Answer:
b=A/h
Step-by-step explanation:
To isolate b, you need to divide both sides by h. Therefore:
A=bh
b=A/h
Hope this helps!
bearing in mind that, on the III Quadrant, sine as well as cosine are both negative, and that hypotenuse is never negative, so, if the sine is -4/5, the negative number must be the numerator, so sin(x) = (-4)/5.
![\bf sin(x)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm\sqrt{9}=a\implies \pm 3=a \\\\\\ \stackrel{III~Quadrant}{-3=a}~\hfill cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28x%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-4%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B5%5E2-%28-4%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B9%7D%3Da%5Cimplies%20%5Cpm%203%3Da%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7BIII~Quadrant%7D%7B-3%3Da%7D~%5Chfill%20cos%28x%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B-3%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}\qquad \leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20tan%5Cleft%28%5Ccfrac%7B%5Ctheta%7D%7B2%7D%5Cright%29%3D%20%5Cbegin%7Bcases%7D%20%5Cpm%20%5Csqrt%7B%5Ccfrac%7B1-cos%28%5Ctheta%29%7D%7B1%2Bcos%28%5Ctheta%29%7D%7D%20%5C%5C%5C%5C%20%5Ccfrac%7Bsin%28%5Ctheta%29%7D%7B1%2Bcos%28%5Ctheta%29%7D%5Cqquad%20%5Cleftarrow%20%5Ctextit%7Blet%27s%20use%20this%20one%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B1-cos%28%5Ctheta%29%7D%7Bsin%28%5Ctheta%29%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
The answer is the square root of 8, or 2root2.
Step-by-step explanation:
Use the distance formula.
d = sqrt((-3-(-5))^2 + (-3-(-1)))
d = sqrt(2^2 + 2^2)
d = sqrt(4 + 4)
d = sqrt8
d = sqrt4 * sqrt2
d = 2 * sqrt2
.79x22+$17.38
Hope this helps and sorry if it is wrong.
