<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>
The First Graph represents the situation.
The runner goes a lot of distance in a short amount of time (Demonstrated by the steep incline of the first portion). When the runner is tying their shoe, the distance remains the same, thus eliminating the second graph and the last graph because the last graph doesn't show the beginning of the problem. Finally, we know that the runner is slower after tying their shoe, this means the incline of the line would be far less as steep (Meaning the slope would be smaller). The only graph that depicts all of these, is the first one.
I Hope this Helps!
-Sinnamin
Answer:
10x -11
Step-by-step explanation:
SX =SM + MX
= 6x-3 + 4x-8
Combine like terms
= 10x -11
Answer: It is not continuous at x=3
1.25 is the slope 5/4 rise over run
