This one too please I really need help

Mathematics

1 answer:

In-s [12.5K]3 years ago

6 0

That answer is correct!

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Trigonometric Identities and Applications? help

Vadim26 [7]

$\bf h=-15cos\left( \frac{2\pi }{5}t \right)\qquad \boxed{h=8}\qquad 8=-15cos\left( \frac{2\pi }{5}t \right)
\\\\\\
\cfrac{8}{-15}=cos\left( \frac{2\pi }{5}t \right)\implies cos^{-1}\left( \frac{8}{-15}\right)=cos^{-1}\left[ cos\left( \frac{2\pi }{5}t \right) \right]
\\\\\\
cos^{-1}\left( \frac{8}{-15}\right)=\cfrac{2\pi }{5}t\implies \cfrac{5}{2\pi }\cdot cos^{-1}\left( \frac{8}{-15}\right)=t\\\\
-------------------------------$

$\bf cos^{-1}\left( \frac{8}{-15}\right)=\stackrel{radians}{\stackrel{II~quadrant}{\approx 2.13}~~,~~\stackrel{III~quadrant}{\approx 4.15}}\qquad \qquad 
t=
\begin{cases}
\frac{5}{2\pi }\cdot 2.13\\\\
\frac{5}{2\pi }\cdot 4.15
\end{cases}$

3 0

3 years ago

Rotate -4,3 90 degrees counter clockwise

Delicious77 [7]

The answer would be (4,3)

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3 years ago

Please help me!!!!!!

FromTheMoon [43]

Answer:

Step-by-step explanation:

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