## Algebra 1: Common Core (15th Edition)

$t^{(-14)}$
$(t^2)^{(-2)} \times (t^2)^{(-5)}$ When solving for the power of a power, we have to multiply the exponents. Solving, we get: $t^{(2\times-2)}\times t^{(2\times-5)}$ = $t^{(-4)} \times t ^{(-10)} = t^{(-4-10)}= t^{-14}$