Check the picture below.
since the vertical distance, namely the y-coordinate, is twice as much as the horizontal, then if the horizontal is "x", the vertical one must be 2x.
let's find the hypotenuse first.
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{2x}\\ \end{cases} \\\\\\ c=\sqrt{x^2+(2x)^2}\implies c=\sqrt{x^2+4x^2}\implies c=\sqrt{5x^2}\implies c=x\sqrt{5} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3D%5Cstackrel%7Badjacent%7D%7Bx%7D%5C%5C%20b%3D%5Cstackrel%7Bopposite%7D%7B2x%7D%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7Bx%5E2%2B%282x%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7Bx%5E2%2B4x%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B5x%5E2%7D%5Cimplies%20c%3Dx%5Csqrt%7B5%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{2~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}{\stackrel{hypotenuse}{~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ \sqrt{5}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \cfrac{2\sqrt{5}}{5}}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B2~~%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B~~%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%5Csqrt%7B5%7D%7D%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Band%20rationalizing%20the%20denominator%7D~%5Chfill%20%7D%7B%5Ccfrac%7B2%7D%7B%5Csqrt%7B5%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%7B5%7D%7D%7B%5Csqrt%7B5%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%7B5%7D%7D%7B%28%5Csqrt%7B5%7D%29%5E2%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%7B5%7D%7D%7B5%7D%7D)
There are 7 gallons of black paint available to motorola. If 5/9 gallons is used each run. To find the amount of runs you can paint, you have to divide the total by the amount for each run.

When dividing fractions, you have to keep, change, and flip.

Your new equation is:

Multiply:

You can paint
or
runs.
Answer:
point 1 (0,-6)
point 2(2,0)
Step-by-step explanation:
Answer:
D
x >= -8
multiply by 2
-X <= 8
divide by -1
X >=-8 (remember flip sign if multiply by negative)
Step-by-step explanation:
Answer:
It is actually choice c
Step-by-step explanation:
I took the test