so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant
![\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx%7D%7B9%7D~~%2C~~%5Cstackrel%7By%7D%7B-3%7D%29%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bhypotenuse%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%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%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%20c%3D%5Csqrt%7B9%5E2%2B%28-3%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B81%2B9%7D%5Cimplies%20c%3D%5Csqrt%7B90%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Sqrt 80n^2 = sqrt 40x2xnxn = sqrt 20x2x2xnxn = sqrt 10x2x2x2xnxn = sqrt 5x2x2x2x2xnxn
Pull out numbers and letters in groups of 2
2x sqrt 5x2x2xnxn
2x2x sqrt 5xnxn
2x2xnx sqrt 5
Multiply the numbers and letters on the outside of the radical
4n x sqrt 5 is the final answer.
From The picture 2 barrels = 73 gallons
Divide by 2 for gallons per barrel:
73/2 = 36.5 gallons per barrel
Multiply gallons per barrel barrels:
36.5 x 0.045 = 1.6425 gallons.
Rounded to 2 decimals = 1.64
Hey there!
ORIGINAL EQUATION:
0.75(2 + 6 x 4 - 2 × 7 - 2)
WORKING WITHIN the PARENTHESES:
2 + 6 x 4 - 2 × 7 - 2
= 2 + 24 - 2 × 7 - 2
= 26 - 2 × 7 - 2
= 26 - 14 - 2
= 12 - 2
= 10
NEW EQUATION:
0.75(10)
SIMPLIFY IT!
7.50 ≈ 7.5
Therefore, your answer should be: 7.5
~Amphitrite1040:)
