Out of the four choices, it would be A
<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>
Answer:
-1.39
Step-by-step explanation:
Revenue and cost as a function of units sold are
and
respectively.
we are have to know for which value or input units are these functions at maximum which translates to for how many units is the revenue maximum and for how many same units is our cost minimum.
8x + 6=4x + 38
4x = 32
x = 8
<B = 4x + 38 = 4(8) + 38 = 70