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Example 2.79. Determine the amount of heat $Q$ generated by the current $I=5+4 t$ in a conductor with resistance $R=40$ over time $t=10$, given that the amount of heat generated per unit time by a constant current flowing through a conductor with constant resistance is equal to the product of the square of the current ... | Solution.
$$
\begin{aligned}
& Q=\int_{t_{1}}^{t_{2}} I^{2}(t) R d t=R \int_{i_{1}}^{t_{2}} I^{2}(t) d t=40 \int_{0}^{10}(5+4 t)^{2} d t= \\
& =40 \int_{0}^{10}(5+4 t)^{2} \frac{1}{4} d(5+4 t)=\left.40 \frac{(5+4 t)^{3}}{3}\right|_{0} ^{10}=303750
\end{aligned}
$$ | 303750 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3.3. Find the level surfaces of the function $y=$ $=\sqrt{36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}}$ and the value of the function at the point $P(1,1,3)$. | Solution. According to the definition of level surfaces, we have: $\sqrt{36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}}=C$, where $C \geqslant 0$. From this, it follows that $36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}=C^{2}$, that is, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=$ $=36-C^{2}$ (obviously, $0 \leqslant C \leqslant 6$). The obtained equation... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3.9. Find the limit of the function $f(x, y)=\left(x^{2}+\right.$ $\left.+y^{2}\right)^{2} x^{2} y^{2}$ as $x \rightarrow 0$ and $y \rightarrow 0$. | Solution. For the calculation of the specified limit, it is more convenient to switch to polar coordinates $x=r \cos \varphi, y=r \sin \varphi$. We obtain
$$
\begin{gathered}
\lim _{\substack{x \rightarrow 0 \\
y \rightarrow 0}}\left(x^{2}+y^{2}\right)^{2 x^{2} y^{2}}=\lim _{r \rightarrow 0}\left(r^{2}\right)^{2 r^{4}... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 4.37. How many times can the trigonometric series $\sum_{k=1}^{\infty} \frac{\cos k x}{k^{4}}$ be differentiated term by term? | Solution. In this example, $a_{k}=\frac{1}{k^{4}}$ and the series with the general term $k^{s} a_{k}$ will converge for $s=1$ and $s=2$ (for $s=3$ we get the harmonic series, and for $s>3$ the general term of the series will be an infinitely large quantity). Therefore, by Theorem 4.28, this series can be term-by-term d... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Expand the function
$$
f(z)=\frac{z}{z^{2}-2 z-3}
$$
into a Taylor series in the neighborhood of the point $z_{0}=0$ using expansion (12), and find the radius of convergence of the series. | Solution. Let's decompose the given function into partial fractions:
$$
\frac{z}{z^{2}-2 z-3}=\frac{1}{4} \frac{1}{z+1}-\frac{3}{4} \frac{1}{z-3}
$$
Transform the right-hand side as follows:
$$
f(z)=\frac{1}{4} \frac{1}{1+z}-\frac{1}{4} \frac{1}{1-\frac{2}{3}}
$$
Using the expansion (12) of the function $\frac{1}{1... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Investigate the convergence of the infinite product
$$
\prod_{k=1}^{\infty}\left(1-\frac{1}{k+1}\right)=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \ldots\left(1-\frac{1}{k+1}\right) \ldots
$$ | Solution. Here all $u_{k}=-\frac{1}{k+1}$ are negative and the series (14)
$$
\sum_{k=1}^{\infty} u_{k}=-\sum_{k=1}^{\infty} \frac{1}{k+1}=-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\right)
$$
obviously diverges.
Then, by Theorem 4, the infinite product (15) diverges.
Remark. Calculating the $n$-th partial pr... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Find the order of the zero $z_{0}=0$ for the function
$$
f(z)=\frac{z^{8}}{z-\sin z}
$$ | Solution. Using the Taylor series expansion of the function $\sin z$ in the neighborhood of the point $z_{0}=0$, we obtain
$$
\begin{aligned}
f(z) & =\frac{z^{8}}{z-\sin z}=\frac{z^{8}}{z-\left(z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\ldots\right)}= \\
& =\frac{z^{8}}{\frac{z^{3}}{3!}-\frac{z^{5}}{5!}+\ldots}=\frac{z^{5}}... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 10. Determine the nature of the singular point $z=1$ of the function
$$
f(z)=\frac{\sin \pi z}{2 e^{z-1}-z^{2}-1}
$$ | Solution. Consider the function
$$
\varphi(z)=\frac{1}{f(z)}=\frac{2 e^{z-1}-z^{2}-1}{\sin \pi z}
$$
The point $z=1$ is a zero of the third order for the numerator
$$
\psi(z)=2 e^{z-1}-z^{2}-1
$$
since
$$
\begin{gathered}
\psi(1)=0 ; \quad \psi^{\prime}(1)=\left.\left(2 e^{z-1}-2 z\right)\right|_{z=1}=0 ; \\
\psi^... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 4. Find the residue of the function
$$
f(z)=z^{3} \cdot \sin \frac{1}{z^{2}}
$$
at its singular point. | Solution. A singular point of the function $f(z)$ is the point $z=0$. It is an essential singular point of the function $f(z)$. Indeed, the Laurent series expansion of the function in the neighborhood of the point $z=0$ is
$$
f(z)=z^{3}\left(\frac{1}{z^{2}}-\frac{1}{3!z^{6}}+\frac{1}{5!z^{10}}-\cdots\right)=z-\frac{1}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Find the residue at the point $z=0$ of the function
$$
f(z)=\frac{\sin 3z - 3 \sin z}{(\sin z - z) \sin z}
$$ | Solution. The point $z=0$ is a zero of both the numerator $\varphi(z)=\sin 3 z-3 \sin z$ and the denominator $\psi(z)=(\sin z-z) \sin z$. Let's determine the orders of these zeros using the Taylor series expansion of $\sin z$ around the point $z=0$:
$$
\sin z=z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\ldots
$$
We have
$$
... | 24 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 9. Find the residue of the function
$$
f(z)=e^{1 / z^{2}} \cos z
$$
at the point $z=0$. | Solution. Since the residue at the point $z=0$ is equal to the coefficient of $z^{-1}$, we immediately obtain that in this case the residue is zero, since the function $f(z)$ is even and its expansion in the neighborhood of the point $z=0$ cannot contain odd powers of $z$.
## Problems for Independent Solution
Find th... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 4. Compute the integral
$$
\int_{|x|=2} \frac{1}{z-1} \sin \frac{1}{z} d z
$$ | Solution. In the circle $|z| \leqslant 2$, the integrand has two singular points $z=1$ and $z=0$. It is easy to establish that $z=1$ is a simple pole, therefore
$$
\operatorname{res}\left(\frac{1}{z-1} \sin \frac{1}{z}\right)=\left.\frac{\sin \frac{1}{z}}{(z-1)^{\prime}}\right|_{z=1}=\sin 1
$$
To determine the nature... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Compute the integral
$$
I=\int_{|z|=2} \frac{d z}{1+z^{4}}
$$ | The poles (finite) of the integrand
$$
f(z)=\frac{1}{1+z^{4}}
$$
are the roots $z_{1}, z_{2}, z_{3}, z_{4}$ of the equation $z^{4}=-1$, which all lie inside the circle $|z|=2$. The function $f(z)=\frac{1}{1+z^{4}}$ has an expansion in the neighborhood of the infinitely distant point
$$
f(z)=\frac{1}{1+z^{4}}=\frac{1... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Find the logarithmic residue of the function
$$
f(z)=\frac{\operatorname{ch} z}{e^{i z}-1}
$$
with respect to the contour $C:|z|=8$. | Solution. We find the zeros $z_{k}$ of the function $f(z)$. For this, we solve the equation $\cosh z=0$ or $e^{z}+e^{-z}=0$. Writing the last equation as $e^{2 z}=-1$, we find
$2 z=\operatorname{Ln}(-1)=(2 k+1) \pi i$, so $z_{k}=\frac{2 k+1}{2} \pi i(k=0, \pm 1, \pm 2, \ldots)$ (all zeros are simple). To find the poles... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Find the logarithmic residue of the function
$$
f(z)=\frac{1+z^{2}}{1-\cos 2 \pi z}
$$
with respect to the circle $|z|=\pi$. | Solution. Setting $1+z^{2}=0$, we find two simple zeros of the function $f(z): a_{1}=-i, a_{2}=i$. Setting $1-\cos 2 \pi z=0$, we find the poles of the function $f(z): z_{n}=n, n=0, \pm 1, \pm 2, \ldots$. The multiplicity of the poles is $k=2$.
In the circle $|z|<\pi$, the function has two simple zeros $a_{1}=-i, a_{2... | -12 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 4. Find the number of roots in the right half-plane $\operatorname{Re} z>0$ of the equation
$$
Q_{5}(z) \equiv z^{5}+z^{4}+2 z^{3}-8 z-1=0
$$ | Solution. By the argument principle, the number of zeros inside the contour $C$ is
$$
N=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} Q_{5}(z)
$$
where the contour $C$ consists of the semicircle $C_{R}:|z|=R, \operatorname{Re} z>0$, and its diameter on the imaginary axis; the radius $R$ is taken to be so large that a... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Find the number of roots of the equation
$$
Q_{7}(z) \equiv z^{7}-2 z-5=0
$$
in the right half-plane. | Solution. We choose the contour $C$ as indicated in Example 4. Then $\Delta_{C_{R}} \operatorname{Arg} Q_{7}(z)=\Delta_{C_{R}} \operatorname{Arg}\left(z^{7}-2 z-5\right)=$
$$
\begin{aligned}
& =\Delta_{C_{R}} \operatorname{Arg}\left[z^{7}\left(1-\frac{2}{z^{6}}-\frac{5}{z^{7}}\right)\right]=7 \Delta_{C_{R}} \operatorn... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Find the number of zeros of the function
$$
F(z)=z^{8}-4 z^{5}+z^{2}-1
$$
inside the unit circle $|z|<1$. | Solution. Let us represent the function $F(z)$ as the sum of two functions $f(z)$ and $\varphi(z)$, which we choose, for example, as follows:
$$
f(z)=-4 z^{5}, \quad \varphi(z)=z^{8}+z^{2}-1
$$
Then on the circle $|z|=1$ we will have
$$
\begin{aligned}
& |f(z)|=\left|-4 z^{5}\right|=4 \\
& |\varphi(z)|=\left|z^{8}+z... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 7. Determine the number of roots of the equation
$$
z^{6}-6 z+10=0
$$
inside the circle $|z|<1$. | Solution. Let, for example, $f(z)=10$ and $\varphi(z)=z^{6}-6 z$. On the circle $|z|=1$ we have
$$
|f(z)|=10, \quad|\varphi(z)|=\left|z^{6}-6 z\right| \leqslant\left|z^{6}\right|+6|z|=7
$$
Thus, in all points of the circle $|z|=1$, the inequality $|f(z)|>|\varphi(z)|$ holds. The function $f(z)=10$ has no zeros inside... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 8. How many roots of the equation
$$
z^{4}-5 z+1=0
$$
lie in the annulus $1<|z|<2 ?$ | Solution. Let $N$ be the number of roots of equation (4) in the ring $1<|\varphi(z)|$, since $|f(z)|=|-5 z|=5,|\varphi(z)|=\left|z^{4}+1\right| \leqslant$ $\left|z^{4}\right|+1=2$. The function $f(z)=-5 z$ has one root in the circle $|z|<1$, and thus $N_{1}=1$.
In the circle $|z|<2$, $|f(z)|>\left|\varphi(z)\right|$, ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 9. Find the number of roots of the equation
$$
z^{2}-a e^{z}=0, \quad \text { where } \quad 0<a<e^{-1}
$$
in the unit circle $|z|<1$. | Solution. Let $f(z)=z^{2}$ and $\varphi(z)=-a e^{z}$. On the circle $|z|=1$ we have
$$
\begin{aligned}
& |f(z)|=\left|z^{2}\right|=1 \\
& |\varphi(z)|=\left|-a e^{z}\right|=a\left|e^{z}\right|=a\left|e^{x+i y}\right|=a e^{x} \leqslant a e|\varphi(z)|$, if $|z|=1$. The function $f(z)=z^{2}$ in the circle $|z|0, \quad \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 10. Find the number of roots of the equation
$$
\lambda-\boldsymbol{z}-e^{-z}=0, \quad \lambda>1
$$
in the right half-plane $\operatorname{Re} z>0$. | Solution. Consider the contour composed of the segment $[-i R, i R]$ and the right semicircle $|z|=R$. Let $f(z)=z-\lambda$ and $\varphi(z)=e^{-z}$. On the segment $[-i R, i R]$, where $z=i y$, we have
$$
\begin{aligned}
& |f(z)|=|i y-\lambda|=\sqrt{\lambda^{2}+y^{2}} \geqslant \sqrt{\lambda^{2}}=\lambda>1 \\
& |\varp... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the approximate value of the smallest characteristic number of the kernel by the Ritz method
$$
K(x, t)=x t ; \quad a=0, b=1
$$ | Solution. As the coordinate system of functions $\psi_{n}(x)$, we choose the system of Legendre polynomials: $\psi_{n}(x)=P_{n}(2 x-1)$. In formula (1), we limit ourselves to two terms, so that
$$
\varphi_{2}(x)=a_{1} \cdot P_{0}(2 x-1)+a_{2} \cdot P_{1}(2 x-1) .
$$
Noting that
$$
\psi_{1} \equiv P_{0}(2 x-1)=1 ; \q... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Using the Kellogg method, calculate the smallest characteristic number of the kernel $K(x, t)=x^{2} t^{2}, 0 \leqslant x, t \leqslant 1$. | Solution. Let $\omega(x)=x$. Then
$$
\begin{aligned}
& \omega_{1}(x)=\int_{0}^{1} x^{2} t^{2} t d t=\frac{x^{2}}{4} \\
& \omega_{2}(x)=\int_{0}^{1} x^{2} t^{4} \frac{1}{4} d t=\frac{1}{4} x^{2} \cdot \frac{1}{5} \\
& \omega_{3}(x)=\int_{0}^{1} \frac{1}{4 \cdot 5} x^{2} t^{4} d t=\frac{1}{4 \cdot 5^{2}} x^{2} \\
& \ldo... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Find the length of the arc of the curve $y=\arcsin \sqrt{x}-\sqrt{x-x^{2}}$ from $x_{1}=0$ to $x_{2}=1$. | Solution. We have $y^{\prime}=\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{1-2 x}{2 \sqrt{x-x^{2}}}=\frac{x}{\sqrt{x-x^{2}}}=$ $=\sqrt{\frac{x}{1-x}}$ $L=\int_{0}^{1} \sqrt{1+y^{\prime 2}} d x=\int_{0}^{1} \sqrt{1+\frac{x}{1-x}} d x=\int_{0}^{1} \frac{1}{\sqrt{1-x}} d x=-\left.2 \sqrt{1-x}\right|_{0} ^{1}=2$.
Answer. $L=2$. | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Calculate the areas of figures bounded by the curves:
a) $\left\{\begin{array}{l}y=x \sqrt{9-x^{2}} \\ y=0(0 \leqslant x \leqslant 3)\end{array}\right.$
b) $\left\{\begin{array}{l}y=2 x-x^{2}+3 \\ y=x^{2}-4 x+3\end{array}\right.$ | Solution. We will construct the corresponding regions (Fig. 3.8) and determine the appropriate limits of integration from them, omitting the systems of inequalities. Thus:
a) $S=\int_{D} \int d x d y=\int_{0}^{3} d x \int_{0}^{x \sqrt{9-x^{2}}} d y=$
$=\left.\int_{0}^{3} d x \cdot y\right|_{0} ^{x \sqrt{9-x^{2}}}=\in... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 7. Calculate the area of the figure bounded above by the lines $x-y+2=0, y=2-\frac{1}{2} x$, and below by the parabola $y=x^{2}-2 x-$ $-8$. | Solution. Referring to Fig. 3.9. The equation of $AC$ is $y=x+2$, and $CB$ is described by the equation $y=2-\frac{1}{2} x$. Then,
$$
S=\int_{-2}^{0} d x \int_{x^{2}-2 x-8}^{x+2} d y+\int_{0}^{2} d x \int_{x^{2}-2 x-8}^{2-x / 2} d y=42
$$
Answer. $S=42$. | 42 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Calculate the mass of the surface $z=x y$, located inside the cylinder $x^{2}+\frac{y^{2}}{4}=1$, if the density is $\rho=\frac{|z|}{\sqrt{1+x^{2}+y^{2}}}$. | Solution. Given the symmetries of the integration region $\sigma$: $x^{2}+\frac{y^{2}}{4} \leqslant 1$, the equation of the surface, and the density function, it is sufficient to compute the integral over one quarter of the region and multiply the result by 4: $m=4 \int_{D} \rho(x, y, z) d \sigma$, where $D$ is the qua... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Calculate the mass of the tetrahedron bounded by the planes $x=0, y=0, z=0$ and $x / 10+y / 8+z / 3=1$, if the density distribution of mass at each point is given by the function $\rho=(1+x / 10+y / 8+z / 3)^{-6}$. | Solution. We have $m=\iint_{W} \int \rho d V$. The triple integral is reduced to a double and a definite one (see point $4^{\circ}$):
$$
m=\iint_{D} d x d y \int_{0}^{z} \frac{d z}{\left(1+\frac{x}{10}+\frac{y}{8}+\frac{z}{3}\right)^{6}}
$$
The upper limit, or the exit point from the region, is the ordinate of the po... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 8. Check the conditions of Green's theorem for the line integral $\int_{L} 2 x y d x + x^{2} d y$ and compute this integral along the parabola $y=\frac{x^{2}}{4}$ from the origin to the point $A(2,1)$. | Solution. We have $P(x, y)=2 x y, Q(x, y)=x^{2}$. These functions are defined, continuous, and differentiable at any point $(x, y)$ in the plane. We have $\frac{\partial P}{\partial y}=2 x, \frac{\partial Q}{\partial x}=2 x$. The conditions of Green's theorem are satisfied. Therefore, the given integral is independent ... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 10. Calculate the integral $I=\oint_{L}\left(x^{2}-y^{2}\right) d x+2 x y d y$, where $L-$ is the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. | Solution. We apply Green's formula and compute the double integral, transitioning to "generalized" polar coordinates. We have: $P=x^{2}-y^{2}, Q=2 x y, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2 y+2 y=4 y$. Therefore,
$$
\begin{aligned}
& \int\left(x^{2}-y^{2}\right) d x+2 x y d y=4 \iint_{D} y d x ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Investigate the convergence of the series $\sum_{n=2}^{\infty} \frac{18}{n^{2}+n-2}$ and, if possible, find its sum. | Solution. We have $\frac{18}{n^{2}+n-2}=\frac{6}{n-1}-\frac{6}{n+2}$. Let's form the partial sum and find its limit. Notice which terms cancel each other out (!). We have:
$$
\begin{aligned}
& S_{n}=u_{2}+u_{3}+\ldots+u_{n}= \\
& =6\left[\left(\frac{1}{1}-\frac{1}{A}\right)+\left(\frac{1}{2}-\frac{1}{b}\right)+\left(\... | 11 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Task 1 (scheduling problem). In how many ways can a daily school schedule be made, consisting of 6 subjects out of 10? | Solution. If we take, for example, BAGRIF (biology, algebra, geometry, Russian, history, physical education) as one of the possible schedules, then any other schedule must differ from it either in the order of subjects or in the subjects themselves, i.e., we are talking about permutations of six different subjects out ... | 151200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5 (Partition problem) In how many ways can 11 items be divided into two groups such that each group has no fewer than three items (selecting one group uniquely determines the other)? | Solution. All possibilities to divide 11 items into two groups are represented by the following decompositions: $11=3+8, 11=4+7$, $11=5+6$ (the possibilities $11=6+5, 11=7+4$ and $11=8+3$ coincide with the previous ones). It remains to understand in how many ways one can choose 3, 4, or 5 items out of 11. Since the ord... | 957 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Find the most probable number of hits in the ring in five throws, if the probability of hitting the ring with the ball in one throw is $p=0.6$. | Solution. We have $n=5 ; p=0.6 ; q=0.4$. For the number $k_{0}$, we obtain the estimate: $5 \cdot 0.6-0.4 \leqslant k_{0} \leqslant 5 \cdot 0.6+0.6$, i.e., $2.6 \leqslant k_{0} \leqslant 3.6$. Since $k_{0}$ is an integer, then $k_{0}=3$.
Direct calculations of $p_{5}(k)$ lead to the values: $p_{5}(0)=$ $=0.01024 ; p_{... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 7. Among some products, the share of top-grade products is $31 \%$. What is the most probable number of top-grade products in a randomly selected batch of 75 products? | Solution. It is known that $p=0.31; q=1-p=0.69; n=75$. By formula (22) we have: $75 \cdot 0.31-0.69 \leqslant k_{0} \leqslant 75 \cdot 0.31+0.31 ; 22.56 \leqslant$ $\leqslant k_{0} \leqslant 23.56 ;$ since $k_{0}$ must be an integer, it follows that $k_{0}=23$.
Answer. 23.
## Exercises
1. Two equally skilled chess p... | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Given a statistical series - the number of days missed due to illness by employees of a laboratory.
| Number of days | 0 | 2 | 3 | 4 | 5 | 7 | 10 | Total |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of employees | 7 | 3 | 5 | 2 | 5 | 6 | 2 | 30 |
Determine the average n... | Solution. Let's determine the sample mean of size $n=30(k=7)$ using formula (1):
$$
\bar{x}_{3}=\frac{1}{30}(7 \cdot 0+3 \cdot 2+5 \cdot 3+2 \cdot 4+5 \cdot 5+6 \cdot 7+2 \cdot 10)=\frac{116}{30}=3.87
$$
We will calculate the variance and standard deviation using formula (2):
$$
\begin{aligned}
\overline{x_{\mathrm{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example. Compute the limit
$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}
$$ | Solution. Here $(2 n+1)^{2}-(n+1)^{2}=3 n^{2}+2 n-$ is a polynomial of the second degree (an infinitely large sequence of order $n^{2}$) and $n^{2}+n+1$ is a polynomial of the second degree (an infinitely large sequence of order $n^{2}$).
1. Factor out $n^{2}$ in the numerator, we get
$$
(2 n+1)^{2}-(n+1)^{2}=n^{2}\l... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Compute the limit
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}
$$ | Solution. The numerator $n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}$ is an infinitely large sequence of order $n^{2}$, and the denominator $(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}$ is an infinitely large sequence of order $n^{2}$.
1. Factor out $n^{2}$ in the numerator, we get
$$
n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}=n^{2}\left(\fr... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Example. Compute the limit
$$
\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}
$$ | Solution. The expression under the limit sign is the ratio of two infinitesimals at the point $x=0$, since
$$
\lim _{x \rightarrow 0}(2 x \sin x)=0, \quad \lim _{x \rightarrow 0}(1-\cos x)=0
$$
The infinitesimals in the numerator and denominator are replaced by equivalent ones:
$$
\begin{array}{ll}
2 x \sin x \sim 2... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Calculate the limit
$$
\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan^{2} 2 x}
$$ | Solution.
1. Since
$$
\lim _{x \rightarrow \pi}[\cos 3 x-\cos x]=0, \quad \lim _{x \rightarrow \pi} \operatorname{tg}^{2} 2 x=0
$$
the expression under the limit sign is a ratio of two infinitesimal functions as $x \rightarrow \pi$. We need to replace these infinitesimal functions with equivalent ones. For this, we ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Compute the limit
$$
\lim _{x \rightarrow 0} \sqrt[3]{x\left(2+\sin \frac{1}{x}\right)+8 \cos x}
$$ | Solution.
1. Since the function $y=\sqrt[3]{x}$ is continuous for all $x$, by passing to the limit under the sign of a continuous function, we get
$$
\lim _{x \rightarrow 0} \sqrt[3]{x\left(2+\sin \frac{1}{x}\right)+8 \cos x}=\sqrt[3]{\lim _{x \rightarrow 0}\left[x\left(2+\sin \frac{1}{x}\right)+8 \cos x\right]}
$$
... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. According to the definition, find the derivative of the function
$$
f(x)=\left[\begin{array}{ll}
1-\cos \left(x \sin \frac{1}{x}\right), & x \neq 0 \\
0, & x=0
\end{array}\right.
$$
at the point $x=0$. | Solution.
1. By definition
$$
f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}=\lim _{x \rightarrow 0} \frac{1-\cos (x \sin (1 / x))-0}{x}
$$
2. Since $\sin (1 / x)$ is a bounded function and $x$ is an infinitesimal function as $x \rightarrow 0$, by the theorem on the product of an infinitesimal function an... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Find the area of the region bounded by the graphs of the functions
$$
y=x^{2}-4 x+3, \quad y=-x^{2}+2 x+3
$$ | ## Solution.
1. Find the abscissas $a$ and $b$ of the points of intersection of the graphs. For this, solve the equation
$$
x^{2}-4 x+3=-x^{2}+2 x+3
$$
We get $a=0, \quad b=3$.
2. Investigate the sign of the function $\varphi=x^{2}-4 x+3-\left(-x^{2}+2 x+3\right)$ on the interval $[a, b]=[0,3]$. For this, assign $x... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Find the sum of the series
$$
\sum_{n=1}^{\infty} \frac{72}{n^{2}+5 n+4}
$$ | Solution.
1. The roots of the denominator $n=-1$ and $n=-4$ differ by an integer, i.e., $n^{2}+5 n+4=(n+1)(n+1+3)$. Therefore, the terms of the sequence of partial sums of the series $\sum_{n=1}^{\infty} a_{n}$ are easily found, as many terms in the expression $S_{n}=a_{1}+a_{2}+\ldots+a_{n}$ cancel each other out.
2.... | 26 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example. Evaluate the double integral
$$
\iint_{D}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d x d y
$$
where the region $D$ is bounded by the lines $x=1, y=x^{3}$ and $y=-\sqrt{x}$. | Solution.
1. Let's define the region $D$ by inequalities. It is obvious that $-\sqrt{x} \leq x^{3}$. Therefore, $-\sqrt{x} \leq y \leq x^{3}$. Since $x$ appears under the square root, $x \geq 0$. For $x$, the possible inequalities are $0 \leq x \leq 1$ or $1 \leq x$. In the second case, the region is unbounded, which ... | 11 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Evaluate the double integral
$$
\iint_{D} \frac{x}{y^{5}} d x d y
$$
where the region $D$ is defined by the inequalities
$$
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3, \quad y \geq \frac{x}{4}, \quad x \geq 0
$$ | SOLUTION.
1. The region $D$ is defined by inequalities in the Cartesian coordinate system:
$$
D=\left\{(x, y): \begin{array}{c}
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3 \\
\\
y \geq \frac{x}{4}, \quad x \geq 0
\end{array}\right\}
$$
2. Since the region $D$ is bounded by ellipses and lines passing through the origin, it ... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the volume of the body bounded by the surfaces
$$
x=17 \sqrt{2 y}, \quad x=2 \sqrt{2 y}, \quad z=1 / 2-y, \quad z=0
$$ | ## Solution.
1. By formula (1) with $f_{2}=1 / 2-y$ and $f_{1}=0$, the desired volume is
$$
V=\iint_{D}\left(\frac{1}{2}-y\right) d x d y
$$
where $D$ is the projection of the body onto the $X O Y$ plane.
2. To find $D$, we define the body using inequalities and eliminate $z$ from them. In this case, the body is de... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the mass of the plate $D$ with surface density $\mu=16 x+9 y^{2} / 2$, bounded by the curves
$$
x=\frac{1}{4}, \quad y=0, \quad y^{2}=16 x \quad(y \geq 0)
$$ | Solution.
1. The mass of the plate $D$ with surface density $\mu=16 x+9 y^{2} / 2$ is determined by the formula
$$
m=\iint_{D}\left(16 x+\frac{9 y^{2}}{2}\right) d x d y
$$
2. We compute the obtained double integral in Cartesian coordinates:
a) define the region $D$ by a system of inequalities:
$$
\left\{\begin{ar... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Find the mass of the plate $D$ with surface density $\mu=x / y^{5}$, bounded by the curves
$$
\frac{x^{2}}{16}+y^{2}=1, \quad \frac{x^{2}}{16}+y^{2}=3, \quad y=\frac{x}{4}, \quad x=0 \quad\left(y \geq \frac{x}{4}, x \geq 0\right)
$$ | Solution.
1. The mass of the plate $D$ with surface density $\mu=x / y^{5}$ is determined by the formula
$$
m=\iint_{D} \frac{x}{y^{5}} d x d y
$$
2. We calculate the obtained double integral:
a) define the region $D$ by inequalities in Cartesian coordinates
$$
D=\left\{(x, y): \begin{array}{c}
1 \leq \frac{x^{2}}... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the volume of the body $\Omega$, bounded by the surfaces
$$
x=17 \sqrt{2 y}, \quad x=2 \sqrt{2 y}, \quad z=\frac{1}{2}-y, \quad z=0
$$ | Solution.
1. Define the region $\Omega$ by inequalities. Since $17 \sqrt{2 y} \geq 2 \sqrt{2 y}$, for $x$ we have the inequalities $2 \sqrt{2 y} \leq x \leq 17 \sqrt{2 y}$. Since $y$ appears under the square root, $y \geq 0$. For $z$, the possible inequalities are $0 \leq z \leq 1 / 2-y$ or $1 / 2-y \leq z \leq 0$. In... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the mass of the body $\Omega$ with density $\mu=2 x$, bounded by the surfaces
$$
x=2 \sqrt{2 y}, \quad x=\sqrt{2 y}, \quad z=1-y, \quad z=0
$$ | Solution.
1. The mass of the body $\Omega$ with density $\mu=2 x$ is determined by the formula
$$
m=\iiint_{\Omega} 2 x d x d y d z
$$
2. Let's define the region $\Omega$ using inequalities. Since $2 \sqrt{2 y} \geq \sqrt{2 y}$, for $x$ we have the inequalities $\sqrt{2 y} \leq x \leq 2 \sqrt{2 y}$. Since $y$ appear... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Find the derivative of the scalar field $u=x z^{2}+2 y z$ at the point $M_{0}(1,0,2)$ along the circle
$$
\left\{\begin{array}{l}
x=1+\cos t \\
y=\sin t-1 \\
z=2
\end{array}\right.
$$ | Solution. The vector equation of the circle has the form
$$
\mathbf{r}(t)=(1+\cos t) \mathbf{i}+(\sin t-1) \mathbf{j}+2 \mathbf{k} .
$$
We find the vector $T$, tangent to it at any point $M$. We have
$$
\left.T=\frac{d r}{d t}=-\sin t \mathbf{i}+\cos t \mathbf{j}\right]
$$
The given point $M_{0}(1,0,2)$ lies in the... | -4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Using the invariant definition, calculate the divergence of the vector $a=x \mathbf{i}$ at the point $O(0,0,0)$, choosing as the surface $\sigma$ surrounding the point $O$, a sphere $\sigma_{\varepsilon}$ of radius $\varepsilon$ centered at this point. | Solution. By the definition of divergence at the given point, we have
$$
\operatorname{div} a(0)=\lim _{\left(\sigma_{k}\right) \rightarrow 0} \frac{\int\left(a, n^{0}\right) d \sigma}{v_{\varepsilon}}
$$
where $v_{\varepsilon}$ is the volume of the ball bounded by the sphere $\sigma_{\varepsilon}$, or
$$
\operatorn... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Show that the vector field $a=\frac{2 \cos \theta}{r^{3}} \mathbf{e}_{r}+\frac{\sin \theta}{r^{3}} \mathbf{e}_{\theta}$
is solenoidal. | Solution. Using formula (5), we will have
$$
\begin{aligned}
& \operatorname{div}=\frac{1}{r^{2}} \frac{\theta}{\partial r}\left(r^{2} \frac{2 \cos \theta}{r^{3}}\right)+\frac{1}{r \sin \theta} \frac{\theta}{\partial \theta}\left(\sin \theta \frac{\sin \theta}{r^{3}}\right)+0= \\
&=\frac{1}{r^{2}}\left(-\frac{2 \cos \... | 0 | Calculus | proof | Yes | Yes | olympiads | false |
Example 12. Calculate the circulation of the vector field given in cylindrical coordinates: $2=\rho \sin \varphi \mathrm{e}_{\rho}+\rho z \mathrm{e}_{\varphi}+\rho^{3} \mathrm{e}_{z}$, along the curve L: $\{\rho=\sin \varphi, z=0,0 \leqslant \varphi \leqslant \pi\}$ directly and using Stokes' theorem. | Solution. Coordinates of the given vector
$$
a_{p}=\rho \sin \varphi, \quad a_{v}=\rho z, \quad a_{n}=\rho^{3}
$$
The contour $L$ represents a closed curve located in the plane $z=0$ (Fig. 41).
1) Direct calculation of circulation.
Substituting the coordinates of the vector into formula (13), we get
$$
\boldsymbol... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
45. How many numbers need to be selected from a table of random numbers to be confident with a probability of at least 0.9 that among them there is at least one even number? | Solution. Let $n$ be the required number of random numbers. Consider the events:
$A_{k}$ - one randomly selected number is even $(k=1,2, \ldots, n)$;
$\overline{A_{k}}$ - one randomly selected number is odd;
$B$ - among $n$ random numbers there will be at least one even;
$\bar{B}$ - among $n$ random numbers there w... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
63. A large batch of car tires contains $1.5\%$ defects. What should be the volume of a random sample so that the probability of finding at least one defective car tire in it would be more than $0.92?$ | S o l u t i o n. Let the volume of the random sample be denoted by $n$. The experiment consists of checking car tires. Event $A$, which may or may not occur in each trial, is that the checked car tire turns out to be defective. The probability $p=0.015$; the probability $q=0.985$. Consider the events:
C - the sample o... | 168 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
64. The probability of failure of each device during testing is 0.2. How many such devices need to be tested to assert with a probability of at least 0.9 that at least two devices will fail? | S o l u t i o n. Let the required number of devices be denoted by $n$. The event $A$, which may or may not occur in each trial, consists of a device failing. The probability $p=0.2$; the probability $q=0.8$.
Consider the events:
$B$ - at least two devices fail;
$\bar{B}$ - fewer than two devices fail.
The event $\b... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
76. When crossing two varieties of peas (with yellow and green seeds), G. Mendel found that in the second generation, the probability of green seeds appearing is 0.25. How many pea seeds need to be taken to expect with a probability of 0.9770 that the relative frequency of green seeds will deviate (in absolute value) f... | Solution. According to the condition $p=0.25 ; \varepsilon=0.02 ;$ we find $q=0.75$. It is known that $P\left(\left|\frac{m}{n}-0.25\right| \leq 0.02\right)=0.997$. We need to find the number of seeds $n$.
In accordance with formula (23) we have
$$
P\left(\left|\frac{m}{n}-p\right| \leq \varepsilon\right) \approx 2 \... | 4107 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
90. Find the mathematical expectation of the random variable $Z=2 X+4 Y+5$, if the mathematical expectations of $X$ and $Y$ are known: $M(X)=3, M(Y)=5$. | Solution. Using the properties of mathematical expectation (formulas (31) - (33)), we get:
$$
\begin{aligned}
& M(Z)=M(2 X+4 Y+5)=2 M(X)+4 M(Y)+5= \\
& =2 \cdot 3+4 \cdot 5+5=31
\end{aligned}
$$ | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
94. Random variables $X$ and $Y$ are independent. The variances of these variables are known: $D(X)=5 ; D(Y)=9$. Find the variance of the random variable $Z=2X-Y+5$. | S o l u t i o n. Using the properties of variance reflected in formulas (40), (41), and (42), we get
$$
D(Z)=D(2 X-Y+5)=2^{2} \cdot D(X)+(-1)^{2} \cdot D(Y)=4 \cdot 5+9=29
$$ | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
118. The random variable $X$ is given by the distribution function:
$$
F(x)=\left\{\begin{array}{ccc}
0 & \text { if } & x \leq -c \\
\frac{1}{2}+\frac{1}{\pi} \arcsin \frac{x}{c} & \text { if } & -c < x \leq c \\
1 & \text { if } & x > c
\end{array}\right.
$$
( arcsine law ).
Find the mathematical expectation of th... | Solution. Let's find the probability density function of the random variable $X: f(x)=F^{\prime}(x)$. For $x \leq -c$ and for $x > c$, the function $f(x)=0$. For $-c < x \leq c$, we have
$$
f(x)=\left(\frac{1}{2}+\frac{1}{\pi} \arcsin \frac{x}{c}\right)^{\prime}=\frac{1}{\pi \sqrt{c^{2}-x^{2}}}
$$
The expected value ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
56. a) $29 \cdot 31$; b) $86^{2}-14^{2}$. | Solution. a) $29 \cdot 31=(30-1)(30+1)=30^{2}-1^{2}=900-1=899$;
b) $86^{2}-14^{2}=(86+14)(86-14)=100 \cdot 72=7200$. | 899 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
120. Calculate $\left(\frac{9}{16}\right)^{-1 / 10}:\left(\frac{25}{36}\right)^{-3 / 2}-\left[\left(\frac{4}{3}\right)^{-1 / 2}\right]^{-2 / 5}\left(\frac{6}{5}\right)^{-3}$. | Solution. We will perform the actions sequentially:
1) $\left(\frac{9}{16}\right)^{-1 / 10}=\left[\left(\frac{3}{4}\right)^{2}\right]^{-1 / 10}=\left(\frac{3}{4}\right)^{-1 / 5}=\left(\frac{4}{3}\right)^{1 / 5}$;
2) $\left(\frac{25}{36}\right)^{-3 / 2}=\left[\left(\frac{5}{6}\right)^{2}\right]^{-3 / 2}=\left(\frac{5}{... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
137. $5^{x}=625$ | Solution. Writing 625 as $5^{4}$, we get $5^{x}=5^{4}$, hence $x=4$. 138. $8^{x}=32$.
Solution. We have $32=2^{5} ; 8^{x}=\left(2^{3}\right)^{x}=2^{3 x}$. Therefore, $2^{3 x}=2^{5}$, hence $3 x=5$, i.e., $x=5 / 3$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
154. $(0,25)^{2-x}=\frac{256}{2^{x+3}}$.
154. $(0.25)^{2-x}=\frac{256}{2^{x+3}}$. | Solution. Let's convert all powers to base $2: 0.25=1/4=2^{-2}$; $256=2^{8}$. Therefore, $\left(2^{-2}\right)^{2-x}=\frac{2^{8}}{2^{x+3}}$. Applying the rule of dividing powers, we have
$$
\begin{gathered}
2^{-4+2 x}=2^{8-x-3} ; 2^{-4+2 x}=2^{5-x} ;-4+2 x=5-x ; 2 x+x=5+4 \\
3 x=9 ; x=3
\end{gathered}
$$
=44 ; 2^{x-3}(8+4-1)=44 ; 2^{x-3} \cdot 11=44
$$
Dividing both sides of the equation by 11, we get
$$
2^{x-3}=4 ; 2^{x-3}=2^{2} ; x-3=2 ; x=5
$$ | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
162. $7^{x}-3 \cdot 7^{x-1}+7^{x+1}=371$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
162. $7^{x}-3 \cdot 7^{x-1}+7^{x+1}=371$. | Solution. The least exponent is $x-1$; therefore, we factor out $7^{x-1}$:
$$
\begin{gathered}
7^{x-1} \cdot\left(7^{1}-3 \cdot 1+7^{12}\right)=371 ; 7^{x-1}(7-3+49)=371 \\
7^{x-1} \cdot 53=371 ; 7^{x-1}=7 ; x-1=1 ; x=2
\end{gathered}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
171. $7^{2 x}-48 \cdot 7^{x}=49$
171. $7^{2 x}-48 \cdot 7^{x}=49$ | Solution. Let $7^{x}=y$, we get the quadratic equation $y^{2}-$ $-48 y-49=0$. Let's solve it. Here $a=1, b=-48, c=-49 ; \quad D=b^{2}-$ $-4 a c=(-48)^{2}-4 \cdot 1(-49)=2304+196=2500 ; \sqrt{D}=50$. Using the formula $y_{1,2}=\frac{-b \pm \sqrt{D}}{2 a}$, we find
$$
y_{1}=\frac{48-50}{2}=\frac{-2}{2}=-1 ; \quad y_{2}=... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
179. $5^{2}=25$. | Solution. Since the base of the power is 5, the exponent (logarithm) is 2, and the power is 25, then $\log _{5} 25=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
231. $\log _{4}(x+3)-\log _{4}(x-1)=2-3 \log _{4} 2$. | solution. Representing the number 2 as the logarithm of 16 to the base 4, we rewrite the given equation as
$$
\log _{4}(x+3)-\log _{4}(x-1)=\log _{4} 16-3 \log _{4} 2
$$
From this, we obtain
$$
\log _{4} \frac{x+3}{x-1}=\log _{4} \frac{16}{8}, \text { or } \frac{x+3}{x-1}=2
$$
We solve this equation:
$$
x+3=2(x-1)... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
319. $\sqrt{5-x}+2=7$.
319. $\sqrt{5-x}+2=7$.
The equation is already in English, so no translation was needed for the mathematical expression. However, if you meant to have the problem solved, here is the solution:
To solve the equation $\sqrt{5-x}+2=7$, we first isolate the square root term:
1. Subtract 2 from b... | Solution. Isolate the radical and square both sides of the equation:
$$
\sqrt{5-x}=7-2 ;(\sqrt{5-x})^{2}=5^{2} ; 5-x=25 ;-x=20 ; x=-20
$$
Perform the check: $\sqrt{5-(-20)}+2=7 ; \sqrt{2} 5+2=7 ; 5+2=7$. Therefore, $x=-20$ is the solution to the equation. | -20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
320. $\sqrt{x-1} \cdot \sqrt{2 x+6}=x+3$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
320. $\sqrt{x-1} \cdot \sqrt{2 x+6}=x+3$. | Solution. First, we perform the multiplication of the roots:
$\sqrt{(x-1)(2 x+6)}=x+3 ; \sqrt{2 x^{2}-2 x+6 x-6}=x+3 ; \sqrt{2 x^{2}+4 x-6}=x+3$.
Now, we square both sides of the equation:
$\left(\sqrt{2 x^{2}+4 x-6}\right)^{2}=(x+3)^{2} ; 2 x^{2}+4 x-6=x^{2}+6 x+9 ; x^{2}-2 x-15=0$.
We solve the quadratic equation... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
321. $\sqrt{2 x+5}+\sqrt{x-1}=8$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
321. $\sqrt{2 x+5}+\sqrt{x-1}=8$. | Solution. Isolate one of the radicals and square both sides of the equation:
$$
(\sqrt{2 x+5})^{2}=(8-\sqrt{x-1})^{2} ; 2 x+5=64-16 \sqrt{x-1}+(x-1)
$$
Move $16 \sqrt{x-1}$ to the left side, and all other terms to the right side:
$$
16 \sqrt{x-1}=64+x-1-2 x-5 ; 16 \sqrt{x-1}=58-x
$$
Square both sides of the equatio... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
385. $(\tan \alpha+\cot \alpha)^{2}-(\tan \alpha-\cot \alpha)^{2}$. | Solution. We will use the formulas for the square of the sum and difference of two numbers:
$$
\begin{gathered}
(\operatorname{tg} \alpha+\operatorname{ctg} \alpha)^{2}-(\operatorname{tg} \alpha-\operatorname{ctg} \alpha)^{2}=\operatorname{tg}^{2} \alpha+2 \operatorname{tg} \alpha \operatorname{ctg} \alpha+\operatorna... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
26. Compute the second-order determinants:
a) $\left|\begin{array}{rr}2 & 5 \\ -3 & -4\end{array}\right|$;
b) $\left|\begin{array}{ll}a^{2} & a b \\ a b & b^{2}\end{array}\right|$. | Solution. a) $\left|\begin{array}{rr}2 & 5 \\ -3 & -4\end{array}\right|=2(-4)-5(-3)=-8+15=7$;
b) $\left|\begin{array}{ll}a^{2} & a b \\ a b & b^{2}\end{array}\right|=a^{2} \cdot b^{2}-a b \cdot a b=a^{2} b^{2}-a^{2} b^{2}=0$.
27-32. Calculate the determinants: | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
33. Calculate the determinants of the third order:
a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|$
b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ | Solution.
a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|=3 \cdot 5 \cdot 3+2 \cdot 3 \cdot 3+2 \cdot 4 \cdot 1-1 \cdot 5 \cdot 3-2 \cdot 2 \cdot 3-$ $-3 \cdot 3 \cdot 4=45+18+8-15-12-36=71-63=8$;
b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=a c b+... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
44. Determinant
$$
D=\left|\begin{array}{rrr}
3 & 1 & 2 \\
-1 & 2 & 5 \\
0 & -4 & 2
\end{array}\right|
$$
expand: a) by the elements of the 1st row; b) by the elements of the 2nd column. | Solution.
a) $D=3\left|\begin{array}{rr}2 & 5 \\ -4 & 2\end{array}\right|-1 \cdot\left|\begin{array}{rr}-1 & 5 \\ 0 & 2\end{array}\right|+2\left|\begin{array}{rr}-1 & 2 \\ 0 & -4\end{array}\right|=3(4+20)-$ $-1(-2-0)+2(4-0)=72+2+8=82$
b) $D=-1 \cdot\left|\begin{array}{rr}-1 & 5 \\ 0 & 2\end{array}\right|+2\left|\begin... | 82 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
45. Calculate the determinant
$$
D=\left|\begin{array}{rrrr}
3 & 0 & 2 & 0 \\
2 & 3 & -1 & 4 \\
0 & 4 & -2 & 3 \\
5 & 2 & 0 & 1
\end{array}\right|
$$ | Solution. We will expand the determinant along the elements of the 1st row (since it contains two zero elements):
$$
\begin{gathered}
D=\left|\begin{array}{rrrr}
3 & 0 & 2 & 0 \\
2 & 3 & -1 & 4 \\
0 & 4 & -2 & 3 \\
5 & 2 & 0 & 1
\end{array}\right|=3\left|\begin{array}{rrr}
3 & -1 & 4 \\
4 & -2 & 3 \\
2 & 0 & 1
\end{ar... | -54 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
142. Calculate $3.27^{3}$. | Solution. We find $3.27 \cdot 3.27 \cdot 3.27 = 34.965 \approx 35.0$. The result is rounded to three significant figures, as the base of the power contains that many significant figures. | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
265. Find $z^{6}$, if $z=-\sqrt{3}+i$. | Solution. Let's write the number $z$ in trigonometric form, considering that $a=-\sqrt{3}, b=1$. We find $r=\sqrt{a^{2}+b^{2}} ; r=\sqrt{3+1}=2$. The point $z$ is located in the second quadrant. We form the ratios
$$
\cos \varphi=a / r=\sqrt{3} / 2, \sin \varphi=b / r=1 / 2
$$
Considering that the point $z$ is locate... | -64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
25. Find the length of the vector: a) $\vec{a}=(5 ; 12) ;$ b) $\vec{b}=(7 ;-1)$. | Solution. According to formula (1), we have:
a) $|\vec{a}|=\sqrt{x^{2}+y^{2}}=\sqrt{25+144}=\sqrt{169}=13$;
b) $|\vec{b}|=\sqrt{x^{2}+y^{2}}=\sqrt{49+1}=\sqrt{50}=5 \sqrt{2}$. | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
40. In an equilateral triangle $ABC$ with a side length of 6 (Fig. 42), find the scalar product of the vectors: a) $\overrightarrow{A B}$ and $\overrightarrow{A C} ;$ b) $\overrightarrow{A B}$ and $\overrightarrow{B C}$. | Solution. a) Since the angle $\varphi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$ (and their directions) is $60^{\circ}$, for the scalar product of these vectors we get
$$
\overrightarrow{A B} \cdot \overrightarrow{A C}=|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cdot \cos B \widehat... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
45. Find the scalar product of vectors $\vec{a}=(3 ; 5)$ and $\vec{b}=(-2 ; 7)$. | Solution. Here $x_{a}=3, x_{b}=-2, y_{a}=5, y_{b}=7$. Using formula (3), we get
$$
\vec{a} \vec{b}=3 \cdot(-2)+5 \cdot 7=-6+35=29
$$
46-51. Find the scalar product of the vectors: | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Determine the value of the function $f(x)=2 x^{2}-1$ at $x=3$. | Solution. Find $f(3)=y_{x=3}=2 \cdot 3^{2}-1=17$. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
92. Show that as $t \rightarrow \infty$ the limit of the variable
$x=\frac{6 t^{3}-9 t+1}{2 t^{3}-3 t}$ is 3. | Solution. We find the difference between the variable $x$ and the number 3:
$$
\begin{gathered}
x-3=\frac{6 t^{3}+9 t+1}{2 t^{3}+3 t}-3=\frac{6 t^{3}+9 t+1-3\left(2 t^{3}+3 t\right)}{2 t^{3}+3 t}=\frac{6 t^{3}+9 t+1-6 t^{3}-9 t}{2 t^{3}+3 t}= \\
=\frac{1}{2 t^{3}+3 t}
\end{gathered}
$$
If $t \rightarrow \infty$, then... | 3 | Calculus | proof | Yes | Yes | olympiads | false |
94. Find $\lim _{x \rightarrow 2}\left(3 x^{2}-2 x\right)$. | Solution. Using properties 1, 3, and 5 of limits sequentially, we get
$$
\begin{gathered}
\lim _{x \rightarrow 2}\left(3 x^{2}-2 x\right)=\lim _{x \rightarrow 2}\left(3 x^{2}\right)-\lim _{x \rightarrow 2}(2 x)=3 \lim _{x \rightarrow 2} x^{2}-2 \lim _{x \rightarrow 2} x= \\
=3\left(\lim _{x \rightarrow 2} x\right)^{2}... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
95. Find $\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}$. | Solution. To apply the property of the limit of a quotient, we need to check whether the limit of the denominator is not zero when $x=4$. Since $\lim _{x \rightarrow 4}(x-3)=$[^3]$=\lim _{x \rightarrow 4} x-\lim 3=4-3=1 \neq 0$, in this case, we can use property 4:
$$
\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}=\fra... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
115. Find $\lim _{x \rightarrow 3} \frac{x^{2}-9}{3-x}$. | Solution. A direct transition to the limit is impossible here, since the limit of the denominator is zero: $\lim _{x \rightarrow 3}(3-x)=3-3=0$. The limit of the dividend is also zero: $\lim _{x \rightarrow 3}\left(x^{2}-9\right)=9-9=0$. Thus, we have an indeterminate form of $0 / 0$. However, this does not mean that t... | -6 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
116. Find $\lim _{x \rightarrow \infty} \frac{3}{x+5}$. | Solution. When $x \rightarrow \infty$, the denominator $x+5$ also tends to infinity, and its reciprocal $\frac{1}{x+5} \rightarrow 0$. Therefore, the product $\frac{1}{x+5} \cdot 3=\frac{3}{x+5}$ tends to zero if $x \rightarrow \infty$. Thus, $\lim _{x \rightarrow \infty} \frac{3}{x+5}=0$.
Identical transformations un... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
117. Find $\lim _{x \rightarrow \infty} \frac{2 x^{3}+x}{x^{3}-1}$. | Solution. Here, the numerator and denominator do not have a limit, as both increase indefinitely. In this case, it is said that there is an indeterminate form of $\infty / \infty$. We will divide the numerator and denominator term by term by $x^{3}$ (the highest power of $x$ in this fraction):
$$
\lim _{x \rightarrow ... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
118. Find $\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}$. | Solution. We will transform this limit into the form (1). For this, we multiply the numerator and the denominator of the fraction by 2, and take the constant factor 2 outside the limit sign. We have
$$
\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}=\lim _{\alpha \rightarrow 0} \frac{2 \sin 2 \alpha}{2 \alph... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
120. Find $\lim _{x \rightarrow 3}\left(x^{2}-7 x+4\right)$. | Solution. To find the limit of the given function, we will replace the argument $x$ with its limiting value:
$$
\lim _{x \rightarrow 3}\left(x^{2}-7 x+4\right)=3^{2}-7 \cdot 3+4=-8
$$ | -8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
135. Find $\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}$. | Solution. As the argument $x$ tends to infinity, we have an indeterminate form of $\infty / \infty$. To resolve it, we divide the numerator and the denominator of the fraction by $x$. Then we get
$$
\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}=\lim _{x \rightarrow \infty} \frac{\sqrt{\frac{x^{2}+4}{x^{2}}}}{1... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
136. Find $\lim _{x \rightarrow \infty} \frac{3 x}{\sqrt{x^{2}-2 x+3}}$. | Solution. The limit transition as $x \rightarrow \infty$ can always be replaced by the limit transition as $\alpha \rightarrow 0$, if we set $\alpha=1 / x$ (the method of variable substitution).
Thus, setting $x=1 / \alpha$ in this case, we find that $\alpha \rightarrow 0$ as $x \rightarrow \infty$. Therefore,
$$
\be... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
137. Find $\lim _{x \rightarrow \infty} \frac{3 x^{2}+5 x+1}{x^{2}-2}$. | Solution. Method I. Dividing the numerator and the denominator by $x^{2}$, we find
$$
\lim _{x \rightarrow \infty} \frac{3 x^{2}+5 x+1}{x^{2}-2}=\lim _{x \rightarrow \infty} \frac{3+\frac{5}{x}+\frac{1}{x^{2}}}{1-\frac{2}{x^{2}}}=\frac{3}{1}=3
$$
Method II. Let $x=1 / a$; then $a \rightarrow 0$ as $x \rightarrow \inf... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
142. Find $\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})$. | Solution. Here, it is required to find the limit of the difference of two quantities tending to infinity (indeterminate form of $\infty-\infty$). By multiplying and dividing the given expression by its conjugate, we get
$$
\sqrt{x+1}-\sqrt{x}=\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}}=\frac... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
145. Find $\lim _{x \rightarrow 0} \frac{1-\cos 8 x}{2 x^{2}}$. | Solution. Transform the numerator to the form $1-\cos 8 x=2 \sin ^{2} 4 x$. Next, we find
$$
\begin{gathered}
\lim _{x \rightarrow 0} \frac{1-\cos 8 x}{2 x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 4 x}{2 x^{2}}=\lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x} \cdot \frac{\sin 4 x}{x}\right)= \\
=\lim _{x \rightar... | 16 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
168. Find the derivative of the function $y=5x$. | Solution. $1^{0} . y_{\mathrm{H}}=5(x+\Delta x)=5 x+5 \Delta x$.
$2^{0} . \Delta y=y_{\mathrm{k}}-y=(5 x+5 \Delta x)-5 x=5 \Delta x$.
$3^{0} . \frac{\Delta y}{\Delta x}=\frac{5 \Delta x}{\Delta x}=5$. | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
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