Chapter 11.6, Problem 5ST

To determine

To approximate: the given square root to the nearest tenth.

Answer to Problem 5ST

  $-\sqrt{81x^8{y}^6}=-9x^4{y}^3$ .

Explanation of Solution

Given:

The given square root is $-\sqrt{81x^8{y}^6}$ .

Calculation:

As the given square root is $-\sqrt{81x^8{y}^6}$ .

Therefore, the given square root to the nearest tenth is

  $\begin{array}{l}-\sqrt{81x^8{y}^6}=-\sqrt{81}\times \sqrt{x^8{y}^6}\\ =-\sqrt{9^2}\times \sqrt{{\left(x^4{y}^3\right)}^2}\left\{\text{simplify}\right\}\\ =-9x^4{y}^3\{\because \sqrt{a^2}=a\}\end{array}$

Hence, the value of the given square root to the nearest tenth is $-\sqrt{81x^8{y}^6}=-9x^4{y}^3$ .