TCAP 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 $(x^{m})^{n} = x^{m \cdot n}\textsf{.}$ 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 $k = 3\textsf{.}$

Substitute $t = 3$: $f(3) = 4e^{9} + 200$. Since $e^{9} \approx 8{,}103$, $f(3) \approx 4(8{,}103) + 200 = 32{,}612 \approx 3 \times 10^{4}$.

The first equation simplifies to $y = \dfrac{3}{2}x - \dfrac{1}{6}\textsf{,}$ and the second simplifies to $y = \dfrac{3}{k}x - \dfrac{5}{k}\textsf{.}$ When $k = 2\textsf{,}$ both slopes are $\dfrac{3}{2}\textsf{,}$ and since $-\dfrac{1}{6}$ does not equal $-\dfrac{5}{2}\textsf{,}$ the lines are parallel with different $y$-intercepts, so the system has no solution.

Rewrite in slope-intercept form: $y = -\dfrac{2}{3}x + \dfrac{4}{3}$ and $y = -\dfrac{2}{3}x + \dfrac{7}{3}$. The slopes are the same, but the $y$-intercepts differ, so the lines are parallel and the system has zero solutions.

Distribute $4$: $4x - 8 + 3x = 13$. Combine like terms: $7x - 8 = 13$. Add $8$ to both sides: $7x = 21$. Divide by $7$: $x = 3\textsf{.}$

$3.2 \times 10^4 = 32{,}000$.

Add $6$ to both sides: $-3y + 6 = 2x$. Divide both sides by $2$: $x = \dfrac{-3y + 6}{2}$.

$9 \times 9 = 81$, so $\sqrt{81} = 9$.