Answer:
yes
Step-by-step explanation:
Answer:
Step-by-step explanation:
<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>
That's false. Take the numbers 1, 3, and 5.
Let a = 1, b = 3, and c = 5.
ab + bc = ac
3 + 15 = 5
18 ≠ 5
Since 18 does not equal 5, this is false. If you were to use the distributive property on ab + bc = ac, you would get
b(a+c) = ac, which doesn't even make sense.
Have a nice day! :)
The option C) y = 15/x
a) y = 15 x is an example of direct variation, with proportionality constant = 15
b) is an example of linear variation (it is the equation of a straight line that does not pass through the origin)
c) y = 15/x => y*x = 15 = constant which is the characteristic of an inverse relation
d) y = x/15 is an example of an inverse relation, with proportion constant = 1/15