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Mathematics

2 answers:

klasskru [66]3 years ago

8 0

Answer:

your epic

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because you are

Gekata [30.6K]3 years ago

6 0

Answer: is it an option

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How solve this <br> x+y=5<br> X-y=7

Misha Larkins [42]

$elimination\ methode\\\\\underline{+\left\{\begin{array}{ccc}x+y=5\\x-y=7\end{array}\right}\ \ \ \ |add\ both\ sides\ of\ the\ equations\\.\ \ \ \ \ \ 2x=12\ \ \ \ \ |divide\ both\ sides\ by\ 2\\.\ \ \ \ \ \ \ \boxed{x=6}\\\\subtitute\ the\ value\ of\ x\ to\ the\ first\ equation:\\\\6+y=5\ \ \ \ |subtract\ 6\ from\ both\ sides\\\boxed{y=-1}\\\\Answer:\boxed{\left\{\begin{array}{ccc}x=6\\y=-1\end{array}\right}$

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3 years ago

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$