FAST Grade 8 Number Sense and Operations. Practice it free.

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 $1\textsf{.}$

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

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

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

The first equation simplifies to $y = -5x + 6\textsf{,}$ and the second simplifies to $y = -\dfrac{10}{m}x + \dfrac{11}{m}\textsf{.}$ When $m = 2\textsf{,}$ $-\dfrac{10}{m} = -5\textsf{,}$ so the lines have the same slope, and since $6$ does not equal $\dfrac{11}{2}\textsf{,}$ the lines are parallel with different $y$-intercepts, so the system has no solution.

$\sqrt{3} \approx 1.732$.

Rewrite in slope-intercept form: $y = \dfrac{1}{2}x - \dfrac{3}{2}$ and $y = \dfrac{1}{2}x - 2$. The slopes are the same, but the $y$-intercepts differ, so the lines are parallel and the system has zero solutions.

Subtract $\dfrac{1}{3}x$ from both sides: $\dfrac{1}{3}x + 5 = 11$. Subtract $5$ from both sides: $\dfrac{1}{3}x = 6$. Multiply both sides by $3$: $x = 18\textsf{.}$