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777dan777 [17]
3 years ago
6

PLEASE HELP PLEASE PLEASE

Mathematics
2 answers:
ch4aika [34]3 years ago
7 0
1) function 1/x, vertical stretch is 2/x
2) reflection over x axis is -2/x,
3) left shift is -2/(x+3)
4) down  4 units -2/(x+3) -4
it should be <span>B. y= -2/(x+3) -4</span>
PolarNik [594]3 years ago
5 0
The answer is b y=-2/(x+3)-4
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Trigonometric question, 30 points, will give brainliest.
zheka24 [161]

hmmm first off let's convert the √3 +i to trigonometric form, and then use De Moivre's root theorem, bearing in mind that √3 and i or 1i are both positive, meaning we're on the I Quadrant.

\bf (\stackrel{a}{\sqrt{3}}~,~\stackrel{b}{1i})\qquad \begin{cases} r=&\sqrt{(\sqrt{3})^2+1^2}\\ &\sqrt{3+1}\\ &2\\ \theta =&tan^{-1}\left( \frac{1}{\sqrt{3}}\right)\\\\ &tan^{-1}\left( \frac{\sqrt{3}}{3} \right)\\ &\frac{\pi }{6} \end{cases}~\hfill \implies ~\hfill 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right]

\bf ~\dotfill\\\\ \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] ~\dotfill

\bf \left[ 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right] \right]^3\implies 2^3\left[ cos\left( 3\cdot \frac{\pi }{6}\right) +i~sin\left( 3\cdot \frac{\pi }{6}\right) \right] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 8\left[cos\left( \frac{\pi }{2} \right) +i~sin\left( \frac{\pi }{2} \right) \right]~\hfill

3 0
4 years ago
Solve the equation. X2=-180
devlian [24]

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Step-by-step explanation:

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7 0
3 years ago
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ICE Princess25 [194]
80/2.5 = 32
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hope it helps (:
4 0
3 years ago
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