10,800 would be the best awnser for you
Answer:
(3, -8) and (2, -10)
Step-by-step explanation:
Given


Required
Select the true coordinate points in (3, -8) (2, 5) (-5, 1) (10, 3) (2, -10)
(3, -8)
x = 3 and y = -8



--- True



--- True
(2, 5)
x = 2 and y = 5



--- False (No need to check the other inequality)
(-5, 1)
x = -5 and y = 1


--- True


--- False
(10, 3)
x = 10 and y = 3


--- False (No need to check the other inequality)
(2, -10)
x = 2 and y = -10


--- True


--- True
Hence, the solution to the inequalities are:
(3, -8) and (2, -10)
The answer is 1/5, or 0.2, because if you're always putting the slip back in, then there will always by 5 slips to choose from, therefore making the probability 1 out of 5. I hope this helps!
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)
