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Dmitriy789 [7]
3 years ago
6

Need some help with these last few problems Thank you for your time and help

Mathematics
1 answer:
konstantin123 [22]3 years ago
3 0
(1) On the unit circle, you have \sin x=-1 when x=-\dfrac\pi2, and more generally this is the case whenever you perform a full rotation about the origin. This means \sin x=-1 any time x=-\dfrac\pi2+2n\pi where n is any integer.

(2) When x=-90^\circ, you have y=f(-90^\circ)=\cos(-90^\circ)=0.

(3) When x=-45^\circ, you have y=f(-45^\circ)=\tan(-45^\circ)=-1.

(4) The law of sines says that

\dfrac{12}{\sin40^\circ}=\dfrac{e}{\sin55^\circ}\implies e\approx15.292

(5) The law of cosines says that

b^2=6.5^2+8^2-2\times6.5\times8\cos37^\circ\implies b\approx4.816
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PLEASE HELPP
agasfer [191]

Answer:

The solution to the system of equations be:

x=3,\:y=-1

Hece, option B is correct.

Step-by-step explanation:

Given the system of equations

\begin{bmatrix}x-3y=6\\ 2x+2y=4\end{bmatrix}

Multiply x − 3y = 6 by 2:      2x-6y=12

\begin{bmatrix}2x-6y=12\\ 2x+2y=4\end{bmatrix}

so

2x+2y=4

-

\underline{2x-6y=12}

8y=-8

so the system of equations becomes

\begin{bmatrix}2x-6y=12\\ 8y=-8\end{bmatrix}

Solve 8y = -8 for y

8y=-8

Divide both sides by 8

\frac{8y}{8}=\frac{-8}{8}

y=-1

\mathrm{For\:}2x-6y=12\mathrm{\:plug\:in\:}y=-1

2x-6\left(-1\right)=12

2x+6=12

subtract 6 from both sides

2x+6-6=12-6

2x=6

Divide both sides by 2

\frac{2x}{2}=\frac{6}{2}

x=3

Therefore, the solution to the system of equations be:

x=3,\:y=-1

Hence, option B is correct.

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