Maine MTA High School (grade 10) Number and Quantity. Practice it free.

$\sqrt{108}=\sqrt{36 \cdot 3}=6\sqrt{3}$ and $\sqrt{147}=\sqrt{49 \cdot 3}=7\sqrt{3}$. Adding like radical terms gives $6\sqrt{3}+7\sqrt{3}=13\sqrt{3}$.

The number of members must divide both $40$ and $100$. Common factors greater than $1$ include $2$, $4$, $5$, $10$, and $20$. Only $10$ is listed: $40 \div 10 = 4$ memory cards and $100 \div 10 = 10$ lens cloths per member.

Expand: $(1+6i)(1-6i)=1-6i+6i-36i^{2}=1-36i^{2}$. Since $i^{2}=-1$, this is $1+36=37\textsf{.}$

$|6 + 8i| = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$.

$\sqrt{36} = 6$ and $\sqrt{-81} = 9i$. So $\sqrt{36} + \sqrt{-81} = 6 + 9i$.

Each term $\dfrac{1}{m}$ and $\dfrac{1}{n}$ is least when its denominator is greatest: $\dfrac{1}{5} + \dfrac{1}{8} = \dfrac{13}{40}$.

The billboard and signal next update together at the least common multiple of $16$ and $20\textsf{.}$ Since the GCF of $16$ and $20$ is $4\textsf{,}$ divide their product by the GCF: $\dfrac{16 \times 20}{4}=80\textsf{.}$ So the next shared update is after $80$ seconds.

Divide each exponent by $\Large 4$: $\Large a^{\tfrac{5}{4}} b^{\tfrac{6}{4}} = a^{\tfrac{5}{4}} b^{\tfrac{3}{2}}$.