Correct answer is 6... NOT 7...
30 students could of been taking both courses and 60 with no course
<span>Answer:
Its too long to write here, so I will just state what I did.
I let P=(2ap,ap^2) and Q=(2aq,aq^2)
But x-coordinates of P and Q differ by (2a)
So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2)
So Q=(2a(p-1), aq^2)
which means, 2aq = 2a(p-1)
therefore, q=p-1
then I subbed that value of q in aq^2
so Q=(2a(p-1), a(p-1)^2)
and P=(2ap,ap^2)
Using these two values, I found the midpoint which was:
M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 )
then x = a(2p-1)
rearranging to make p the subject
p= (x+a)/2a</span>
23+8X is the answer to this problem
(8,1);((-2,-6)
2•8-3•1>=12
16-3>=12
13>=12
2•(-2)-3(-6)>=12
-4+18>=12
14>=12
