If a submarine goes $2 k^{3}+14 k^{2}+10 k-56$ miles in $2 k+8$ hours, what is its rate of speed?
A) $\mathrm{k}^{2}+3 \mathrm{k}-7 \mathrm{mph}$
B) $\mathrm{k}^{3}+3 \mathrm{k}^{2}-7 \mathrm{kmph}$
C) $\mathrm{k}^{2}-3 \mathrm{k}+7 \mathrm{mph}$
D) $\mathrm{k}^{2}+11 \mathrm{k}-7 \mathrm{mph}$

Question

Quizplus · Accepted Answer

The rate of speed is found by dividing the distance by the time. Dividing $2 k^{3}+14 k^{2}+10 k-56$ by $2 k+8$ gives $k^{2}+3 k-7$, which matches choice A.

If a Submarine Goes 2k3+14k2+10k−562 k^{3}+14 k^{2}+10 k-562k3+14k2+10k−56 Miles In 2k+82 k+82k+8

If a Submarine Goes $2 k^{3}+14 k^{2}+10 k-56$ Miles In $2 k+8$