Question

Please any math experts helps thank you

Answer 1

Answer:

$\frac{t^2-16}{t^2-8t+16}$

We can factorise the numerator which the difference of squares formula:

$x^2-y^2=(x+y)(x-y)$

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$\frac{t^2-4^2}{t^2-8t+16} =\frac{(t+4)(t-4)}{t^2-8t+16}$

Next, factorise the denominator:

$\frac{t^2-4^2}{t^2-8t+16} =\frac{(t+4)(t-4)}{(t-4)^2}$

$\frac{(t+4)(t-4)}{(t-4)(t-4)}$

Cancel out the common factor of (t-4):

$\frac{t+4}{t-4}$

Answer 2

Answer:

$\frac{t+4}{t-4}$

Step-by-step explanation:

=)