NJSLA Grade 8 Expressions and Equations. Practice it free.

The Power of a Power Property says that a power raised to another power is simplified by multiplying the exponents, written $(y^{m})^{n} = y^{m \cdot n}\textsf{.}$ Here $y^{5}$ is raised to the second power, so $(y^{5})^{2} = y^{5 \cdot 2} = y^{10}\textsf{.}$

Substitute $t = 4$: $f(4) = 5e^{12} + 200$. Since $e^{12} \approx 162{,}755$, $f(4) \approx 5(162{,}755) + 200 = 813{,}975 \approx 8 \times 10^{5}$.

Both equations have slope $4$ but different $y$-intercepts ($-1$ and $3$), so the lines are parallel and the system has no solution.

Rewrite in slope-intercept form: $y = \dfrac{5}{3}x - \dfrac{4}{3}$ and $y = -2x + 7$. The slopes $\dfrac{5}{3}$ and $-2$ are different, so the lines intersect at exactly one point.

Subtract $3x$ from both sides: $4x - 3 = 17$. Add $3$ to both sides: $4x = 20$. Divide by $4$: $x = 5\textsf{.}$

$\dfrac{9 \times 10^5}{3 \times 10^2} = 3 \times 10^3 = 3{,}000$.

Divide coefficients and subtract exponents: $3 \div 6 = 0.5$ and $5-2=3$, giving $0.5 \times 10^{3}$. Rewrite as $5 \times 10^{-1} \times 10^{3}=5 \times 10^{2}\textsf{.}$

Subtract $8$ from both sides: $4y - 8 = x$. So $x = 4y - 8$.