AK STAR Grades 6-8 Expressions & Equations. Practice it free.

Let $x$ be the total servings. Breakfast, lunch, and snacks account for $\dfrac{1}{5}x + \dfrac{1}{4}x + \dfrac{1}{10}x = \dfrac{11}{20}x$. Dinner and desserts total $27 + 18 = 45$, so $\dfrac{9}{20}x = 45$ and $x = 100$.

The constants $15$ and $-7$ are like terms: $15 - 7 = 8$. The term $r^{2}$ stays the same, so the expression is equivalent to $r^{2} + 8\textsf{.}$

Distribute 5: $5 \cdot 9x + 5 \cdot -5 = 45x - 25$.

$3^{4} = 3 \cdot 3 \cdot 3 \cdot 3 = 81\textsf{.}$

The Negative Exponent Property says that a nonzero base raised to a negative exponent equals the reciprocal of that base raised to the positive exponent, written $a^{-n} = \dfrac{1}{a^{n}}\textsf{.}$ Applying the property, $a^{-2} = \dfrac{1}{a^{2}}\textsf{.}$

Substitute $t = 5$: $f(5) = e^{15}$. Since $e^{15} \approx 3{,}269{,}017$, $f(5) \approx 3 \times 10^{6}$.

The base $6$ appears $3$ times as a factor, so $6 \times 6 \times 6 = 6^{3}$.

Both equations have slope $2$ but different $y$-intercepts ($3$ and $-1$), so the lines are parallel and the system has no solution.