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mylaaaa79

19.12.2020 •

Mathematics

Solve the system and choose the type of solution the system has. [y=−3x+6y=x+2]

The system has one solution (1,3).

The system has no solution. The lines are parallel.

The system has infinitely many solutions. The equations are the same line.

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Ответ:

aronandalp3yep6

09.07.2022

$\theta=\dfrac{11\pi}6$ behaves the same as $\theta=-\dfrac\pi6$ , since $-\dfrac\pi6+2\pi=\dfrac{11\pi}6$ . Recall that $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$ . Then

$\sin\dfrac{11\pi}6=-\sin\dfrac\pi6=-\dfrac12$

$\cos\dfrac{11\pi}6=\cos\dfrac\pi6=\dfrac{\sqrt3}2$

$\tan\dfrac{11\pi}6=\dfrac{\sin\frac{11\pi}6}{\cos\frac{11\pi}6}=-\dfrac1{\sqrt3}$

$\csc\dfrac{11\pi}6=\dfrac1{\sin\frac{11\pi}6}=-2$

$\sec\dfrac{11\pi}6=\dfrac1{\cos\frac{11\pi}6}=\dfrac2{\sqrt3}$

$\cot\dfrac{11\pi}6=\dfrac1{\tan\frac{11\pi}6}=-\sqrt3$

- - -

$\theta=-\dfrac{7\pi}4$ behaves the sames as $\theta=\dfrac\pi4$ , since $\dfrac\pi4-2\pi=-\dfrac{7\pi}4$ . Then

$\sin\left(-\dfrac{7\pi}4}\right)=\sin\dfrac\pi4=\dfrac1{\sqrt2}$

$\cos\left(-\dfrac{7\pi}4\right)=\cos\dfrac\pi4=\dfrac1{\sqrt2}$

$\tan\left(-\dfrac{7\pi}4\right)=1$

$\csc\left(-\dfrac{7\pi}4\right)=\sqrt2$

$\sec\left(-\dfrac{7\pi}4\right)=\sqrt2$

$\cot\left(-\dfrac{7\pi}4\right)=1$

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