Two Answers: Angle 1, Angle 4
Adjacent angles share a common segment, line, or ray. Think of two adjacent rooms sharing a common wall.
Linear, because the points (1,8) (2,6) (3,4) etc are decreasing at a constant rate of -2.
This is a binomial distribution with n = 5, p = 0.55, q = 1 - 0.55 = 0.45, x = 0, 1, 2, 3
P(x) = nCx p^x q^(n - x)
P(x ≤ 3) = 1 - P(x > 3) = 1 - [P(4) + P(5)]
P(4) = 5C4 x (0.55)^4 x (0.45) = 0.2059
P(5) = 5C5 x (0.55)^5 x 1 = 0.0503
P(x ≤ 3) = 1 - (0.2059 + 0.0503) = 1 - 0.2562 = 0.7438
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)
see the attachments below for the steps :)
The simplified expression is as follows:
d(-3x^2 + 9x - 14)
