges.
D: Each face of a parallelepiped is a rectangle. | To analyze each statement about a parallelepiped systematically:
- **Statement A**: A parallelepiped, by definition, is a three-dimensional figure formed by six parallelograms. By visualizing or constructing a model, we can count the vertices (corners) where the edges meet. We find that there are 4 vertices on the top... | \text{C: A parallelepiped has 12 edges.} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,281 |
A six-digit number is formed by repeating a three-digit number, such as 256256, or 678678, etc. This type of number cannot be divided by ( )
A: 11
B: 101
C: 13
D: 1001 | **Analysis:** A six-digit number formed by repeating a three-digit number means that this type of number can definitely be divided by this three-digit number. Let's divide the six-digit number by the three-digit number to find the number in question.
$256256 \div 256 = 1001$,
$678678 \div 678 = 1001$,
Let the thr... | \text{B} | Number Theory | MCQ | Yes | Yes | cn_k12 | false | 549,282 |
Ma Xiaohu's home is $1800$ meters away from school. One day, Ma Xiaohu went to school from home. After leaving for $10$ minutes, his father found out that he had forgotten his math textbook and immediately took the textbook to catch up with him. He caught up with Ma Xiaohu $200$ meters away from the school. It is known... | To solve this problem, let's follow the steps and calculations closely related to the given solution:
### Step 1: Fill in the Table
Given that Ma Xiaohu's home is $1800$ meters away from the school and he was caught up by his father $200$ meters away from the school, we can deduce that:
- The distance Ma Xiaohu trav... | 80 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,283 |
Define a function $f(x)$ on $\mathbb{R}$ that satisfies: $f'(x) > 1 - f(x)$, $f(0) = 6$, where $f'(x)$ is the derivative of $f(x)$. The solution set for the inequality $e^x f(x) > e^x + 5$ (where $e$ is the base of the natural logarithm) is | Let's define $g(x) = e^x f(x) - e^x$ (for $x \in \mathbb{R}$).
Then, we have $g'(x) = e^x f(x) + e^x f'(x) - e^x = e^x [f(x) + f'(x) - 1]$.
Since $f'(x) > 1 - f(x)$,
it follows that $f(x) + f'(x) - 1 > 0$,
which implies $g'(x) > 0$.
Therefore, $y = g(x)$ is monotonically increasing over its domain.
Given $e^x f(x... | (0, +\infty) | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,284 |
Let $\{a\_n\}$ be a geometric sequence with a common ratio $q > 1$. If $a_{2016}$ and $a_{2017}$ are the roots of the equation $4x^2-8x+3=0$, then $a_{2018}+a_{2019}=$ \_\_\_\_\_\_. | Since $\{a\_n\}$ is a geometric sequence with a common ratio $q > 1$, and $a_{2016}$ and $a_{2017}$ are the roots of the equation $4x^2-8x+3=0$,
We can find the roots of the equation: $a_{2016}=\frac{1}{2}$ and $a_{2017}=\frac{3}{2}$.
Hence, the common ratio $q=\frac{a_{2017}}{a_{2016}}=\frac{\frac{3}{2}}{\frac{1}{2}... | 18 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,285 |
Define the operation $a \otimes b =
\begin{cases}
a & \text{if } a \geq b, \\
b & \text{if } a < b
\end{cases}$. Let the function $f(x) = x \otimes (2-x)$.
(1) Use algebraic methods to prove: The graph of the function $f(x)$ is symmetric about the line $x=1$.
(2) Let $g(x) = m^{2}x + 2 + m$, if $f(e^{x}) \leq g(x... | (1) First, we express $f(x)$ in terms of the absolute value function:
$$f(x)=x\otimes(2-x)=
\begin{cases}
2-x & \text{if } x \geq 1, \\
x & \text{if } x < 1
\end{cases}
= 1 - |1-x|.$$
Let $(x_0, y_0)$ be any point on the curve $y=f(x)$. Then we have
$$f(2-x_0) = 1 - |2 - x_0 - 1| = 1 - |1 - x_0| = 1 - |x_0 - 1| = y_0 =... | [-1, +\infty) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,286 |
Given that the function $f\left(x\right)$ defined on $R$ is monotonically increasing on $\left[0,+\infty \right)$, and the function $f\left(x\right)-1$ is an odd function, then the solution set of $f\left(3x+4\right)+f\left(1-x\right) \lt 2$ is ______. | Given that the function $f(x)$ defined on $\mathbb{R}$ is monotonically increasing on $[0,+\infty)$, and the function $f(x)-1$ is an odd function, we aim to find the solution set of $f(3x+4)+f(1-x) < 2$.
Step 1: Understanding the properties of $f(x)$
- Since $f(x)-1$ is an odd function, we know that $f(-x)-1 = -(f(x)-... | \{x|x < -\frac{5}{2}\} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,287 |
Given the equations of the two altitudes in triangle $ABC$ are $2x-3y+1=0$ and $x+y=0$, with vertex $A$ at $(1,2)$, find:
(1) The equation of the line on which side $BC$ lies;
(2) The area of $\triangle ABC$. | (1) Since point $A(1,2)$ is not on the two altitude lines $2x-3y+1=0$ and $x+y=0$,
the slopes of the lines on which sides $AB$ and $AC$ lie are respectively $-\frac{3}{2}$ and $1$.
Substituting into the point-slope form, we get: $y-2=-\frac{3}{2}(x-1)$, $y-2=x-1$
Therefore, the equations of the lines on which sid... | \frac{45}{2} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,288 |
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