KSA Grade 8 The Number System. Practice it free.

$10 \times 10 \times 10 = 1{,}000$, so $\sqrt[3]{1{,}000} = 10$.

The Zero Exponent Property says that any nonzero base raised to the zero power equals one, written $y^{0} = 1\textsf{.}$ The stem states that $y \neq 0$, so the property applies and the expression simplifies to $1\textsf{.}$

Substitute $t = 4$: $f(4) = 4e^{8} + 50$. Since $e^{8} \approx 2{,}981$, $f(4) \approx 4(2{,}981) + 50 = 11{,}974 \approx 1 \times 10^{4}$.

$\sqrt{3} \approx 1.732$.

Multiply coefficients ($2 \cdot 3 = 6$) and add exponents: $3 + 4 = 7$, so $n = 7$.

Multiply coefficients and add exponents: $2.5 \cdot 4 = 10$ and $6+(-3)=3$, giving $10 \times 10^{3}$. Rewrite as $1 \times 10^{1} \times 10^{3}=1 \times 10^{4}\textsf{.}$

$8 \times 8 = 64$, so $\sqrt{64} = 8$.

The Power of a Power Property says that a power raised to another power is simplified by multiplying the exponents, written $(x^{m})^{n} = x^{m \cdot n}\textsf{.}$ So $(3^{4})^{2} = 3^{4 \cdot 2} = 3^{8}\textsf{.}$ The Quotient of Powers Property then subtracts the exponents, written $\dfrac{x^{m}}{x^{n}} = x^{m-n}\textsf{.}$ So $\dfrac{3^{8}}{3^{5}} = 3^{8-5} = 3^{3}$ and $k = 3\textsf{.}$