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values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+x+2}{(x+2) x^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+x+2}{(x+2) x^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+x+2}{(x+2) x^{3}}=\frac{A}{x+2}+\frac{B_{1}}{x}+\frac{B_{2}}{x^{2}}+\frac{B_{3}}{x^{3}}= \\
& =\frac{A x^{3}+B_{1}(x+... | \ln|x+2|-\frac{1}{2x^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,207 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{3 x^{3}+9 x^{2}+10 x+2}{(x-1)(x+1)^{3}} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}+9 x^{2}+10 x+2}{(x-1)(x+1)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{3 x^{3}+9 x^{2}+10 x+2}{(x-1)(x+1)^{3}}=\frac{A}{x-1}+\frac{B_{1}}{x+1}+\frac{B_{2}}{(x+1)^{2}}+\frac{B_... | 3\cdot\ln|x-1|-\frac{1}{2(x+1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,208 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}+x+1}{(x+1) x^{3}} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+x+1}{(x+1) x^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+x+1}{(x+1) x^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x}+\frac{B_{2}}{x^{2}}+\frac{B_{3}}{x^{3}}= \\
& =\frac{A x^{3}+B... | 2\cdot\ln|x+1|-\frac{1}{2x^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,209 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}+6 x^{2}+7 x+4}{(x+2)(x+1)^{3}} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+6 x^{2}+7 x+4}{(x+2)(x+1)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+6 x^{2}+7 x+4}{(x+2)(x+1)^{3}}=\frac{A}{x+2}+\frac{B_{1}}{x+1}+\frac{B_{2}}{(x+1)^{2}}+\frac{B_{3... | 2\cdot\ln|x+2|-\frac{1}{2(x+1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,210 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}+6 x^{2}+5 x}{(x+2)(x+1)^{3}} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+6 x^{2}+5 x}{(x+2)(x+1)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+6 x^{2}+5 x}{(x+2)(x+1)^{3}}=\frac{A}{x+2}+\frac{B_{1}}{x+1}+\frac{B_{2}}{(x+1)^{2}}+\frac{B_{3}}{(... | 2\cdot\ln|x+2|+\frac{1}{2(x+1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,211 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}+6 x^{2}+7 x}{(x-2)(x+1)^{3}} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+6 x^{2}+7 x}{(x-2)(x+1)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+6 x^{2}+7 x}{(x-2)(x+1)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+1}+\frac{B_{2}}{(x+1)^{2}}+\frac{B_{3}}{(... | 2\cdot\ln|x-2|-\frac{1}{2(x+1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,212 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}+6 x^{2}+5 x+4}{(x-2)(x+1)^{3}} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+6 x^{2}+5 x+4}{(x-2)(x+1)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+6 x^{2}+5 x+4}{(x-2)(x+1)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+1}+\frac{B_{2}}{(x+1)^{2}}+\frac{B_{3... | 2\cdot\ln|x-2|+\frac{1}{2(x+1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,213 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+4 x+24}{(x-2)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+4 x+24}{(x-2)(x+2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+4 x+24}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}... | \ln|x-2|+\frac{4}{(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,214 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+14 x+4}{(x-2)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+14 x+4}{(x-2)(x+2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+14 x+4}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}... | \ln|x-2|-\frac{1}{(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,215 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+18 x-4}{(x-2)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+18 x-4}{(x-2)(x+2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+18 x-4}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}... | \ln|x-2|-\frac{3}{(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,216 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+10 x+12}{(x-2)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+10 x+12}{(x-2)(x+2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+10 x+12}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3... | \ln|x-2|+\frac{1}{(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,217 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-6 x^{2}+14 x-4}{(x-2)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}-6 x^{2}+14 x-4}{(x-2)(x+2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+14 x-4}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}... | \frac{1}{8}\cdot\ln|x-2|+\frac{7}{8}\cdot\ln|x+2|+\frac{17x+18}{2(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,218 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-6 x^{2}+14 x-4}{(x+2)(x-2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}-6 x^{2}+14 x-4}{(x+2)(x-2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+14 x-4}{(x+2)(x-2)^{3}}=\frac{A}{x+2}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3}}... | \ln|x+2|-\frac{1}{(x-2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,219 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+15 x+2}{(x-2)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+15 x+2}{(x-2)(x+2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+15 x+2}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}... | \ln|x-2|-\frac{3}{2(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,220 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}-6 x^{2}+7 x-4}{(x-2)(x-1)^{3}} d x
$$ | ## Solution
## Method 1
$$
\int \frac{2 x^{3}-6 x^{2}+7 x-4}{(x-2)(x-1)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}-6 x^{2}+7 x-4}{(x-2)(x-1)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x-1}+\frac{B_{2}}{(x-1)^{2... | 2\cdot\ln|x-2|-\frac{1}{2(x-1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,221 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}-6 x^{2}+7 x}{(x+2)(x-1)^{3}} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}-6 x^{2}+7 x}{(x+2)(x-1)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}-6 x^{2}+7 x}{(x+2)(x-1)^{3}}=\frac{A}{x+2}+\frac{B_{1}}{x-1}+\frac{B_{2}}{(x-1)^{2}}+\frac{B_{3}}{(... | 2\cdot\ln|x+2|-\frac{1}{2(x-1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,222 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}-10 x+52}{(x-2)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}-10 x+52}{(x-2)(x+2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}-10 x+52}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3... | \ln|x-2|+\frac{11}{(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,223 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-6 x^{2}+13 x-6}{(x-2)(x+2)^{3}} d x
$$
One of the users on the website believes that this problem is solved incorrectly or has errors (You can help the project by correcting and supplementing it. | ## Solution
$$
\int \frac{x^{3}-6 x^{2}+13 x-6}{(x-2)(x+2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+13 x-6}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}... | \frac{1}{16}\cdot\ln|x-2|+\frac{15}{16}\cdot\ln|x+2|+\frac{33x+34}{4(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,224 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+13 x+6}{(x-2)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+13 x+6}{(x-2)(x+2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+13 x+6}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}... | \ln|x-2|-\frac{1}{2(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,225 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-1 ; 7 ;-4\}$
$p=\{-1 ; 2 ; 1\}$
$q=\{2 ; 0 ; 3\}$
$r=\{1 ; 1 ;-1\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 2p-q+3r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,226 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{-1 ; 4 ; 2\}$
$b=\{3 ;-2 ; 6\}$
$c_{1}=2 a-b$
$c_{2}=3 b-6 a$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
It is not hard to notice that $c_{2}=-3(2 a-b)=-3 c_{1}$ for any $a$ and $b$.
That is,
$c_{1}=-\frac{1}{3} \cdot c_{2}$
, which means the v... | c_{1}=-\frac{1}{3}\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,227 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(3 ; 3 ;-1), B(1 ; 5 ;-2), C(4 ; 1 ; 1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(1-3 ; 5-3 ;-2-(-1))=(-2 ; 2 ;-1)$
$\overrightarrow{A C}=(4-3 ; 1-3 ; 1-(-1))=(1 ;-2 ; 2)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrighta... | -\frac{8}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,228 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=p+4q$
$b=2p-q$
$|p|=7$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{3}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(p+4 q) \times(2 p-q)=2 \cdot p \times p-p \times q+4 \cdot 2 \cdot q \times p-4 ... | 63\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,229 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{3 ; 7 ; 2\}$
$b=\{-2 ; 0 ;-1\}$
$c=\{2 ; 2 ; 1\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
3 & 7 & 2 \\
-2 & 0 & -1 \\
2 & 2 & 1
\end{array}\right|= \\
& =3 \cdot\left|\b... | -2\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,230 |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(14 ; 4 ; 5) \)
\( A_{2}(-5 ;-3 ; 2) \)
\( A_{3}(-2 ;-6 ;-3) \)
\( A_{4}(-2 ; 2 ;-1) \) | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-5-14 ;-3-4 ; 2-5\}=\{-19 ;-7 ;-3\} \\
& A_{1} A_{3}=\{-2-14 ;-6-4 ;-3-5\}=\{-16 ;-10 ;-8\} \\
& A_{1} A_{4}=\{-2-14 ; 2-4 ;-1-5\}=\{-16 ;-2 ;-6\}
\end{aligned}
$$
In accordance with the geometric meaning ... | 112\frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,231 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(0 ;-3 ; 1)$
$M_{2}(-4 ; 1 ; 2)$
$M_{3}(2 ;-1 ; 5)$
$M_{0}(-3 ; 4 ;-5)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-0 & y-(-3) & z-1 \\
-4-0 & 1-(-3) & 2-1 \\
2-0 & -1-(-3) & 5-1
\end{array}\right|=0
$$
Perform transformations:
$\left|\begin{array}{ccc}x & y+3 & z-1 \\ -4 & 4 & 1 \\ 2 & 2 & 4\end{array}\rig... | \frac{90}{\sqrt{194}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,232 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-7 ; 0 ; 3)$
$B(1 ;-5 ;-4)$
$C(2 ;-3 ; 0)$ | ## Solution
Let's find the vector $\overrightarrow{B C}$:
$$
\overrightarrow{B C}=\{2-1 ;-3-(-5) ; 0-(-4)\}=\{1 ; 2 ; 4\}
$$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& (x-(-7)... | x+2y+4z-5=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,233 |
## problem statement
Find the angle between the planes:
$2 x-y+5 z+16=0$
$x+2 y+3 z+8=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{2 ;-1 ; 5\}$
$\overrightarrow{n_{2}}=\{1 ; 2 ; 3\}$
The angle $\phi_{\text {between the planes is determined by the formula: }}$
$\cos \phi=\frac{\left... | \arccos\sqrt{\frac{15}{28}}\approx4257^{\}7^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,234 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; 0 ; z)$
$B(6 ;-7 ; 1)$
$C(-1 ; 2 ; 5)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(6-0)^{2}+(-7-0)^{2}+(1-z)^{2}}=\sqrt{36+49+1-2 z+z^{2}}=\sqrt{z^{2}-2 z+86} \\
& A C=\sqrt{(-1-0)^{2}+(2-0)^{2}+(5-z)^{2}}=\sqrt{1+4+25-10 z+z^{2}}=\sqrt{z^{2}-10 z+30}
\end{aligned}
$$
Since according to the problem $A B=A C$, the... | A(0;0;-7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,235 |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(2; -3; 1)$
$a: x + y - 2z + 2 = 0$
$k = \frac{5}{2}$ | ## Solution
When transforming similarity with the center at the origin of the coordinate plane
$a: A x+B y+C z+D=0$ and the coefficient $k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x+y-2 z+5=0$
Substitute the coordinates of point $A$ into the ... | 2\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,236 |
## problem statement
Write the canonical equations of the line.
$x-y-z-2=0$
$x-2 y+z+4=0$ | ## Solution
Canonical equations of a line:
$$
\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}
$$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction ... | \frac{x-8}{-3}=\frac{y-6}{-2}=\frac{z}{-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,237 |
## problem statement
Find the point of intersection of the line and the plane.
$\frac{x+2}{1}=\frac{y-2}{0}=\frac{z+3}{0}$
$2 x-3 y-5 z-7=0$ | ## Solution
Let's write the parametric equations of the line.
$\frac{x+2}{1}=\frac{y-2}{0}=\frac{z+3}{0}=t \Rightarrow$
$\left\{\begin{array}{l}x=-2+t \\ y=2 \\ z=-3\end{array}\right.$
Substitute into the equation of the plane:
$2(-2+t)-3 \cdot 2-5 \cdot(-3)-7=0$
$-4+2 t-6+15-7=0$
$2 t-2=0$
$t=1$
Find the coor... | (-1,2,-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,238 |
Condition of the problem
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$$
\begin{aligned}
& M(3 ;-3 ;-1) \\
& \frac{x-6}{5}=\frac{y-3.5}{4}=\frac{z+0.5}{0}
\end{aligned}
$$ | ## Solution
We find the equation of the plane that is perpendicular to the given line and passes through point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector:
$\vec{n}=\vec{s}=\{5 ; 4 ; 0\}$
Then the equation of the desired plane is:
$5 \cdo... | M^{\}(-1;2;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,239 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+2)^{4}-(n-2)^{4}}{(n+5)^{2}+(n-5)^{2}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+2)^{4}-(n-2)^{4}}{(n+5)^{2}+(n-5)^{2}}=\lim _{n \rightarrow \infty} \frac{\left((n+2)^{2}-(n-2)^{2}\right) \cdot\left((n+2)^{2}+(n-2)^{2}\right)}{(n+5)^{2}+(n-5)^{2}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(n^{2}+4 n+4-n^{2}+4 n-4\right)\le... | +\infty | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,240 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n^{7}+5}-\sqrt{n-5}}{\sqrt[7]{n^{7}+5}+\sqrt{n-5}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n^{7}+5}-\sqrt{n-5}}{\sqrt[7]{n^{7}+5}+\sqrt{n-5}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n^{7}+5}-\sqrt{n-5}\right)}{\frac{1}{n}\left(\sqrt[7]{n^{7}+5}+\sqrt{n-5}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{n^{5}+... | \infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,241 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{\left(n^{4}+1\right)\left(n^{2}-1\right)}-\sqrt{n^{6}-1}}{n}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{\left(n^{4}+1\right)\left(n^{2}-1\right)}-\sqrt{n^{6}-1}}{n}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{\left(n^{4}+1\right)\left(n^{2}-1\right)}-\sqrt{n^{6}-1}\right)\left(\sqrt{\left(n^{4}+1\right)\left(n^{2}-1\right)}+\sqrt{n^{6}-1... | -\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,242 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{n^{2}-6 n+5}{n^{2}-5 n+5}\right)^{3 n+2}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n^{2}-6 n+5}{n^{2}-5 n+5}\right)^{3 n+2}=\lim _{n \rightarrow \infty}\left(\frac{n^{2}-5 n+5}{n^{2}-6 n+5}\right)^{-3 n-2}= \\
& =\lim _{n \rightarrow \infty}\left(\frac{n^{2}-6 n+5+n}{n^{2}-6 n+5}\right)^{-3 n-2}=\lim _{n \rightarrow \infty}\lef... | e^{-3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,244 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{\text {): }}$
$f(x)=-4 x^{2}-6, x_{0}=1$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,246 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{x^{4}-1}{2 x^{4}-x^{2}-1}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{x^{4}-1}{2 x^{4}-x^{2}-1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{(x-1)\left(x^{3}+x^{2}+x+1\right)}{(x-1)\left(2 x^{3}+2 x^{2}+x+1\right)}= \\
& =\lim _{x \rightarrow 1} \frac{x^{3}+x^{2}+x+1}{2 x^{3}+2 x^{2}+x+1}=\frac{1^{3}+1^{2}+1+1}{2... | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,247 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt[3]{27+x}-\sqrt[3]{27-x}}{\sqrt[3]{x^{2}}+\sqrt[5]{x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt[3]{27+x}-\sqrt[3]{27-x}}{\sqrt[3]{x^{2}}+\sqrt[3]{x}}= \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt[3]{27+x}-\sqrt[3]{27-x})\left(\sqrt[3]{(27+x)^{2}}+\sqrt[3]{27+x} \cdot \sqrt[3]{27-x}+\sqrt[3]{(27-x)^{2}}\right)}{\left(x^{\frac{2}{3}}+x^{\frac{1}{3}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,248 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{\sin \left(\pi\left(\frac{x}{2}+1\right)\right)}$ | ## Solution
Let's use the substitution of equivalent infinitesimals:
$e^{4 x}-1 \sim 4 x$, as $x \rightarrow 0(4 x \rightarrow 0)$
$\sin \left(\frac{\pi x}{2}\right) \sim \sin \frac{\pi x}{2}$, as $x \rightarrow 0\left(\frac{\pi x}{2} \rightarrow 0\right)$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac... | -\frac{8}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,249 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow-2} \frac{\operatorname{tg} \pi x}{x+2}$ | ## Solution
Substitution:
$x=y-2 \Rightarrow y=x+2$
$x \rightarrow-2 \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow-2} \frac{\tan \pi x}{x+2}=\lim _{y \rightarrow 0} \frac{\tan \pi(y-2)}{(y-2)+2}=$
$=\lim _{y \rightarrow 0} \frac{\tan(\pi y-2 \pi)}{y}=\lim _{y \rightarrow 0} \frac{\tan \pi y}{y}=$
Us... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,250 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\left(x^{3}-\pi^{3}\right) \sin 5 x}{e^{\sin ^{2} x}-1}$ | ## Solution
Substitution:
$x=y+\pi \Rightarrow y=x-\pi$
$x \rightarrow \pi \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} \frac{\left(x^{3}-\pi^{3}\right) \sin 5 x}{e^{\sin ^{2} x}-1}=\lim _{y \rightarrow 0} \frac{\left((y+\pi)^{3}-\pi^{3}\right) \sin 5(y+\pi)}{e^{\sin ^{2}(y+\... | -15\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,251 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{5^{2 x}-2^{3 x}}{\sin x+\sin x^{2}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{5^{2 x}-2^{3 x}}{\sin x+\sin x^{2}}=\lim _{x \rightarrow 0} \frac{\left(25^{x}-1\right)-\left(8^{x}-1\right)}{\sin x+\sin x^{2}}=$
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 25}\right)^{x}-1\right)-\left(\left(e^{\ln 8}\right)^{x}-1\right)}{\sin x... | \ln\frac{25}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,252 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 3} \frac{\sqrt[3]{5+x}-2}{\sin \pi x}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y+3 \Rightarrow y=x-3 \\
& x \rightarrow 3 \Rightarrow y \rightarrow 0
\end{aligned}
$$
## We get:
$\lim _{x \rightarrow 3} \frac{\sqrt[3]{5+x}-2}{\sin \pi x}=\lim _{y \rightarrow 0} \frac{\sqrt[3]{5+(y+3)}-2}{\sin \pi(y+3)}=$
$=\lim _{y \rightarrow 0} \frac{\sqrt[3... | -\frac{1}{12\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,253 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{1+\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)^{\frac{1}{\sin x^{3}}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{1+\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)^{\frac{1}{\sin x^{3}}}=$
$$
\begin{aligned}
& =\lim _{x \rightarrow 0}\left(e^{\ln \left(\frac{1+\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)}\right)^{\frac{1}{\sin x^{3}}}= \\
& =\lim _{x \rightarrow 0}... | e^{\frac{5}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,254 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\arcsin x}{x}\right)^{\frac{2}{x+5}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{\arcsin x}{x}\right)^{\frac{2}{x+5}}=\left(\lim _{x \rightarrow 0} \frac{\arcsin x}{x}\right)^{\lim _{x \rightarrow 0} \frac{2}{x+5}}=$ $=\left(\lim _{x \rightarrow 0} \frac{\arcsin x}{x}\right)^{\frac{2}{0+5}}=\left(\lim _{x \rightarrow 0} \frac{\arcsin x}{x}\right)^{\f... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,255 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(\frac{2-x}{x}\right)^{\frac{1}{\ln (2-x)}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(\frac{2-x}{x}\right)^{\frac{1}{\ln (2-x)}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{2-x}{x}\right)}\right)^{\frac{1}{\ln (2-x)}}= \\
& =\lim _{x \rightarrow 1} e^{\frac{1}{\ln (2-x)} \cdot \ln \left(\frac{2-x}{x}\right)}=\exp \left\{\lim _{x \rightar... | e^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,256 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{x^{3}-1}{x-1}\right)^{\frac{1}{x^{2}}}$ | $$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(\frac{x^{3}-1}{x-1}\right)^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 1}\left(\frac{(x-1)\left(x^{2}+x+1\right)}{x-1}\right)^{\frac{1}{x^{2}}}= \\
& =\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)^{\frac{1}{x^{2}}}=\left(1^{2}+1+1\right)^{\frac{1}{1^{2}}}=3^{1}=3
\end{aligne... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,257 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\cos (2 \pi x)}{2+\left(e^{\sqrt{x-1}}-1\right) \operatorname{arctg} \frac{x+2}{x-1}}$ | Solution
Since $\operatorname{arctg} \frac{x+2}{x-1}$ is bounded, and $e^{\sqrt{x-1}}-1 \rightarrow 0$, as $x \rightarrow 1$, then
$\left(e^{\sqrt{x-1}}-1\right) \operatorname{arctg} \frac{x+2}{x-1} \rightarrow 0 \quad$, as $x \rightarrow 1$
Then:
$\lim _{x \rightarrow 1} \frac{\cos (2 \pi x)}{2+\left(e^{\sqrt{x-1}... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,258 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-1 ; 7 ; 0\}$
$p=\{0 ; 3 ; 1\}$
$q=\{1 ;-1 ; 2\}$
$r=\{2 ;-1 ; 0\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 2p-q | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,259 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{4 ; 2 ; 9\}$
$b=\{0 ;-1 ; 3\}$
$c_{1}=4 b-3 a$
$c_{2}=4 a-3 b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=4 b-3 a=\{4 \cdot 0-3 \cdot 4 ; 4 \cdot(-1)-3 \cdot 2 ; 4 \cdot 3-3 \cdot 9\}=\{-12 ;-10 ;-15\} \\
& c_{2... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,260 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(0 ; 3 ;-6), B(9 ; 3 ; 6), C(12 ; 3 ; 3)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(9-0 ; 3-3 ; 6-(-6))=(9 ; 0 ; 12)$
$\overrightarrow{A C}=(12-0 ; 3-3 ; 3-(-6))=(12 ; 0 ; 9)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrigh... | 0.96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,261 |
## Problem Statement
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=3 p+q$
$b=p-3 q$
\[
\begin{aligned}
|p| & =7 \\
|q| & =2
\end{aligned}
\]
\[
(\widehat{p, q})=\frac{\pi}{4}
\] | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(3 p+q) \times(p-3 q)=3 \cdot p \times p+3 \cdot(-3) \cdot p \times q+q \times p-... | 70\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,262 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{3 ; 0 ; 3\}$
$b=\{8 ; 1 ; 6\}$
$c=\{1 ; 1 ;-1\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
3 & 0 & 3 \\
8 & 1 & 6 \\
1 & 1 & -1
\end{array}\right|= \\
& =3 \cdot\left|\be... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,263 |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(-1 ; 2 ; 4) \)
\( A_{2}(-1 ;-2 ;-4) \)
\( A_{3}(3 ; 0 ;-1) \)
\( A_{4}(7 ;-3 ; 1) \) | ## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-1-(-1) ;-2-2 ;-4-4\}=\{0 ;-4 ;-8\} \\
& \vec{A}_{1} A_{3}=\{3-(-1) ; 0-2 ;-1-4\}=\{4 ;-2 ;-5\} \\
& \overrightarrow{A_{1} A_{4}}=\{7-(-1) ;-3-2 ; 1-4\}=\{8 ;-5 ;-3\}
\end{aligned}
$$
According to the geometric mean... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,264 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$$
\begin{aligned}
& M_{1}(14 ; 4 ; 5) \\
& M_{2}(-5 ;-3 ; 2) \\
& M_{3}(-2 ;-6 ;-3) \\
& M_{0}(-1 ;-8 ; 7)
\end{aligned}
$$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-14 & y-4 & z-5 \\ -5-14 & -3-4 & 2-5 \\ -2-14 & -6-4 & -3-5\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-14 & y-4 & z-5 \\
-19 & -7 & -3 \\
-16 &... | 3\sqrt{\frac{13}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,265 |
## Task Condition
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(0 ; 7 ;-9)$
$B(-1 ; 8 ;-11)$
$C(-4 ; 3 ;-12)$ | ## Solution
Let's find the vector $\overrightarrow{B C}:$
$\overrightarrow{B C}=\{-4-(-1) ; 3-8 ;-12-(-11)\}=\{-3 ;-5 ;-1\}$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& -3 \cdo... | -3x-5y-z+26=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,266 |
## Task Condition
Find the angle between the planes:
$x+4 y-z+1=0$
$2 x+y+4 z-3=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ; 4 ;-1\}$
$\overrightarrow{n_{2}}=\{2 ; 1 ; 4\}$
The angle $\phi$ between the planes is determined by the formula:
$\cos \phi=\frac{\left(\overr... | 845^{\}4^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,267 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(x ; 0 ; 0)$
$B(3 ; 5 ; 6)$
$C(1 ; 2 ; 3)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(3-x)^{2}+(5-0)^{2}+(6-0)^{2}}=\sqrt{9-6 x+x^{2}+25+36}=\sqrt{x^{2}-6 x+70} \\
& A C=\sqrt{(1-x)^{2}+(2-0)^{2}+(3-0)^{2}}=\sqrt{1-2 x+x^{2}+4+9}=\sqrt{x^{2}-2 x+14}
\end{aligned}
$$
Since by the condition of the problem $A B=A C$, t... | A(14;0;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,268 |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(2; -5; -1)$
$a: 5x + 2y - 3z - 9 = 0$
$k = \frac{1}{3}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0$ and the coefficient $k$, the plane transitions to
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 5 x+2 y-3 z-3=0$
Substitute the coordinates of point $A$ into the equati... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,269 |
## Task Condition
Write the canonical equations of the line.
$$
\begin{aligned}
& x+5 y+2 z-5=0 \\
& 2 x-5 y-z+5=0
\end{aligned}
$$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vecto... | \frac{x}{5}=\frac{y-1}{5}=\frac{z}{-15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,270 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x+1}{-2}=\frac{y}{0}=\frac{z+1}{3}$
$x+4 y+13 z-23=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x+1}{-2}=\frac{y}{0}=\frac{z+1}{3}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-1-2 t \\
y=0 \\
z=-1+3 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$(-1-2 t)+4 \cdot 0+13(-1+3 t)-23=0$
$-1-... | (-3;0;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,271 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(1 ; 0 ;-1)$
$2 y+4 z-1=0$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$.
Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{0 ; 2 ; 4\}$
Then the equation of the desired line is:
$\... | M^{\}(1;1;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,272 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{6} x \cos ^{2} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \sin ^{6} x \cos ^{2} x d x=\left|\sin x \cos x=\frac{1}{2} \sin 2 x, \sin ^{2} x=\frac{1}{2}(1-\cos 2 x)\right|= \\
& =\int_{0}^{\pi} 2^{4} \cdot \frac{1}{2^{2}} \sin ^{2} 2 x \cdot \frac{1}{2^{2}}(1-\cos 2 x)^{2} d x=\int_{0}^{\pi} \sin ^{2} 2 x\left(1-2 \c... | \frac{5\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,274 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \sin ^{4} x \cos ^{4} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{4} x \cos ^{4} x d x=\int_{0}^{2 \pi}(\sin x \cdot \cos x)^{4} d x=\int_{0}^{2 \pi}\left(\frac{\sin 2 x}{2}\right)^{4} d x=\frac{1}{16} \int_{0}^{2 \pi} \sin ^{4} 2 x d x= \\
& =\frac{1}{16} \int_{0}^{2 \pi} \frac{1}{4} \cdot(1-\cos 4 x)^{2} d x=\frac{1}{64} \in... | \frac{3\pi}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,275 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \sin ^{2}\left(\frac{x}{4}\right) \cos ^{6}\left(\frac{x}{4}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{2}\left(\frac{x}{4}\right) \cos ^{6}\left(\frac{x}{4}\right) d x=\int_{0}^{2 \pi} \frac{1-\cos \frac{x}{2}}{2}\left(\frac{1+\cos \frac{x}{2}}{2}\right)^{3} d x=\int_{0}^{2 \pi} \frac{1-\cos \frac{x}{2}\left(1+\cos \frac{x}{2}\right)^{3}}{2} d x= \\
& =\frac{1}{2... | \frac{5}{64}\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,276 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \cos ^{8}\left(\frac{x}{2}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \cos ^{8}\left(\frac{x}{2}\right) d x=\left|\cos ^{2}\left(\frac{x}{2}\right)=\frac{1}{2}(1+\cos x)\right|=\int_{0}^{\pi} 2^{4} \cdot \frac{1}{2^{4}}(1+\cos x)^{4} d x= \\
& =\int_{0}^{\pi}\left(1+4 \cos x+6 \cos ^{2} x+4 \cos ^{3} x+\cos ^{4} x\right) d x= \... | \frac{35\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,277 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi / 2}^{\pi} 2^{4} \cdot \sin ^{6} x \cos ^{2} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\pi / 2}^{\pi} 2^{4} \cdot \sin ^{6} x \cos ^{2} x d x=\left|\sin x \cos x=\frac{1}{2} \sin 2 x, \sin ^{2} x=\frac{1}{2}(1-\cos 2 x)\right|= \\
& =\int_{\pi / 2}^{\pi} 2^{4} \cdot \frac{1}{2^{2}} \sin ^{2} 2 x \cdot \frac{1}{2^{2}}(1-\cos 2 x)^{2} d x=\int_{\pi / 2}^{\pi} \sin ^{... | \frac{5\pi}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,279 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{4} x \cos ^{4} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \sin ^{4} x \cos ^{4} x d x=2^{4} \cdot \int_{0}^{\pi}(\sin x \cdot \cos x)^{4} d x=2^{4} \cdot \int_{0}^{\pi}\left(\frac{\sin 2 x}{2}\right)^{4} d x=\int_{0}^{\pi} \sin ^{4} 2 x d x= \\
& =\int_{0}^{\pi} \frac{1}{2^{2}} \cdot(1-\cos 4 x)^{2} d x=\frac{1}{4} ... | \frac{3\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,280 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \sin ^{2} x \cos ^{6} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{2} x \cos ^{6} x d x=\left|\sin x \cos x=\frac{1}{2} \sin 2 x, \cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|= \\
& =\int_{0}^{2 \pi} \frac{1}{2^{2}} \sin ^{2} 2 x \cdot \frac{1}{2^{2}}(1+\cos 2 x)^{2} d x=\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \sin ^{2} 2 x\left(1+... | \frac{5\pi}{2^{6}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,281 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \cos ^{8}\left(\frac{x}{4}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \cos ^{8}\left(\frac{x}{4}\right) d x=\left|\cos ^{2}\left(\frac{x}{4}\right)=\frac{1}{2}\left(1+\cos \left(\frac{x}{2}\right)\right)\right|=\int_{0}^{2 \pi} \frac{1}{2^{4}}\left(1+\cos \left(\frac{x}{2}\right)\right)^{4} d x= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \p... | \frac{35\pi}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,282 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{8}\left(\frac{x}{2}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \sin ^{8}\left(\frac{x}{2}\right) d x=\left|\sin ^{2}\left(\frac{x}{2}\right)=\frac{1}{2}(1-\cos x)\right|=\int_{0}^{\pi} 2^{4} \cdot \frac{1}{2^{4}}(1-\cos x)^{4} d x= \\
& =\int_{0}^{\pi}\left(1-4 \cos x+6 \cos ^{2} x-4 \cos ^{3} x+\cos ^{4} x\right) d x= \... | \frac{35\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,283 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\pi}^{0} 2^{8} \sin ^{6} x \cos ^{2} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{-\pi}^{0} 2^{8} \sin ^{6} x \cos ^{2} x d x=2^{8} \int_{-\pi}^{0} \frac{(1-\cos 2 x)^{3}}{2^{3}} \cdot \frac{(1+\cos 2 x)}{2} d x= \\
& =2^{4} \int_{-\pi}^{0}(1+\cos 2 x)(1-\cos 2 x)(1-\cos 2 x)^{2} d x=2^{4} \int_{-\pi}^{0}\left(1-\cos ^{2} 2 x\right)\left(1-2 \cos 2 x+\cos ^{2}... | 10\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,284 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi / 2}^{2 \pi} 2^{8} \cdot \sin ^{4} x \cos ^{4} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\pi / 2}^{2 \pi} 2^{8} \cdot \sin ^{4} x \cos ^{4} x d x=\int_{\pi / 2}^{2 \pi} 2^{4} \cdot(2 \sin x \cdot \cos x)^{4} d x=\int_{\pi / 2}^{2 \pi} 2^{4} \cdot \sin ^{4} 2 x d x= \\
& =16 \int_{\pi / 2}^{2 \pi} \sin ^{4} 2 x d x=16 \int_{\pi / 2}^{2 \pi} \frac{1}{4} \cdot(1-\cos 4 ... | 9\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,285 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{2} x \cos ^{6} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \sin ^{2} x \cos ^{6} x d x=\left|\sin x \cos x=\frac{1}{2} \sin 2 x, \cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|= \\
& =\int_{0}^{\pi} 2^{4} \cdot \frac{1}{2^{2}} \sin ^{2} 2 x \cdot \frac{1}{2^{2}}(1+\cos 2 x)^{2} d x=\int_{0}^{\pi} \sin ^{2} 2 x\left(1+2 \c... | \frac{5\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,286 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \cos ^{8} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \cos ^{8} x d x=\left|\cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|=\int_{0}^{2 \pi} \frac{1}{2^{4}}(1+\cos 2 x)^{4} d x= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi}\left(1+4 \cos 2 x+6 \cos ^{2} 2 x+4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\frac{1}{2^{4}} \cdot \... | \frac{35\pi}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,287 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \sin ^{8}\left(\frac{x}{4}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{8}\left(\frac{x}{4}\right) d x=\left|\sin ^{2}\left(\frac{x}{4}\right)=\frac{1}{2}\left(1-\cos \left(\frac{x}{2}\right)\right)\right|=\int_{0}^{2 \pi} \frac{1}{2^{4}}\left(1-\cos \left(\frac{x}{2}\right)\right)^{4} d x= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \p... | \frac{35\pi}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,288 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{6}\left(\frac{x}{2}\right) \cos ^{2}\left(\frac{x}{2}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \sin ^{6}\left(\frac{x}{2}\right) \cos ^{2}\left(\frac{x}{2}\right) d x=\left|\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)=\frac{1}{2} \sin x, \sin ^{2}\left(\frac{x}{2}\right)=\frac{1}{2}(1-\cos x)\right|= \\
& =\int_{0}^{\pi} 2^{4} \cdot \fra... | \frac{5\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,289 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\pi / 2}^{0} 2^{8} \cdot \sin ^{4} x \cos ^{4} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{-\pi / 2}^{0} 2^{8} \cdot \sin ^{4} x \cos ^{4} x d x=\int_{-\pi / 2}^{0} 2^{4} \cdot(2 \sin x \cdot \cos x)^{4} d x=\int_{-\pi / 2}^{0} 2^{4} \cdot \sin ^{4} 2 x d x= \\
& =16 \int_{-\pi / 2}^{0} \sin ^{4} 2 x d x=16 \int_{-\pi / 2}^{0} \frac{1}{4} \cdot(1-\cos 4 x)^{2} d x=4 \i... | 3\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,290 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi / 2}^{\pi} 2^{8} \cdot \sin ^{2} x \cos ^{6} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\pi / 2}^{\pi} 2^{8} \sin ^{2} x \cos ^{6} x d x=2^{8} \int_{\pi / 2}^{\pi} \frac{1-\cos 2 x}{2} \cdot \frac{(1+\cos 2 x)^{3}}{8} d x= \\
& =2^{4} \int_{\pi / 2}^{\pi / 2}(1-\cos 2 x)(1+\cos 2 x)\left(1+2 \cos 2 x+\cos ^{2} 2 x\right) d x= \\
& =2^{4} \int_{\pi / 2}^{\pi}\left(1-... | 5\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,291 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \cos ^{8} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \cos ^{8} x d x=\left|\cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|=\int_{0}^{\pi} 2^{4} \cdot \frac{1}{2^{4}}(1+\cos 2 x)^{4} d x= \\
& =\int_{0}^{\pi}\left(1+4 \cos 2 x+6 \cos ^{2} 2 x+4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\int_{0}^{\pi}(1+4 \cos 2 x... | \frac{35\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,292 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \sin ^{8} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{8} x d x=\left|\sin ^{2} x=\frac{1}{2}(1-\cos 2 x)\right|=\int_{0}^{2 \pi} \frac{1}{2^{4}}(1-\cos 2 x)^{4} d x= \\
& =\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi}\left(1-4 \cos 2 x+6 \cos ^{2} 2 x-4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\frac{1}{2^{4}} \cdot \... | \frac{35\pi}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,293 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \sin ^{6}\left(\frac{x}{4}\right) \cos ^{2}\left(\frac{x}{4}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{6}\left(\frac{x}{4}\right) \cos ^{2}\left(\frac{x}{4}\right) d x=\left|\sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{4}\right)=\frac{1}{2} \sin \left(\frac{x}{2}\right), \sin ^{2}\left(\frac{x}{4}\right)=\frac{1}{2}\left(1-\cos \left(\frac{x}{2}\right)\right... | \frac{5\pi}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,294 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \sin ^{4}\left(\frac{x}{2}\right) \cos ^{4}\left(\frac{x}{2}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \sin ^{4}\left(\frac{x}{2}\right) \cos ^{4}\left(\frac{x}{2}\right) d x=2^{4} \cdot \int_{0}^{\pi}\left(\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)\right)^{4} d x=2^{4} \cdot \int_{0}^{\pi}\left(\frac{\sin x}{2}\right)^{4} d x= \\
& =\int_{0}^{\pi} ... | \frac{3\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,295 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\pi / 2}^{0} 2^{8} \cdot \sin ^{2} x \cos ^{6} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{-\pi / 2}^{0} 2^{8} \sin ^{2} x \cos ^{6} x d x=2^{8} \int_{-\pi / 2}^{0} \frac{1-\cos 2 x}{2} \cdot \frac{(1+\cos 2 x)^{3}}{8} d x= \\
= & 2^{4} \int_{-\pi / 2}^{0}(1-\cos 2 x)(1+\cos 2 x)(1+\cos 2 x)^{2} d x= \\
= & 2^{4} \int_{-\pi / 2}^{0}\left(1-\cos ^{2} 2 x\right)\left(1+2... | 5\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,296 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi / 2}^{2 \pi} 2^{8} \cdot \cos ^{8} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\pi / 2}^{2 \pi} 2^{8} \cdot \cos ^{8} x d x=\left|\cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|=2^{8} \cdot \int_{\pi / 2}^{2 \pi} \frac{1}{2^{4}}(1+\cos 2 x)^{4} d x= \\
& =2^{4} \cdot \int_{\pi / 2}^{2 \pi}\left(1+4 \cos 2 x+6 \cos ^{2} 2 x+4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d ... | 105\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,297 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{8} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \sin ^{8} x d x=\int_{0}^{\pi} 2^{4}\left(\sin ^{2} x\right)^{4} d x=\int_{0}^{\pi} 2^{4}\left(\frac{1-\cos 2 x}{2}\right)^{4} d x=\int_{0}^{\pi}(1-\cos 2 x)^{4} d x= \\
& =\int_{0}^{\pi}\left(1-4 \cos 2 x+6 \cos ^{2} 2 x-4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \... | \frac{35\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,298 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \sin ^{6} x \cos ^{2} x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{6} x \cos ^{2} x d x=\left|\sin x \cos x=\frac{1}{2} \sin 2 x, \sin ^{2} x=\frac{1}{2}(1-\cos 2 x)\right|= \\
& =\int_{0}^{2 \pi} \frac{1}{2^{2}} \sin ^{2} 2 x \cdot \frac{1}{2^{2}}(1-\cos 2 x)^{2} d x=\frac{1}{2^{4}} \cdot \int_{0}^{2 \pi} \sin ^{2} 2 x\left(1-... | \frac{5\pi}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,299 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \sin ^{4}\left(\frac{x}{4}\right) \cos ^{4}\left(\frac{x}{4}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{4}\left(\frac{x}{4}\right) \cos ^{4}\left(\frac{x}{4}\right) d x=\int_{0}^{2 \pi}\left(\sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{4}\right)\right)^{4} d x=\int_{0}^{2 \pi}\left(\frac{\sin \left(\frac{x}{2}\right)}{2}\right)^{4} d x= \\
& =\frac{1}{2^{4}} ... | \frac{3\pi}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,300 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{2}\left(\frac{x}{2}\right) \cos ^{6}\left(\frac{x}{2}\right) d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \cdot \sin ^{2}\left(\frac{x}{2}\right) \cos ^{6}\left(\frac{x}{2}\right) d x=\left|\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)=\frac{1}{2} \sin x, \cos ^{2}\left(\frac{x}{2}\right)=\frac{1}{2}(1+\cos x)\right|= \\
& =\int_{0}^{\pi} 2^{4} \cdot \fra... | \frac{5\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,301 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\pi / 2}^{0} 2^{8} \cdot \cos ^{8} x d x
$$ | ## Solution
$\int_{-\pi / 2}^{0} 2^{8} \cdot \cos ^{8} x d x=\left|\cos ^{2} x=\frac{1}{2}(1+\cos 2 x)\right|=2^{8} \cdot \int_{-\pi / 2}^{0} \frac{1}{2^{4}}(1+\cos 2 x)^{4} d x=$
$=2^{4} \cdot \int_{-\pi / 2}^{0}\left(1+4 \cos 2 x+6 \cos ^{2} 2 x+4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x=$
$=2^{4} \cdot \int_{-\pi /... | 35\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,302 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi} \sin ^{4} 3 x \cos ^{4} 3 x d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2 \pi} \sin ^{4} 3 x \cos ^{4} 3 x d x=\int_{0}^{2 \pi} \frac{1}{2^{4}} \cdot(2 \sin 3 x \cdot \cos 3 x)^{4} d x=\int_{0}^{2 \pi} \frac{1}{2^{4}} \cdot \sin ^{4} 6 x d x= \\
& =\frac{1}{16} \cdot \int_{0}^{2 \pi} \sin ^{4} 6 x d x=\frac{1}{16} \cdot \int_{0}^{2 \pi} \frac{1}{... | \frac{3\pi}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,303 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{7 n+4}{2 n+1}, a=\frac{7}{2}$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{14 n+8-14 n-7}{2(2 n+1)}\right| \\
& \left.\frac{1}{2(2 n+1)} \right\rvert\, \\
& \frac{1}{2(2 n+1)} \\
& 2 n+1>\frac{1}{2 \varepsilon} ;=> ... | N(\varepsilon)=[\frac{1+2\varepsilon}{4\varepsilon}] | Algebra | proof | Yes | Yes | olympiads | false | 46,304 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n^{3}+1}-\sqrt{n-1}}{\sqrt[3]{n^{3}+1}-\sqrt{n-1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n^{3}+1}-\sqrt{n-1}}{\sqrt[3]{n^{3}+1}-\sqrt{n-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n^{3}+1}-\sqrt{n-1}\right)}{\frac{1}{n}\left(\sqrt[3]{n^{3}+1}-\sqrt{n-1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{n+\fra... | \infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,306 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\frac{1+3+5+7+\ldots+(2 n-1)}{n+1}-\frac{2 n+1}{2}\right)
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{1+3+5+7+\ldots+(2 n-1)}{n+1}-\frac{2 n+1}{2}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{n+1} \cdot \frac{(1+(2 n-1)) n}{2}-\frac{2 n+1}{2}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{n+1} \cdot \frac{2 n^{2}}{2}-\frac{2... | -\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,308 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ ):
$$
\lim _{x \rightarrow-2} \frac{3 x^{2}+5 x-2}{x+2}=-7
$$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). The number
_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$$
f(x)=3 x^{2}-3, x_{0}=4
$$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$. $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x-x_{0}\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,311 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\left(x^{2}+3 x+2\right)^{2}}{x^{3}+2 x^{2}-x-2}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{\left(x^{2}+3 x+2\right)^{2}}{x^{3}+2 x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+2)^{2}(x+1)^{2}}{\left(x^{2}+x-2\right)(x+1)}= \\
& =\lim _{x \rightarrow-1} \frac{(x+2)^{2}(x+1)}{x^{2}+x-2}=\frac{(-1+2)^{2}(-1+1)}{(-1)^{2}+(-1)... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,312 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\sqrt{x-1}}{\sqrt[3]{x^{2}-1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\sqrt{x-1}}{\sqrt[3]{x^{2}-1}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{\sqrt{x-1} \sqrt{x+1}}{\sqrt[3]{x^{2}-1} \sqrt{x+1}}= \\
& =\lim _{x \rightarrow 1} \frac{\sqrt{x^{2}-1}}{\sqrt[3]{x^{2}-1} \sqrt{x+1}}=\lim _{x \rightarrow 1} \frac{\s... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,313 |
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