<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:
13
Step-by-step explanation:
there is a rule that says that if you add any 2 sides, it should be more than the third. 9+21 is greater than 13. 21+13 is greater than 9. 13 is the only number which if you add 9 to it it will be greater than 21. 12 will equal 21, but it still wont work.
hope this helps :))
Number 2 is 4x^5/3y^5 and 14 is 3125x^10
A because (-5)^2 = 25 + 2(5) =35-1=34
Parallel graphs have the same slope. perpendicular has opposite reciprocal slopes. These equations have slopes of -2/5 and 25 so they are neither.
