If p varies directly as the square of q, and q varies directly as the square of

Question

If p varies directly as the square of q, and q varies directly as the square of r, what effect will doubling r have on the value of P?

I'm struggling with this section of the chapter. My dad thinks the relationships are:

p/q2 and q/r2. So when r is doubled, q effectively becomes 1/2, and then when that is applied to the other relationship, p is effectively doubled. So is that a) correct, and why? Not really getting the core concept of this stuff and a good explanation step by step would be great.

Thanks

Explanation / Answer

p = q^2 q = r^2 if r =2 then q = 2^2 = 4; and p would equal 4^2 = 16 if r = 4 then q = 16; and p = 256 if r = 8 then q = 64; p = 4096 if r = 16 then q = 256; p = 65536 so by doubling r, the value of p is multiplied by 16 this is because if you have 2r; then q = (2r)^2 which equal 4(r^2); then p = (4(r^2))^2 which means tha p = 16(r^2) plug any number into r and find q and p then double it and find q and p the 2nd p should equal the 1st p multiplied by 16 each time

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