ATLAS Algebra I Number and Quantity. Practice it free.

Expand: $(4-i)^{2}=16-8i+i^{2}$. Since $i^{2}=-1$, this is $16-8i-1=15-8i\textsf{.}$

$i^{3} = i^{2} \cdot i = (-1)(i) = -i$.

$\sqrt{64} = 8$ and $\sqrt{-1} = i$. So $\sqrt{64} + \sqrt{-1} = 8 + i$.

The number of staff members must divide both $24$ and $36$. Common factors greater than $1$ include $2$, $3$, $4$, $6$, and $12$. Only $12$ appears among the choices: $24 \div 12 = 2$ photos and $36 \div 12 = 3$ pins per person.

Minimize $\dfrac{m}{n}$ with $m = 1$ and $n = 7$, and minimize $\dfrac{1}{p}$ with $p = 12$: $\dfrac{1}{7} \cdot \dfrac{1}{12} = \dfrac{1}{84}$.

The next day both jobs occur is the least common multiple of $10$ and $14\textsf{.}$ Since the GCF of $10$ and $14$ is $2\textsf{,}$ divide their product by the GCF: $\dfrac{10 \times 14}{2}=70\textsf{.}$ So they coincide again in $70$ days.

Multiply by conversion factors so the old units cancel: $20\,\dfrac{\text{m}}{\text{s}} \times \dfrac{1\,\text{km}}{1{,}000\,\text{m}} \times \dfrac{3{,}600\,\text{s}}{1\,\text{h}} = 72\,\dfrac{\text{km}}{\text{h}}$.

Expand: $(5-3i)(-2+i)=-10+5i+6i-3i^{2}$. Since $i^{2}=-1$, this is $-10+11i+3=-7+11i\textsf{.}$