RISE Grade 8 Expressions and Equations. Practice it free.

The Power of a Product Property says that a product raised to a power equals each factor raised to that power, written $(ab)^{n} = a^{n} \cdot b^{n}\textsf{.}$ Here $(3 \cdot 7)^{4} = 3^{4} \cdot 7^{4}$, so $a = 4\textsf{.}$

Substitute $t = 4$: $f(4) = 2e^{12} + 1{,}000$. Since $e^{12} \approx 162{,}755$, $f(4) \approx 2(162{,}755) + 1{,}000 = 326{,}510 \approx 3 \times 10^{5}$.

Both equations have slope $-3$ but different $y$-intercepts ($2$ and $5$), so the lines are parallel and the system has no solution.

Rewrite in slope-intercept form: $y = 2x - 3$ and $y = 2x - 3$. The slopes and $y$-intercepts match, so the equations describe the same line and the system has infinitely many solutions.

Subtract $2x$ from both sides: $3x + 4 = 19$. Subtract $4$ from both sides: $3x = 15$. Divide by $3$: $x = 5\textsf{.}$

Coefficients multiply to $9$; exponents add: $2 + 3 = 5$, so $n = 5$.

Multiply coefficients and add exponents: $5 \cdot 4 = 20$ and $-3+(-2)=-5$, giving $20 \times 10^{-5}$. Rewrite $20=2 \times 10^{1}$, so $2 \times 10^{1} \times 10^{-5}=2 \times 10^{-4}\textsf{.}$

Subtract $4$ from both sides: $5y - 4 = -2x$. Divide both sides by $-2$: $x = \dfrac{5y - 4}{-2} = \dfrac{-5y + 4}{2}$.