cosine is adjacent/hypotenuse, however the hypotenuse is simply a radius unit, and thus is never negative, so in the -(2/3) the negative must be the numerator, -2.

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{hypotenuse}{3}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{3^2-(-2)^2}=b\implies \pm\sqrt{5}=b\implies \stackrel{\textit{II Quadrant}}{+\sqrt{5}=b}~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{5}}}{\stackrel{adjacent}{-2}} \\\\\\ ~\hspace{34em}$

6 0

4 years ago

Please help evaluate?

Dominik [7]

3.

4^3 is equal to 64. Therefor log4(64) must be equal to 3.

3 0

3 years ago

When the domain is -1, the range is ______. Show your work to find the range.

lys-0071 [83]

we can't find the range of a function only with its domain , we must have its graph or its equation to say anything about it , with your information we are only able to say , its range has just a member.

5 0

3 years ago