Algebra II?

Question

If p varies inversely as the square root of q, and p = 12 when q = 36, find p when q = 16.

Answer 1

p = k / SQRT(q)
12 = k / SQRT(36) = k / 6, so k=72.

p = 72 / SQRT(16) = 72/4 = 18

Answer 2

let us assume the symbol ~ as the sign of proportionality for our convenience.
thus,
p~1/(q^1/2)
or, p (q^1/2)=k (k is the constant of proportianality)
Now,
p=12 when q=36
therefore,
12 * (36^1/2)=k
or, 12*6=k (b'cos 6^2 = 36)
or, k=72
therefore, when q = 16
p * (16)^1/2=72
or, p * 4 = 72
or, p = 72/4 = 18
therefore p = 18.
Check whether i am right or not. bye.

Answer 3

Since we are dealing with indirect variation, we use the formula:

p=k/√q [k is a constant]

Plugging in the first values of p and q to find the value of k:

12=k/√36

k=72

Plugging in the new value of q and that of k:

p=72/√16

p=18

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Answer 4

p would be 18

Answer 5

p = k / √q
12 = k / 6
k = 72
p = 72 / √16
p = 18