MO MAP Algebra I Number and Quantity. Practice it free.

$\sqrt{28}=2\sqrt{7}$, $\sqrt{63}=3\sqrt{7}$, and $\sqrt{7}=\sqrt{7}$. Adding gives $2\sqrt{7}+3\sqrt{7}+\sqrt{7}=6\sqrt{7}$.

A square root divides each exponent by $\Large 2$: $\Large x^{\tfrac{10}{2}} y^{\tfrac{8}{2}} = x^5 y^4$.

The product is $12 \cdot 64 = 768$. Factor $768 = 256 \cdot 3$, where $256$ is the largest perfect-square factor. So $\sqrt{768} = \sqrt{256 \cdot 3} = 16\sqrt{3}$, and $3$ is square-free.

$16^{1/2} = \sqrt{16} = 4$.

$9800 = 70^2 \cdot 2$, so $\sqrt{9800} = 70\sqrt{2}$.

The number of team members must divide both totals evenly. The common factors of $50$ and $75$ greater than $1$ are $5$ and $25$. Only $25$ is listed: $50 \div 25 = 2$ packs and $75 \div 25 = 3$ cables per member.

Minimize the numerator with $m = 2$ and $n = 1$, and maximize the denominator with $p = 10$: $\dfrac{2 + 1}{10} = \dfrac{3}{10}$.

The next shared meeting day is the least common multiple of $6$ and $9\textsf{.}$ Since the GCF of $6$ and $9$ is $3\textsf{,}$ divide their product by the GCF: $\dfrac{6 \times 9}{3}=18\textsf{.}$ So they meet together again in $18$ school days.