Question

If y varies directly as x and y is 400 when x is r and y is r when x is 4, what is the numeric constant of variation in this relation?

Answer 1

First, you need the change the given equation into standard form.
2x+7y=-15
7y = -2x - 15
y = (-2/7)x - 15/7
The linear equation you are finding has a form like this also y = ax + b
In order to be perpendicular, the a of this equation you are finding and the a in the given equation must multiply together and give you - 1, so you have
(-2/7)a = -1
a = 7/2
So now you have the a, let's plug it in and have y = (7/2)x + b
Now you have the point (-1/2,-2) on this line, -1/2 is x and -2 is y, so you plug it in
-2 = (7/2)*(-1/2) + b
-2 = (-7/4) + b
-2 + 7/4 = b
-1/4 = b
TA-DAH
Your answer is y = (7/2)x - 1/4

Answer 2

y=kx
when x=r,y=400
400=kr ...(1)
when x=4,y=r
r=4k
from (1)
400=k*4k
k^2=100
k=10,-10
whenk=10,r=4*10=40
when k=-10,r=4*-10=-40

Answer 3

thank you @surjithayer i am on an exam and couldnt think of the answer. lol, last question, and couldnt think of it