Algebra Question

Question

a varies directly as the square of b and inversely as the square of c. If a=131 when b=4 and c=7, find a if b=3 and c=9. Round your answer to two decimal places if necessary.

Answer 1

So let's break this problem down by sentence:

a varies directly as the square of b and inversely as the square of c.

a = k * ( b² / c² )

reasoning:

the sentence doesn't tell you the direct formula of a with respect to b and c so you need to add a constant to account for any result in which a does not directly equal b² / c². This constant is denoted as k.

now the sentence states that a varies directly to square of b. that means " a α b² "

it also mentions that a varies inversely to square of c. that means " a α 1/c² ". IF the problem said a varies directly to square of c, then that would mean " a α c² ".

Now the problem doesn't do a good job of saying that the formula for a includes both b & c in one formula, but you can assume that for this problem because you're solving for one value of a. So combining both relationships into one formula (along with the constant k we talked about in the beginning), you get the formula for a:

a = k * ( b² / c² )

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Next sentence!

If a=131 when b=4 and c=7, find a if b=3 and c=9.

So they give us enough information to solve for the constant k that we established as the formula for a.

a = k * ( b² / c²) -> substitute known values for a, b, and c

131 = k * ( 4² / 7² ) -> solve for k

k = 401.1875

Now that we know the value of k, we can solve for a when b=3 and c=9

a = k * ( b² / c² ) -> substitute known values for k, b, and c

a = 401.1875 * ( 3² / 9² )

a = 44.57638889

Rounding the answer to two decimal places if necessary -> FINAL ANSWER: 44.58