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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
139. $16^{x}=1 / 4$.
139. $16^{x}=1 / 4$.
(Note: The translation is the same as the original text because it is a mathematical equation, which is universal and does not change in translation.) | Solution. Since $16=2^{4}, 1 / 4=2^{-2}$, the equation will take the form $2^{4 x}=2^{-2}$, from which $4 x=-2$, i.e., $x=-1 / 2$. | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,744 |
140. $\sqrt{5^{x}}=\sqrt[3]{25}$
140. $\sqrt{5^{x}}=\sqrt[3]{25}$ | Solution. We have $\sqrt{5^{x}}=5^{x / 2} ; \sqrt[3]{2} 5=\sqrt[3]{5^{2}}=5^{2 / 3} ;$ therefore, $5^{x / 2}=5^{2 / 3} ; x / 2=2 / 3 ; x=4 / 3$. | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,745 |
141. $5,2^{(x+2)(x+3)}=1$.
141. $5,2^{(x+2)(x+3)}=1$. | Solution. Any non-zero number to the power of zero equals one; therefore, we can write $1=5.2^{0}$. Thus, $5.2^{(x-2)(x+3)}=5.2^{0}$, from which $(x-2)(x+3)=0$. According to the property of multiplication, $x-2=0$ or $x+3=0$, i.e., $x=2, x=-3$. | 2,-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,746 |
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
$$ | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,748 |
162. $7^{x}-3 \cdot 7^{x-1}+7^{x+1}=371$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
162. $7^{x}-3 \cdot 7^{x-1}+7^{x+1}=371$. | Solution. The least exponent is $x-1$; therefore, we factor out $7^{x-1}$:
$$
\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}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,749 |
171. $7^{2 x}-48 \cdot 7^{x}=49$
171. $7^{2 x}-48 \cdot 7^{x}=49$ | Solution. Let $7^{x}=y$, we get the quadratic equation $y^{2}-$ $-48 y-49=0$. Let's solve it. Here $a=1, b=-48, c=-49 ; \quad D=b^{2}-$ $-4 a c=(-48)^{2}-4 \cdot 1(-49)=2304+196=2500 ; \sqrt{D}=50$. Using the formula $y_{1,2}=\frac{-b \pm \sqrt{D}}{2 a}$, we find
$$
y_{1}=\frac{48-50}{2}=\frac{-2}{2}=-1 ; \quad y_{2}=... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,750 |
172. $5 \cdot 5^{2x}-6 \cdot 5^{x}+1=0$ | Solution. Let $5^{x}=y;$ then we get $5 y^{2}-6 y+1=0$. Here $a=5, \quad b=-6, \quad c=1 ; \quad D=b^{2}-4 a c=36-4 \cdot 5 \cdot 1=36-20=16$, $\sqrt{D}=4$. Therefore, $y_{1}=\frac{6-4}{10}=\frac{2}{10}=\frac{1}{5}, y_{2}=\frac{6+4}{10}=\frac{10}{10}=1$.
Since $y_{1}=1 / 5, y_{2}=1$, we have $5^{x}=1 / 5,5^{x}=1 ;$ th... | -1,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,751 |
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 | 31,752 |
184. $\log _{10} 1000=3$.
184. $\log _{10} 1000=3$. | Solution. Here the base of the power is 10, the exponent is 3, and the number being logarithmed is 1000. Therefore, $10^{3}=1000$. | 10^{3}=1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,753 |
189. a) $\log _{2} 16$
b) $\log _{6} 36$
c) $\log _{8} 1$. | Solution. a) Here we need to find such an exponent $x$ that $2^{x}=16$. Solving this equation, we get $2^{x}=2^{4}$, hence $x=4$. Therefore, $\log _{2} 16=4$.
b) From the equation $6^{x}=36$ we find $6^{x}=6^{2}$, i.e., $x=2$. Thus, $\log _{6} 36=2$.
c) In this case, we have the equation $8^{x}=1$. This is only possi... | 4,2,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,754 |
199. a) $\log _{4} x=-3$;
b) $\log _{x} \frac{1}{8}=\frac{3}{2}$. | Solution. a) By the definition of logarithm, we write $4^{-3}=x$, from which $x=\frac{1}{64}$.
6) According to the definition of logarithm, we obtain the equation $x^{3 / 2}=\frac{1}{8}$. Since $\frac{1}{8}=2^{-3}$ and $x^{3 / 2}=\sqrt{x^{3}}$, the equation becomes $\sqrt{x^{3}}=2^{-3}$. Squaring both sides:
$$
\left... | \frac{1}{64} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,755 |
210. $x=\frac{a b}{c^{3}}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
210. $x=\frac{a b}{c^{3}}$. | Solution. Applying theorem 2 first, and then theorems 1 and 3, we get
$$
\log x=\log (a b)-\log \left(c^{3}\right)=\log a+\log b-3 \log c .
$$
Here and in the following examples, we do not write the base of the logarithm, as the obtained equalities are valid for any base. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,756 | |
212. $x=\frac{a^{2}(a+b)^{3}}{(a-b)^{2} c^{3}}$.
212. $x=\frac{a^{2}(a+b)^{3}}{(a-b)^{2} c^{3}}$. | Solution. Applying theorems 2,1 and 3, we get
$\log x=\log \left[a^{2}(a+b)^{3}\right]-\log \left[(a-b)^{2} c^{3}\right]=\log a^{2}+\log (a+b)^{3}-\log (a-b)^{2}-$
$$
-\log c^{3}=2 \log a+3 \log (a+b)-2 \log (a-b)-3 \log c
$$ | 2\log+3\log(+b)-2\log(-b)-3\log | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,757 |
221. $\log x = \log a + \log b - \log c$. | Solution. By the converse statements of Theorems 1 and 2, we write $\log x=\log \frac{a b}{c}$, from which $x=\frac{a b}{c}$. | \frac{}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,758 |
222. $\log x=3 \log a+2 \log (a+b)-\frac{1}{2} \log c$. | Solution. According to the converse theorems $3,4,1$ and 2, we obtain
$$
\begin{gathered}
\log x=\log a^{3}+\log (a+b)^{2}-\log \sqrt{\bar{c}}=\log \frac{a^{3}(a+b)^{2}}{\sqrt{\bar{c}}} ; \log x=\log \frac{a^{3}(a+b)}{\sqrt{\bar{c}}} ; \\
x=\frac{a^{3}(a+b)}{\sqrt{c}} .
\end{gathered}
$$ | \frac{^{3}(+b)}{\sqrt{}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,759 |
223. $\log x=\frac{1}{3}(\log a+\log b)-\frac{1}{2} \log (a+c)$. | Solution. Using the converse statements of Theorems 1, 3, 4, and 2, we have
$$
\begin{gathered}
\log x=\frac{1}{3} \log (a b)-\frac{1}{2} \log (a+c)=\log \sqrt[3]{a b}-\log \sqrt{a+c}=\log \frac{\sqrt[3]{a b}}{\sqrt{a+c}} ; \\
\log x=\log \frac{\sqrt[3]{a b}}{\sqrt{a+c}} ; \quad x=\frac{\sqrt[3]{a b}}{\sqrt{a+c}}
\end... | \frac{\sqrt[3]{}}{\sqrt{+}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,760 |
229. $\log _{3}(12 x+4)-\log _{3}(x-7)=\log _{3} 9$. | Solution. Let's write the given equation in the form
$$
\log _{3} \frac{12 x+4}{x-7}=\log _{3} 9
$$
Since the logarithms and their bases are equal, the numbers being logarithmed are also equal:
$$
\frac{12 x+4}{x-7}=9
$$
Assuming $x-7 \neq 0$, we bring the fraction to a common denominator and solve the resulting eq... | Nosolution | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,761 |
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:
$$
x+3=2(x-1)... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,762 |
232. $\frac{1}{12} \lg ^{2} x=\frac{1}{3}-\frac{1}{4} \lg x$. | Solution. This is a logarithmic equation reducible to a quadratic. By setting $\lg x=z$, we obtain the equation
$$
\frac{1}{12} z^{2}=\frac{1}{3}-\frac{1}{4} z \text { or } z^{2}+3 z-4=0
$$
Here $a=1, b=3, c=-4, \quad D=b^{2}-4 a c=9-4 \cdot 1(-4)=9+16=25$; $\sqrt{D}=5$. Using the formula $z=\frac{-b \pm \sqrt{D}}{2 ... | x_{1}=0.0001,x_{2}=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,763 |
253. Extract the root: а) $\sqrt{4 \cdot 9}$; б) $\sqrt{\frac{49}{36}}$; в) $\sqrt[3]{a^{6}}$; г) $\sqrt{9 a^{2}}$. | Solution. a) According to rule I, $\sqrt{4 \cdot 9}=\sqrt{4} \cdot \sqrt{9}=2 \cdot 3=6$.
b) Using rule II, we find $\sqrt{\frac{49}{36}}=\frac{\sqrt{49}}{\sqrt{36}}=\frac{7}{6}$.
c) By rule III, $\sqrt[3]{a^{6}}=a^{2}$.
d) According to rules I, III, $\sqrt{9 a^{2}}=\sqrt{9} \cdot \sqrt{a^{2}}=3 a$. 254-257. Extract... | 6,\frac{7}{6},^{2},3a | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,764 |
266. a) $\sqrt{8} ;$ b) $\sqrt[3]{a^{8}} ;$ c) $\sqrt[3]{16 x^{4}}$. | Solution. a) $\sqrt{8}=\sqrt{4 \cdot 2}=\sqrt{4} \cdot \sqrt{2}=2 \sqrt{2}$;
6) $\sqrt[3]{a^{8}}=\sqrt[3]{a^{6} \cdot a^{2}}=\sqrt[3]{a^{6}} \cdot \sqrt[3]{a^{2}}=a \sqrt[3]{a^{2}}$
c) $\sqrt[3]{16 x^{4}}=\sqrt[3]{8 x^{3} \cdot 2 x}=\sqrt[3]{8 x^{3}} \cdot \sqrt[3]{2 x}=2 x \sqrt[3]{2 x}$. | )2\sqrt{2};b)\sqrt[3]{^2};)2x\sqrt[3]{2x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,765 |
288. $2 \sqrt{8}-7 \sqrt{18}+5 \sqrt{72}-\sqrt{50}$. | Solution. Transform the roots: $2 \sqrt{8}=2 \sqrt{4 \cdot 2}=4 \sqrt{2} ; \quad 7 \sqrt{18}=$ $=7 \sqrt{9 \cdot 2}=7 \cdot 3 \sqrt{2}=21 \sqrt{2} ; \quad 5 \sqrt{72}=5 \sqrt{36 \cdot 2}=5 \cdot 6 \sqrt{2}=30 \sqrt{2} ; \quad \sqrt{50}=$ $=\sqrt{25 \cdot 2}=5 \sqrt{2}$.
Substitute the obtained expressions and combine ... | 8\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,768 |
289. a) $\sqrt[\sqrt{a}]{a} \cdot \sqrt[3]{a} ;$ b) $\left(\sqrt[5]{x^{2}} \cdot \sqrt{x}\right)^{10} ;$ c) $\left(\sqrt[3]{a^{2} b}\right)^{4}.$ | Solution. a) We have $\sqrt{a}=\sqrt[6]{a^{3}} ; \quad \sqrt[3]{a}=\sqrt[6]{a^{2}} ; \quad$ therefore, $\sqrt{a} \cdot \sqrt[3]{a}=\sqrt[6]{a^{3}} \cdot \sqrt[6]{a^{2}}=\sqrt[6]{a^{3}}$.
b) Raise the radicands to the power:
$$
\left(\sqrt[5]{x^{2}} \cdot \sqrt{x}\right)^{10}=\sqrt[5]{x^{20}} \cdot \sqrt{x^{10}}=x^{4}... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,769 | |
308. a) $\frac{1}{\sqrt{2}}$;
b) $\frac{3}{\sqrt{40}}$;
c) $\frac{2}{\sqrt[5]{3^{2}}}$;
d) $\frac{2}{3+\sqrt{5}}$. | Solution. a) to get rid of the root, multiply the numerator and denominator by $\sqrt{2}$ :
$$
\frac{1}{\sqrt{2}}=\frac{1 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}=\frac{\sqrt{2}}{\sqrt{4}}=\frac{\sqrt{2}}{2}
$$
b) Multiplying the numerator and denominator by $\sqrt{10}$ and considering that $\sqrt{40}=$ $=\sqrt{4 \cd... | \frac{\sqrt{2}}{2},\frac{3\sqrt{10}}{20},\frac{2}{3}\sqrt[5]{27},\frac{3-\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,770 |
318. $\sqrt{x^{2}-1}=\sqrt{3}$. | Solution. Let's square both sides of the equation:
$$
\left(\sqrt{x^{2}-1}\right)^{2}=(\sqrt{3})^{2} ; x^{2}-1=3 ; x^{2}=4 ; x_{1}=2, x_{2}=-2
$$
We have obtained two solutions. Let's check each of them: if $x=2$, then $\sqrt{4-1}=\sqrt{3}$, i.e., $\sqrt{3}=\sqrt{3}$; if $x=-2$, then $\sqrt{4-1}=\sqrt{3}$, i.e., $\sq... | x_1=2,x_2=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,771 |
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 | 31,772 |
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 | 31,773 |
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 | 31,774 |
336. Write in degree measure the angles: a) $\pi / 6$; b) $\pi / 8$; c) $3 \pi / 4$. | Solution. Applying formula (1), we get:
a) $\alpha=\frac{180^{\circ}(\pi / 6)}{\pi}=\frac{180^{\circ}}{6}=30^{\circ}$;
b) $\alpha=\frac{180^{\circ}(\pi / 8)}{\pi}=\frac{180^{\circ}}{8}=22^{\circ} 30^{\prime}$;
c) $\alpha=\frac{180^{\circ}(3 \pi / 4)}{\pi}=\frac{180^{\circ} \cdot 3 \pi}{\pi \cdot 4}=135^{\circ}$. | 30,2230^{\},135 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,775 |
345. $\sin 35^{\circ} ; \cos 167^{\circ} ; \tan 3 ; \cot(-1.5)$. | Solution. Since $35^{\circ}$ is an angle in the I quadrant, then $\sin 35^{\circ}>0$; next, $167^{\circ}$ is an angle ending in the II quadrant, and thus $\cos 167^{\circ}<0$; 3 radians is an angle in the II quadrant, so $\operatorname{tg} 3<0$; finally, $-1.5$ radians is an angle ending in the IV quadrant (measured al... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,778 |
358. $3 \sin \frac{\pi}{3}-2 \cos \frac{\pi}{6}+3 \tan \frac{\pi}{3}-4 \cot \frac{\pi}{2}$. | Solution. From the table, we take the values $\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} ; \cos \frac{\pi}{6}=$ $=\frac{\sqrt{3}}{2} ; \operatorname{tg} \frac{\pi}{3}=\sqrt{3}, \operatorname{ctg} \frac{\pi}{2}=0$ and substitute them into the given expression:
$$
3 \cdot \frac{\sqrt{3}}{2}-2 \cdot \frac{\sqrt{3}}{2}+3 \sqrt... | \frac{7\sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,779 |
359. $4 a^{2} \sin ^{4} \frac{\pi}{6}-6 a b \tan^{2} \frac{\pi}{6}+\left(b \cot \frac{\pi}{4}\right)^{2}$. | Solution. Since $\sin \frac{\pi}{6}=\frac{1}{2}, \operatorname{tg} \frac{\pi}{6}=\frac{\sqrt{3}}{3}, \operatorname{ctg} \frac{\pi}{4}=1$, we get $4 a^{2} \sin ^{4} \frac{\pi}{6}-6 a b \operatorname{tg}^{2} \frac{\pi}{6}+\left(b \operatorname{ctg} \frac{\pi}{4}\right)^{2}=4 a^{2}\left(\frac{1}{2}\right)^{4}-6 a b\left(\... | \frac{^{2}}{4}-2+b^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,780 |
371. Calculate: a) $\sin 110^{\circ}$; b) $\operatorname{tg} 945^{\circ}$; c) $\cos \frac{25 \pi}{4}$. | Solution. a) The period of the function $y=\sin x$ is $360^{\circ}$; therefore, we can omit the integer number of periods:
$$
\sin 1110^{\circ}=\sin \left(360^{\circ} \cdot 3+30^{\circ}\right)=\sin 30^{\circ}=\frac{1}{2}
$$
b) Since the period of the function $y=\operatorname{tg} x$ is $180^{\circ}$, then
$$
\operat... | \frac{1}{2},1,\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,781 |
378. Calculate: a) $\sin \left(-60^{\circ}\right)$; b) $\cos \left(-45^{\circ}\right)$; c) $\operatorname{tg}\left(-945^{\circ}\right)$. | Solution. a) Since $\sin x$ is an odd function, then $\sin \left(-60^{\circ}\right)=$ $=-\sin 60^{\circ}$. Therefore, $\sin \left(-60^{\circ}\right)=-\sin 60^{\circ}=-\sqrt{3} / 2$.
6) The function $\cos x$ is even; therefore, the minus sign can be omitted, i.e., $\cos \left(-45^{\circ}\right)=\cos 45^{\circ}$. Theref... | -\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{2},-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,782 |
384. $\frac{1-\cos ^{2} x}{1-\sin ^{2} x}+\operatorname{tg} x \operatorname{ctg} x$.
384. $\frac{1-\cos ^{2} x}{1-\sin ^{2} x}+\tan x \cot x$. | Solution. Applying the formulas $1-\cos ^{2} \alpha=\sin ^{2} \alpha, 1-\sin ^{2} \alpha=$ $=\cos ^{2} \alpha$ and identities IV and VI, we find
$$
\frac{1-\cos ^{2} x}{1-\sin ^{2} x}+\operatorname{tg} x \operatorname{ctg} x=\frac{\sin ^{2} x}{\cos ^{2} x}+1=\operatorname{tg}^{2} x+1=\frac{1}{\cos ^{2} x}
$$ | \frac{1}{\cos^{2}x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,783 |
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:
$$
\begin{gathered}
(\operatorname{tg} \alpha+\operatorname{ctg} \alpha)^{2}-(\operatorname{tg} \alpha-\operatorname{ctg} \alpha)^{2}=\operatorname{tg}^{2} \alpha+2 \operatorname{tg} \alpha \operatorname{ctg} \alpha+\operatorna... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,784 |
386. $\sqrt{\sin ^{2} \alpha(1+\operatorname{ctg} \alpha)+\cos ^{2} \alpha(1+\operatorname{tg} \alpha)}$.
386. $\sqrt{\sin ^{2} \alpha(1+\cot \alpha)+\cos ^{2} \alpha(1+\tan \alpha)}$. | $$
\begin{gathered}
\sqrt{\sin ^{2} \alpha(1+\operatorname{ctg} \alpha)+\cos ^{2} \alpha(1+\operatorname{tg} \alpha)}= \\
=\sqrt{\sin ^{2} \alpha+\sin ^{2} \alpha \operatorname{ctg} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha \operatorname{tg} \alpha}= \\
=\sqrt{\sin ^{2} \alpha+\cos ^{2} \alpha+\sin ^{2} \alpha \frac{\co... | \sin\alpha+\cos\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,785 |
395. Find the values of $\cos x, \operatorname{tg} x, \operatorname{ctg} x$, if it is known that $\sin x=-3 / 5, 0<x<3 \pi / 2$. | Solution. Since the angle $x$ ends in the III quadrant, then $\cos x < 0$ and $\operatorname{ctg} x > 0$.
Using the value of $\sin x$ and the formula $\cos ^{2} x=1-\sin ^{2} x$, we find $\cos ^{2} x=1-\left(-\frac{3}{5}\right)^{2}=1-\frac{9}{25}=\frac{16}{25}$, from which $\cos x=-\frac{4}{5} ; \operatorname{tg} x=\f... | \cosx=-\frac{4}{5},\operatorname{tg}x=\frac{3}{4},\operatorname{ctg}x=\frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,786 |
396. Find the values of $\sin x, \cos x, \operatorname{ctg} x$, if it is known that $\operatorname{tg} x=-8 / 15,3 \pi / 2<x<2 \pi$. | Solution. First, find $\operatorname{ctg} x = -15 / 8$. Since the angle $x$ ends in the IV. quadrant, we conclude that $\sin x < 0$. According to identity V, we have $1 + \operatorname{tg}^{2} x = \frac{1}{\cos ^{2} x}$, from which $\cos ^{2} x = \frac{1}{1 + \operatorname{tg}^{2} x}$. Therefore, $\cos ^{2} x = \frac{1... | \sinx=-\frac{8}{17},\cosx=\frac{15}{17},\operatorname{ctg}x=-\frac{15}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,787 |
402. Simplify the expression
$$
\frac{\cos \alpha \cos \beta-\cos (\alpha+\beta)}{\cos (\alpha-\beta)-\sin \alpha \sin \beta}
$$ | Solution. Applying formulas (5) and (6), we get
$$
\begin{gathered}
\frac{\cos \alpha \cos \beta - \cos (\alpha + \beta)}{\cos (\alpha - \beta) - \sin \alpha \sin \beta} = \frac{\cos \alpha \cos \beta - \cos \alpha \cos \beta + \sin \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta - \sin \alpha \sin ... | \operatorname{tg}\alpha\operatorname{tg}\beta | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,788 |
403. Simplify the expression
$$
\frac{\sin 11^{\circ} \cos 15^{\circ}+\sin 15^{\circ} \cos 11^{\circ}}{\sin 18^{\circ} \cos 12^{\circ}+\sin 12^{\circ} \cos 18^{\circ}}
$$ | Solution. The numerator and denominator are expanded expressions of the sine of a sum according to formula (3). Therefore, $\frac{\sin 11^{\circ} \cos 15^{\circ}+\sin 15^{\circ} \cos 11^{\circ}}{\sin 18^{\circ} \cos 12^{\circ}+\sin 12^{\circ} \cos 18^{\circ}}=\frac{\sin \left(11^{\circ}+15^{\circ}\right)}{\sin \left(18... | 2\sin26 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,789 |
404. Prove the identity
$$
\frac{\sin \left(\frac{\pi}{6}+\alpha\right)+\sin \left(\frac{\pi}{6}-\alpha\right)}{\sin \left(\frac{\pi}{4}+\alpha\right)+\sin \left(\frac{\pi}{4}-\alpha\right)}=\frac{\sqrt{2}}{2}
$$ | Solution. Using formulas (3) and (4), we transform the left side of the identity:
$$
\begin{gathered}
\frac{\sin \left(\frac{\pi}{6}+\alpha\right)+\sin \left(\frac{\pi}{6}-\alpha\right)}{\sin \left(\frac{\pi}{4}+\alpha\right)+\sin \left(\frac{\pi}{4}-\alpha\right)}=\frac{\sin \frac{\pi}{6} \cos \alpha+\sin \alpha \cos... | \frac{\sqrt{2}}{2} | Algebra | proof | Yes | Yes | olympiads | false | 31,790 |
415. Calculate $\sin 210^{\circ}$. | Solution. Represent $210^{\circ}$ as $180^{\circ}+30^{\circ}$. Applying the $1^{\circ}$ and $2^{\circ}$ rules and considering that the angle $210^{\circ}$ ends in the III quadrant, we find
$$
\sin 210^{\circ}=\sin \left(180^{\circ}+30^{\circ}\right)=-\sin 30^{\circ}=-1 / 2
$$ | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,791 |
416. Calculate $\cos 300^{\circ}$. | Solution. Since $300^{\circ}=270^{\circ}+30^{\circ}$ and the given angle ends in the IV quadrant, then $\cos 300^{\circ}=\cos \left(270^{\circ}+30^{\circ}\right)=\sin 30^{\circ}=1 / 2$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,792 |
417. Calculate $\sin \frac{53 \pi}{6}$. | Solution. We have $\frac{53 \pi}{6}=8 \frac{5}{6}$ cycles. By omitting the whole number of periods, we get
$$
\sin \frac{53 \pi}{6}=\sin \frac{5 \pi}{6}=\sin \left(\pi-\frac{\pi}{6}\right)=\sin \frac{\pi}{6}=\frac{1}{2}
$$ | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,793 |
418. Calculate $\operatorname{tg}\left(-300^{\circ}\right)$. | Solution. The function $y=\tan x$ is odd, so $\tan\left(-300^{\circ}\right)=$ $=-\tan 300^{\circ}$. Since $300^{\circ}=270^{\circ}+30^{\circ}$ and the angle $300^{\circ}$ ends in the IV quadrant, then
$$
\tan\left(-300^{\circ}\right)=-\tan 300^{\circ}=-\tan\left(270^{\circ}+30^{\circ}\right)=\cot 30^{\circ}=\sqrt{3} .... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,794 |
419. Calculate $\sin \left(-\frac{5 \pi}{3}\right)+\cos \left(-\frac{5 \pi}{4}\right)+\operatorname{tg}\left(-\frac{11 \pi}{6}\right)+$ $+\operatorname{ctg}\left(-\frac{4 \pi}{3}\right)$. | Solution. Using first the properties of even and odd functions, and then the reduction formulas, we find
$$
\sin \left(-\frac{5 \pi}{3}\right)+\cos \left(-\frac{5 \pi}{4}\right)+\operatorname{tg}\left(-\frac{11 \pi}{6}\right)+\operatorname{ctg}\left(-\frac{4 \pi}{3}\right)=
$$
$$
\begin{gathered}
=-\sin \frac{5 \pi}{... | \frac{\sqrt{3}-\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,795 |
433. $2 \sin \alpha \cos \alpha\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)$. | Solution. According to formulas (9) and (10), we have $2 \sin \alpha \cos \alpha\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)=\sin 2 \alpha \cos 2 \alpha$.
Multiplying and dividing the product by 2, we get
$$
\sin 2 \alpha \cos 2 \alpha=\frac{2 \sin 2 \alpha \cos 2 \alpha}{2}=\frac{\sin 4 \alpha}{2} \text {. }
$$ | \frac{\sin4\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,796 |
434.
$$
\frac{1-\cos 2 \alpha+\sin 2 \alpha}{1+\cos 2 \alpha+\sin 2 \alpha}
$$ | Solution. Applying formulas (13), (12), and (9) sequentially, factoring out the common factor, and transforming, we find
$\frac{(1-\cos 2 \alpha)+\sin 2 \alpha}{(1+\cos 2 \alpha)+\sin 2 \alpha}=\frac{2 \sin ^{2} \alpha+2 \sin \alpha \cos \alpha}{2 \cos ^{2} \alpha+2 \sin \alpha \cos \alpha}=\frac{2 \sin \alpha(\sin \a... | \operatorname{tg}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,797 |
## 6. Formulas for Adding Like Functions
Formulas for adding like trigonometric functions allow the transformation of the sum and difference of functions into the product of these functions. They have the following form:
$$
\begin{aligned}
& \sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta... | Solution. We will use formula (20):
$$
\begin{gathered}
\sin 75^{\circ}+\sin 15^{\circ}=2 \sin \frac{75^{\circ}+15^{\circ}}{2} \cos \frac{75^{\circ}-15^{\circ}}{2}=2 \sin 45^{\circ} \cos 30^{\circ}= \\
=2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{2}
\end{gathered}
$$ | \frac{\sqrt{6}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,798 |
448. $\frac{\sin 7 \alpha-\sin 5 \alpha}{\sin 7 \alpha+\sin 5 \alpha}$. | Solution. Applying formulas (20) and (21), we get
$$
\frac{\sin 7 \alpha-\sin 5 \alpha}{\sin 7 \alpha+\sin 5 \alpha}=\frac{2 \sin \frac{7 \alpha-5 \alpha}{2} \cos \frac{7 \alpha+5 \alpha}{2}}{2 \sin \frac{7 \alpha+5 \alpha}{2} \cos \frac{7 \alpha-5 \alpha}{2}}=\frac{\sin \alpha \cos 6 \alpha}{\sin 6 \alpha \cos \alpha... | \operatorname{tg}\alpha\operatorname{ctg}6\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,799 |
449. $\frac{\cos \alpha - \cos 3 \alpha + \cos 5 \alpha - \cos 7 \alpha}{\sin \alpha + \sin 3 \alpha + \sin 5 \alpha + \sin 7 \alpha}$. | Solution. Grouping the terms and applying formulas (23) and (20), we factor out the common factors and simplify the fraction:
$$
\begin{gathered}
\frac{(\cos \alpha-\cos 3 \alpha)+(\cos 5 \alpha-\cos 7 \alpha)}{(\sin \alpha+\sin 3 \alpha)+(\sin 5 \alpha+\sin 7 \alpha)}= \\
=\frac{\left(-2 \sin \frac{\alpha-3 \alpha}{2... | \operatorname{tg}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,800 |
459. a) $\sin x=\frac{\sqrt{3}}{2}$; b) $\cos x=\frac{\sqrt{2}}{2}$; c) $\operatorname{tg} x=3$; d) $\operatorname{ctg} x=-1$. | Solution. a) According to formula (24), we get
$$
x=(-1)^{k} \arcsin \frac{\sqrt{3}}{2}+\pi k=(-1)^{k} \frac{\pi}{3}+\pi k, \quad k \in \mathbf{Z}
$$
b) By formula (25) we find
$$
x= \pm \arccos \frac{\sqrt{2}}{2}+2 \pi k= \pm \frac{\pi}{4}+2 \pi k, \quad k \in \mathbf{Z}
$$
c) In accordance with formula (26) we ha... | )(-1)^{k}\frac{\pi}{3}+\pik,\quadk\in{Z};\quadb)\\frac{\pi}{4}+2\pik,\quadk\in{Z};\quad)\operatorname{arcctg}3+\pik,\quadk\in{Z};\quad) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,802 |
469. $2 \sin ^{2} x-3 \sin x-2=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
469. $2 \sin ^{2} x-3 \sin x-2=0$. | Solution. Let $\sin x=y$, we get the quadratic equation $2 y^{2}-3 y-2=0$. Here $a=2, b=-3, c=-2, \quad D=b^{2}-4 a c=(-3)^{2}-$ $-4 \cdot 2 \cdot(-2)=9+16=25, \sqrt{D}=5$. Therefore, $y_{1}=\frac{3-5}{4}=-\frac{1}{2}, \quad y_{2}=$ $=\frac{3+5}{4}=2$.
Thus, $y_{1}=-\frac{1}{2}$ and $y_{2}=2$, i.e., $\sin x=-\frac{1}{... | x_{1}=(-1)^{k+1}\frac{\pi}{6}+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,803 |
470. $\sin ^{2} x+2 \sin x \cos x=3 \cos ^{2} x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
470. $\sin ^{2} x+2 \sin x \cos x=3 \cos ^{2} x$. | Solution. This is a homogeneous equation in terms of $\sin x$ and $\cos x$; dividing all its terms by $\cos ^{2} x \neq 0$, we get
$$
\frac{\sin ^{2} x}{\cos ^{2} x}+\frac{2 \sin x \cos x}{\cos ^{2} x}-\frac{3 \cos ^{2} x}{\cos ^{2} x}=0 ; \quad \operatorname{tg}^{2} x+2 \operatorname{tg} x-3=0
$$
Letting $\operatorn... | x_{1}=-\operatorname{arctg}3+\pik,x_{2}=\pi/4+\pik,k\in{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,804 |
471. $\cos x + \cos 3x + \cos 5x = 0$. | Solution. Grouping the first and last terms and applying the cosine sum formula, we get
$$
(\cos x+\cos 5 x)+\cos 3 x=0 ; 2 \cos 3 x \cos x+\cos 3 x=0
$$
Therefore, $\cos 3 x(2 \cos x+1)=0$, from which $\cos 3 x=0$ or $2 \cos x+1=0$. Solving the equation $\cos 3 x=0$, we find $3 x= \pm \arccos 0+2 \pi k= \pm \frac{\p... | x_{1}=\\frac{\pi}{6}+\frac{2\pik}{3},k\in{Z},\quadx_{2}=\\frac{2\pi}{3}+2\pik,k\in{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,805 |
1. Add matrices $A$ and $B$, if:
a) $A=\left(\begin{array}{rr}2 & 4 \\ -1 & 3\end{array}\right), \quad B=\left(\begin{array}{rr}-1 & 3 \\ 1 & -4\end{array}\right)$;
b) $A=\left(\begin{array}{rrr}1 & 2 & -3 \\ 2 & -4 & 5\end{array}\right), \quad B=\left(\begin{array}{rrr}2 & -4 & 1 \\ 3 & 0 & 2\end{array}\right)$;
c)... | Solution. a) Here $A$ and $B$ are square matrices of the second order. By adding their corresponding elements, we get
$$
C=A+B=\left(\begin{array}{rr}
2-1 & 4+3 \\
-1+1 & 3-4
\end{array}\right)=\left(\begin{array}{rr}
1 & 7 \\
0 & -1
\end{array}\right)
$$
b) Here $A$ and $B$ are rectangular matrices of type $2 \times... | \begin{pmatrix}(\begin{pmatrix}1&7\\0&-1\end{pmatrix})\\(\begin{pmatrix}3&-2&-2\\5&-4&7\end{pmatrix})\\(\begin{pmatrix}-3&4&19\\0&0&0\\6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,806 |
8. Find the linear combination $3 A-2 B$, if
$$
A=\left(\begin{array}{rrr}
2 & -4 & 0 \\
-1 & 5 & 1 \\
0 & 3 & -7
\end{array}\right), \quad B=\left(\begin{array}{rrr}
4 & -1 & -2 \\
0 & -3 & 5 \\
2 & 0 & -4
\end{array}\right)
$$ | Solution. First, we find the product of $A$ by $k_{1}=3$ and $B$ by $k_{2}=-2$:
$$
3 A=\left(\begin{array}{rrr}
6 & -12 & 0 \\
-3 & 15 & 3 \\
0 & 9 & -21
\end{array}\right), \quad-2 B=\left(\begin{array}{rrr}
-8 & 2 & 4 \\
0 & 6 & -10 \\
-4 & 0 & 8
\end{array}\right)
$$
Now let's find the sum of the obtained matrices... | (\begin{pmatrix}-2&-10&4\\-3&21&-7\\-4&9&-13\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,807 |
12. Find the product of matrices $A$ and $B$, if
$$
A=\left(\begin{array}{lll}
3 & 1 & 1 \\
2 & 1 & 2 \\
1 & 2 & 3
\end{array}\right), \quad B=\left(\begin{array}{rrr}
1 & 1 & -1 \\
2 & -1 & 1 \\
1 & 0 & 1
\end{array}\right)
$$ | Solution. Let's find each element of the product matrix:
$$
\begin{aligned}
& c_{11}=a_{11} b_{11}+a_{12} b_{21}+a_{13} b_{31}=3 \cdot 1+1 \cdot 2+1 \cdot 1=6 \\
& c_{12}=a_{11} b_{12}+a_{12} b_{22}+a_{13} b_{32}=3 \cdot 1+1 \cdot(-1)+1 \cdot 0=2 \\
& c_{13}=a_{11} b_{13}+a_{12} b_{23}+a_{13} b_{33}=3 \cdot(-1)+1 \cdo... | (\begin{pmatrix}6&2&-1\\6&1&1\\8&-1&4\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,808 |
16. Find the product $A B$, if
$$
A=\left(\begin{array}{rrr}
0 & -1 & 2 \\
2 & 1 & 1 \\
3 & 0 & 1 \\
3 & 7 & 1
\end{array}\right), \quad B=\left(\begin{array}{ll}
3 & 1 \\
2 & 1 \\
1 & 0
\end{array}\right)
$$ | Solution.
$$
A B=\left(\begin{array}{ll}
0 \cdot 3+(-1) \cdot 2+2 \cdot 1 & 0 \cdot 1+(-1) \cdot 1+2 \cdot 0 \\
2 \cdot 3+1 \cdot 2+1 \cdot 1 & 2 \cdot 1+1 \cdot 1+1 \cdot 0 \\
3 \cdot 3+0 \cdot 2+1 \cdot 1 & 3 \cdot 1+0 \cdot 1+1 \cdot 0 \\
3 \cdot 3+7 \cdot 2+1 \cdot 1 & 3 \cdot 1+7 \cdot 1+1 \cdot 0
\end{array}\rig... | (\begin{pmatrix}0&-1\\9&3\\10&3\\24&10\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,809 |
26. Compute the second-order determinants:
a) $\left|\begin{array}{rr}2 & 5 \\ -3 & -4\end{array}\right|$;
b) $\left|\begin{array}{ll}a^{2} & a b \\ a b & b^{2}\end{array}\right|$. | Solution. a) $\left|\begin{array}{rr}2 & 5 \\ -3 & -4\end{array}\right|=2(-4)-5(-3)=-8+15=7$;
b) $\left|\begin{array}{ll}a^{2} & a b \\ a b & b^{2}\end{array}\right|=a^{2} \cdot b^{2}-a b \cdot a b=a^{2} b^{2}-a^{2} b^{2}=0$.
27-32. Calculate the determinants: | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,810 |
33. Calculate the determinants of the third order:
a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|$
b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ | Solution.
a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|=3 \cdot 5 \cdot 3+2 \cdot 3 \cdot 3+2 \cdot 4 \cdot 1-1 \cdot 5 \cdot 3-2 \cdot 2 \cdot 3-$ $-3 \cdot 3 \cdot 4=45+18+8-15-12-36=71-63=8$;
b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=a c b+... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,811 |
40. Write down all the minors of the determinant
$$
D=\left|\begin{array}{rrr}
-1 & 2 & 0 \\
3 & 7 & -1 \\
5 & 4 & 2
\end{array}\right|
$$ | Solution.
$$
\begin{gathered}
M_{11}=\left|\begin{array}{rr}
7 & -1 \\
4 & 2
\end{array}\right| ; \quad M_{12}=\left|\begin{array}{rr}
3 & -1 \\
5 & 2
\end{array}\right| ; \quad M_{13}=\left|\begin{array}{ll}
3 & 7 \\
5 & 4
\end{array}\right| \\
M_{21}=\left|\begin{array}{cc}
2 & 0 \\
4 & 2
\end{array}\right| ; \quad ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,812 |
42. Find the algebraic complements of the elements $a_{13}, a_{21}, a_{31}$ of the determinant
$$
D=\left|\begin{array}{rrr}
-1 & 2 & 3 \\
2 & 0 & -3 \\
3 & 2 & 5
\end{array}\right|
$$ | $$
\begin{gathered}
A_{13}=(-1)^{1+3}\left|\begin{array}{ll}
2 & 0 \\
3 & 2
\end{array}\right|=\left|\begin{array}{ll}
2 & 0 \\
3 & 2
\end{array}\right|=4-0=4 \\
A_{21}=(-1)^{2+1}\left|\begin{array}{ll}
2 & 3 \\
2 & 5
\end{array}\right|=-\left|\begin{array}{cc}
2 & 3 \\
2 & 5
\end{array}\right|=-(10-6)=-4 ; \\
A_{31}=(... | A_{13}=4,A_{21}=-4,A_{31}=-6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,813 |
44. Determinant
$$
D=\left|\begin{array}{rrr}
3 & 1 & 2 \\
-1 & 2 & 5 \\
0 & -4 & 2
\end{array}\right|
$$
expand: a) by the elements of the 1st row; b) by the elements of the 2nd column. | Solution.
a) $D=3\left|\begin{array}{rr}2 & 5 \\ -4 & 2\end{array}\right|-1 \cdot\left|\begin{array}{rr}-1 & 5 \\ 0 & 2\end{array}\right|+2\left|\begin{array}{rr}-1 & 2 \\ 0 & -4\end{array}\right|=3(4+20)-$ $-1(-2-0)+2(4-0)=72+2+8=82$
b) $D=-1 \cdot\left|\begin{array}{rr}-1 & 5 \\ 0 & 2\end{array}\right|+2\left|\begin... | 82 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,814 |
45. Calculate the determinant
$$
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 | 31,815 |
52. Find the matrix inverse of the matrix
$$
A=\left(\begin{array}{rr}
2 & -1 \\
4 & 3
\end{array}\right)
$$ | Solution. $1^{0}$. Find the determinant of matrix $A$:
$$
D=\left|\begin{array}{cc}
2 & -1 \\
4 & 3
\end{array}\right|=2 \cdot 3-(-1) \cdot 4=6+4=10
$$
Since $D \neq 0$, the matrix is non-singular and, therefore, an inverse matrix exists.
$2^{\circ}$. Find the algebraic complements of each element: $A_{11}=(-1)^{1+1... | (\begin{pmatrix}3/10&1/10\\-2/5&1/5\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,816 |
53. Find the matrix inverse of the matrix
$$
A=\left(\begin{array}{rrr}
1 & 2 & 3 \\
0 & -1 & 2 \\
3 & 0 & 7
\end{array}\right)
$$ | Solution. $1^{0}$. Find the determinant of matrix $A$:
$$
\begin{aligned}
& D=\left|\begin{array}{rrr}
1 & 2 & 3 \\
0 & -1 & 2 \\
3 & 0 & 7
\end{array}\right|=1 \cdot(-1) \cdot 7+2 \cdot 2 \cdot 3+0 \cdot 0 \cdot 3-3 \cdot(-1) \cdot 3- \\
& -2 \cdot 0 \cdot 7-1 \cdot 2 \cdot 0=-7+12+9=14
\end{aligned}
$$
Since $D \ne... | (\begin{pmatrix}-7&-14&7\\6&-2&-2\\3&6&-1\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,817 |
60. Solve the matrix equation
$$
\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right) x=\binom{7}{17}
$$ | Solution. $1^{0}$. We will find the inverse matrix $A^{-1}$.
Let's find the determinant of matrix $A$:
$$
D=\left|\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right|=1 \cdot 4-2 \cdot 3=4-6=-2 \neq 0
$$
Calculate the algebraic complements of each element of matrix $A$:
$A_{11}=(-1)^{1+1} \cdot 4=4, A_{12}=(-1)^{1+2}... | \binom{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,818 |
61. Solve the matrix equation
$$
\left(\begin{array}{rrr}
3 & -1 & 0 \\
-2 & 1 & 1 \\
2 & -1 & 4
\end{array}\right) X=\left(\begin{array}{r}
5 \\
0 \\
15
\end{array}\right)
$$ | Solution. $1^{0}$. Find the inverse matrix $A^{-1}$.
Calculate the determinant of matrix $A$:
$$
\begin{gathered}
D=\left|\begin{array}{rrr}
3 & -1 & 0 \\
-2 & 1 & 1 \\
2 & -1 & 4
\end{array}\right|=3 \cdot 1 \cdot 4+(-1) \cdot 1 \cdot 2+(-2) \cdot(-1) \cdot 0- \\
-2 \cdot 1 \cdot 0-(-1) \cdot 1 \cdot 3-(-2) \cdot(-1... | (\begin{pmatrix}2\\1\\3\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,819 |
66. Solve the system of equations using the matrix method
$$
\left\{\begin{array}{cc}
x_{1}+2 x_{2} & =10 \\
3 x_{1}+2 x_{2}+x_{3} & =23 \\
x_{2}+2 x_{3} & =13
\end{array}\right.
$$ | Solution. Let's form the matrix equation $A X=B$, where
$$
A=\left(\begin{array}{lll}
1 & 2 & 0 \\
3 & 2 & 1 \\
0 & 1 & 2
\end{array}\right), X=\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right), B=\left(\begin{array}{l}
10 \\
23 \\
13
\end{array}\right)
$$
and solve it using the specified method. We f... | x_{1}=4,x_{2}=3,x_{3}=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,820 |
71. Solve the system of equations
$$
\left\{\begin{array}{l}
5 x+3 y=12 \\
2 x-y=7
\end{array}\right.
$$ | Solution. Let's calculate the determinant of the system $\Delta$ and the determinants $\Delta_{x}$ and $\Delta_{y}:$
$$
\Delta=\left|\begin{array}{rr}
5 & 3 \\
2 & -1
\end{array}\right|=-11, \Delta_{x}=\left|\begin{array}{rr}
2 & 3 \\
7 & -1
\end{array}\right|=-33 ; \Delta_{y}=\left|\begin{array}{rr}
5 & 12 \\
2 & 7
\... | (3,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,821 |
73. Solve the system of equations
$$
\left\{\begin{array}{l}
2 x-3 y=11 \\
6 x-9 y=33
\end{array}\right.
$$ | solution. We find
$\Delta=\left|\begin{array}{ll}2 & -3 \\ 6 & -9\end{array}\right|=0, \Delta_{x}=\left|\begin{array}{ll}11 & -3 \\ 33 & -9\end{array}\right|=0 ; \Delta_{y}=\left|\begin{array}{ll}2 & 11 \\ 6 & 33\end{array}\right|=0$.
This system has an infinite number of solutions (the coefficients of the unknowns a... | \inftyinitesolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,823 |
82. Using the Gaussian method, solve the system of equations
$$
\left\{\begin{array}{l}
3 x+2 y-z=4 \\
2 x-y+3 z=9 \\
x-2 y+2 z=3
\end{array}\right.
$$ | Solution. Let's rearrange the third equation to the first position:
$$
\left\{\begin{array}{l}
x-2 y+2 z=3 \\
3 x+2 y+z=4 \\
2 x-y+3 z=9
\end{array}\right.
$$
Write the augmented matrix:
$$
\left(\begin{array}{rrr|r}
1 & -2 & 2 & 3 \\
3 & 2 & -1 & 4 \\
2 & -1 & 3 & 9
\end{array}\right)
$$
To get $a_{21}=a_{31}=0$ i... | (1;2;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,824 |
1. Find the true absolute error of the number $a_{0}=$ $=245.2$, if $a=246$. | Solution. We have $\left|a-a_{0}\right|=|245.2-246|=0.8$.
2-9. Find the true absolute errors of the numbers: | 0.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,826 |
10. Write the number $a_{0}=9.3 \pm 0.5$ using a double inequality. | Solution. $9.3-0.5 \leqslant a_{0} \leqslant 9.3+0.5 ; 8.8 \leqslant a_{0} \leqslant 9.8$.
11-18. Write the numbers in the form of a double inequality: | 8.8\leqslanta_{0}\leqslant9.8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 31,827 |
19. Find the significant and doubtful digits of the number $a=945.673 \pm$ $\pm 0.03$. | Solution. Here $a=945,673, \Delta a=0.03$. The digit 6 represents the tenths place, i.e., the unit of this place can be written as: 0.1. Compare this unit with the error of the number; since $0.1>0.03$, the absolute error of the number does not exceed (in this case, is less than) the unit of the place where the digit 6... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 31,828 |
32. Write the number correctly:
a) $a=0.075 \pm 0.000005$
b) $a=746000000 \pm 5000$ | Solution. a) Since the error of the number does not exceed 0.00001, the number should be written in the form $a=0.07500$.
b) Here the first correct digit is the ten-thousands digit, since the error of the number does not exceed 10000. Therefore, the number should be written in the form $a=74600 \cdot 10^{4}$.
33-40. ... | =0.07500,\,=74600\cdot10^{4} | Other | math-word-problem | Yes | Yes | olympiads | false | 31,829 |
41. Indicate the absolute error of the approximate number: a) $a=2175000$; b) $a=173 \cdot 10^{4}$. | Solution. a) Since all zeros are written out, the zeros in the hundreds, tens, and units places are correct digits. Therefore, the absolute error of the number does not exceed the unit of the smallest place where the correct digits stand, i.e., $\Delta a=1$.
b) According to Rule III, zeros are replaced by $10^{4}$, wh... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 31,830 |
103. Find the limit of the relative error of the number $a=$ $=142.5$, if $\Delta a=0.05$. | Solution. $\quad \mathbf{\varepsilon}_{a}=\frac{0.05}{142.5} \cdot 100 \% =0.00034 \cdot 100 \% =0.03 \%$.
104-111. Determine the limits of the relative errors of the following numbers:
$\begin{array}{ll}\text { 104. } a=6.93 ; \Delta a=0.02 . & \text { 105. } a=12.79 ; \Delta a=2 .\end{array}$
$\begin{array}{ll}\te... | 0.03 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,832 |
112. Find the limit of the absolute error of the number $a=$ $=1348$, if $\varepsilon_{a}=0.04 \%$. | Solution. Let's write the relative error boundary as $0.04\% = 0.0004$. To find the boundary of the absolute error of the number $a$, we will use the formula $\Delta a = |a| \varepsilon_{a}$, from which $\Delta a = 1348 \cdot 0.0004 = 0.539 \approx$ $\approx 0.5$. Therefore, $\Delta a = 0.5$ and the number can be writt... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,833 |
119. Add the approximate numbers:
$$
14.5+113.76+12.783+11.2161
$$ | Solution. We round all numbers to the least precise number $(14.5)$, leaving an extra digit, and perform the addition:
$$
14.5+113.76+12.78+11.22=152.26
$$
Rounding the extra digit, we get the answer: 152.3.
120-131. Perform operations with approximate numbers: | 152.3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,834 |
133. Find the sum of $318,7864+211,1246+76,1613+106,1914$ with an accuracy of 0.01. | Solution. Round all numbers, leaving a spare digit:
$$
318.786+211.125+76.161+106.191=712.263
$$
Round the spare digit and get the answer: 712.26.
134-136. Find with an accuracy of 0.01: | 712.26 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,835 |
140. Find the product of two approximate numbers: $0.3862 \times$ $\times 0.85$. | Solution. Round the first number, leaving one extra digit, since the second number contains two significant figures. Thus,
$$
0.3862 \cdot 0.85=0.386 \cdot 0.85=0.3281 \approx 0.33
$$ | 0.33 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,836 |
141. Calculate $x=\frac{2.48 \cdot 0.3665}{5.643}$. | Solution. The number 2.48 has the least number of significant figures, which is 3; therefore, we round the other numbers to three significant figures (0.367 and 5.64). Consequently,
$$
x=\frac{2.48 \cdot 0.367}{5.64}=0.16137 \approx 0.161
$$ | 0.161 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,837 |
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 | 31,838 |
150. Find: $i^{28} ; i^{33} ; i^{135}$. | Solution. We have $28=4 \cdot 7$ (no remainder); $33=4 \cdot 8+1$; $135=$ $=4 \cdot 33+3$. Accordingly, we get $i^{28}=1$; $i^{33}=i$; $i^{135}=-i$.
151-157. Calculate: | i^{28}=1;i^{33}=i;i^{135}=-i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,839 |
158. Find $x$ and $y$ from the equality:
a) $3 y + 5 x i = 15 - 7 i$;
b) $(2 x + 3 y) + (x - y) i = 7 + 6 i$. | Solution. a) According to the condition of equality of complex numbers, we have $3 y=15, 5 x=-7$. From this, $x=-7 / 5, y=5$.
b) From the condition of equality of complex numbers, it follows that
$$
\left\{\begin{array}{r}
2 x+3 y=7 \\
x-y=6
\end{array}\right.
$$
Multiplying the second equation by 3 and adding the r... | -\frac{7}{5},5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,840 |
165. Given complex numbers $z_{1}=2+3 i, z_{2}=5-7 i$. Find: a) $z_{1}+z_{2}$; b) $z_{1}-z_{2}$; c) $z_{1} z_{2}$. | Solution. a) $z_{1}+z_{2}=(2+3 i)+(5-7 i)=2+3 i+5-7 i=(2+$ $+5)+(3 i-7 i)=7-4 i$
b) $z_{1}-z_{2}=(2+3 i)-(5-7 i)=2+3 i-5+7 i=(2-5)+(3 i+7 i)=$ $=-3+10 i$
c) $z_{1} z_{2}=(2+3 i)(5-7 i)=10-14 i+15 i-21 i^{2}=10-14 i+15 i+$ $+21=(10+21)+(-14 i+15 i)=31+i$, (here it is taken into account that $\left.i^{2}=-1\right)$. 16... | )7-4i,b)-3+10i,)31+i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,841 |
182. Perform the operations: a) $(2+3 i)^{2} ;$ b) $(3-5 i)^{2}$; c) $(5+$ $+3 i)^{3}$ | Solution. $\quad$ a) $\quad(2+3 i)^{2}=4+2 \cdot 2 \cdot 3 i+9 i^{2}=4+12 i-9=-5+$ $+12 i$
b) $(3-5 i)^{2}=9-2 \cdot 3 \cdot 5 i+25 i^{2}=9-30 i-25=-16-30 i$;
c) $(5+3 i)^{3}=125+3 \cdot 25 \cdot 3 i+3 \cdot 5 \cdot 9 i^{2}+27 i^{3}$;
since $i^{2}=-1$, and $i^{3}=-i$, we get $(5+3 i)^{3}=125+225 i-135-$ $-27 i=-10+1... | -5+12i,-16-30i,-10+198i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,842 |
191. Perform the operations: a) $(5+3 i)(5-3 i) ;$ b) $(2+5 i)(2-$ $-5 i)$; c) $(1+i)(1-i)$. | Solution. a) $(5+3 i)(5-3 i)=5^{2}-(3 i)^{2}=25-9 i^{2}=25+9=34$;
b) $(2+5 i)(2-5 i)=2^{2}-(5 i)^{2}=4+25=29$;
c) $(1+i)(1-i)=1^{2}-i^{2}=1+1=2$.
192-197. Perform the operations; | 34,29,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,843 |
198. Perform the division:
a) $\frac{2+3 i}{5-7 i}$;
b) $\frac{3+5 i}{2+6 i}$. | Solution. a) We have
$$
\frac{2+3 i}{5-7 i}=\frac{(2+3 i)(5+7 i)}{(5-7 i)(5+7 i)}
$$
Let's perform the multiplication for the dividend and divisor separately:
$$
\begin{gathered}
(2+3 i)(5+7 i)=10+14 i+15 i+21 i^{2}=-11+29 i ; \\
(5-7 i)(5+7 i)=25-49 i^{2}=25+49=74
\end{gathered}
$$
Thus,
$$
\frac{2+3 i}{5-7 i}=\f... | \frac{9}{10}-\frac{1}{5}i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,844 |
217. Solve the equation: a) $x^{2}-6 x+13=0$; b) $9 x^{2}+12 x+$ $+29=0$ | Solution. a) Let's find the discriminant using the formula $D=b^{2}-4 a c$. Since $a=1, b=-6, c=13$, then $D=(-6)^{2}-4 \cdot 1 \cdot 13=36-52=$ $=-16 ; \sqrt{D}=\sqrt{-16}=\sqrt{16 \cdot(-1)}=4 i$. The roots of the equation are found using the formulas $x_{1}=\frac{-b-\sqrt{D}}{2 a} ; x_{2}=\frac{-b+\sqrt{D}}{2 a}$ :
... | x_1=3-2i,x_2=3+2i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,845 |
223. Write the complex number $z=1+i$ in trigonometric form. | Solution. $1^{0}$. Since $a=1, b=1$, then $r=|z|=\sqrt{1^{2}+1^{2}}=\sqrt{2}$. $2^{0}$. Let's represent the number $z$ geometrically (Fig. 10). We see that the number $z$ corresponds to the point $Z$, lying in the first quadrant, and the vector $\vec{z}$.
$3^{6}$. We form the ratios $\cos \varphi=a / r$ and $\sin \var... | 1+i=\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,847 |
224. Write the number $z=-2+2 i \sqrt{3}$ in trigonometric form. | Solution. $1^{0}$. Here $a=-2, b=2 \sqrt{3}$. Therefore,
$$
r=\sqrt{a^{2}+b^{2}}=\sqrt{(-2)^{2}+(2 \sqrt{3})^{2}}=\sqrt{4+12}=\sqrt{16}=4
$$
$2^{0}$. Let's represent the number $z$ geometrically (Fig. 11). We see that the number $z$ corresponds to the point $Z$, lying in the II quadrant, and the vector $\overrightarr... | -2+2i\sqrt{3}=4(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,848 |
225. Write the purely imaginary number \( z = -3i \) in trigonometric form. | Solution. $1^{0}$. Write the given number as $z=0-3i$. Thus, $a=0, b=-3$, from which
Fig. 12
$$
r=\sqrt{a^{2}+b^{2}}=\sqrt{0+(-3)^{2}}=\sqrt{9}=3
$$
$2^{0}$. The point corresponding to t... | 3(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,849 |
233. Write the number $z=e^{a}(\cos b+i \sin b)$ in exponential form. | Solution. From the given trigonometric form of the number, we establish that $r=e^{a}$ and $\varphi=b$. Substituting these values into the exponential form of the number $z=r e^{i \phi}$, we get $z=e^{a} e^{b i}=e^{a+b i}$. | e^{+bi} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,850 |
234. Write the number $z=-5 i$ in trigonometric and exponential forms. | Solution. To represent the number $z$ in the form $z=r(\cos \varphi+$ $+i \sin \varphi$ ) and $z=r e^{i \varphi}$, we need to find the modulus and argument of the number $z$. Here $a=$ $=0, b=-5$; then $r=\sqrt{0+(-5)^{2}}=5$; $\varphi=3 \pi / 2$, since the point $z$ lies on the imaginary axis of the complex plane. Kno... | 5(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,851 |
235. Write the number $z=3-3 i \sqrt{3}$ in trigonometric and exponential forms. | Solution. Since $a=3, b=-3 \sqrt{3}$,
then $r=\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+(-3 \sqrt{3})^{2}}=$ $=\sqrt{9+9 \cdot 3}=6$. Geometrically, we determine that the number $z$ corresponds to the point $Z$, which lies in the IV quadrant (Fig. 13). We form the ratios
 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,852 |
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