How do you simplify 7x^3(2x^2)^2/(10x^5)?

It is $7 {x}^{3} {\left(2 {x}^{2}\right)}^{2} / \left(10 {x}^{5}\right) = 7 {x}^{3} \cdot \frac{4 {x}^{4}}{10 {x}^{5}} = \frac{28 \cdot {x}^{7}}{10 {x}^{5}} = \frac{14}{5} \cdot {x}^{2}$

Oct 2, 2015

The answer is $\frac{14 {x}^{2}}{5}$.

Explanation:

Simplify $7 {x}^{3} {\left(2 {x}^{2}\right)}^{2} / \left({10}^{5}\right)$ to $\frac{\left(7 {x}^{3}\right) {\left(2 {x}^{2}\right)}^{2}}{10 {x}^{5}}$ .

$\frac{\left(7 {x}^{3}\right) {\left(2 {x}^{2}\right)}^{2}}{10 {x}^{5}}$

Apply the exponent rule ${\left({a}^{m}\right)}^{n} = {a}^{m \cdot n}$

$\frac{\left(7 {x}^{3}\right) \left(4 {x}^{4}\right)}{10 {x}^{5}}$

Apply the exponent rule ${a}^{m} \times {a}^{n} = {a}^{m + n}$.

$\frac{28 {x}^{3 + 4}}{10 {x}^{5}} =$

$\frac{28 {x}^{7}}{10 {x}^{5}}$

Apply the exponent rule ${a}^{m} / {a}^{n} = {a}^{m - n}$.

$\frac{28 {x}^{7 - 5}}{10} =$

$\frac{28 {x}^{2}}{10}$

Reduce $\frac{28}{10}$ to $\frac{14}{5} =$.

$\frac{14 {x}^{2}}{5}$