well, is noteworthy that an x-intercept is when y = 0 or namely is a solution or root of the quadratic, so we know then that the x-intercepts or solutions are at (-1,0) and (3,0), that simply means that

![\bf -8=a(2)(-2)\implies -8=-4a\implies \cfrac{-8}{-4}=a\implies \boxed{2=a} \\\\[-0.35em] ~\dotfill\\\\ y=2(x+1)(x-3)\implies y=2(\stackrel{\mathbb{FOIL}}{x^2-2x-3})\implies y=2x^2-4x-6](https://tex.z-dn.net/?f=%5Cbf%20-8%3Da%282%29%28-2%29%5Cimplies%20-8%3D-4a%5Cimplies%20%5Ccfrac%7B-8%7D%7B-4%7D%3Da%5Cimplies%20%5Cboxed%7B2%3Da%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20y%3D2%28x%2B1%29%28x-3%29%5Cimplies%20y%3D2%28%5Cstackrel%7B%5Cmathbb%7BFOIL%7D%7D%7Bx%5E2-2x-3%7D%29%5Cimplies%20y%3D2x%5E2-4x-6)
Answer:
12
Step-by-step explanation:
Set x=0
when you do that, you will find the point where y=12 since x^2+8x=0, but adding the 12 moves up the y-component by 12 units
Answer:
Step-by-step explanation:
Suppose we think of an alphabet X to be the Event of the evidence.
Also, if Y be the Event of cheating; &
Y' be the Event of not involved in cheating
From the given information:



Thus, 
P(Y') = 1 - 0.01
P(Y') = 0.99
The probability of cheating & the evidence is present is = P(YX)



The probabilities of not involved in cheating & the evidence are present is:


(b)
The required probability that the evidence is present is:
P(YX or Y'X) = 0.006 + 0.000099
P(YX or Y'X) = 0.006099
(c)
The required probability that (S) cheat provided the evidence being present is:
Using Bayes Theorem



Step 1
<u>Find the value of x</u>
we know that
-------> by corresponding angles, because the given lines are parallel
Solve for x

Step 2
<u>Find the value of y</u>
we know that

substitute the value of x and solve for y





therefore
<u>the answer is the option B</u>
