Question

What is t squared minus 36 over 3 divided by t squared plus 6t over 9t

Answer 1

Answer: 3t-18

Step-by-step explanation:

$\frac{\frac{t^2-36}{3}}{\frac{t^2+6t}{9t} }$

The first step is to multiply by the reciprocal of the denominator. The reciprocal of a fraction is its inverted fraction which will make it equal to 1.

$\frac{\frac{t^2-36}{3}}{\frac{t^2+6t}{9t} }*\frac{9t}{t^2+6t}$

This will eliminate the denominator,leaving the fraction like;

$\frac{t^2-36}{3}*\frac{9t}{t^2+6t}$

You can simplify 9 and 3;

9/3=3

3/3=1

$t^2-36*\frac{3t}{t^2+6t}$

Multiply;

$\frac{3t^3-108t}{t^2+6t}$

We can factor both numerator and denominator. The common factor in the numerator is 3t. And, the common factor in the denominator is t.

$\frac{3t(t^2-36)}{t(t+6)}$

Now we can simplify t and t.

$\frac{3(t^2-36)}{t+6}$

$t^2$ and $36$ are both perfect squares with a negative (-) sign in the middle, in this case, we can extract both squares and multiply them by themselves being one positive and the other negative.

$\frac{3[(t+6)(t-6)]}{t+6}$

Since t + 6 and t - 7 are being multiplied, it is possible to simplify with the denominator. Leaving the fraction like;

$3(t-6)$

Multiply;

$3t-18$