Question

(−5)3x7(yz)4 / (3)2x2y8z2

Answer 1

Answer:

$\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}$

Step-by-step explanation:

Given:

The expression to simplify is given as:

$\frac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}$

In order to simplify this, we have to use the law of indices.

1. $(ab)^m=a^mb^m$

So, $(yz)^4=y^4z^4$

Substitute this value in the above expression. This gives,

$=\dfrac{(-5)3x^7y^4z^4}{(3)2x^2y^8z^2}\\\\\\=\dfrac{-15x^7y^4z^4}{6x^2y^8z^2}......(-5\times 3=15\ and\ 3\times 2=6)$

Now, we use another law of indices.

2. $\frac{a^m}{a^n}=a^{m-n}$

So,  $\frac{x^7}{x^2}=x^{7-2}=x^5,\frac{y^4}{y^8}=y^{4-8}=y^{-4}, \frac{z^4}{z^2}=z^{4-2}=z^2$

Substitute these values in the above expression. This gives,

$=\frac{-15}{6}\times x^5\times y^{-4}\times z^2\\\\=\frac{-5x^5y^{-4}z^2}{2}$

Finally, we further simplify it using the law $a^{-m}=\frac{1}{a^m}$

So, $y^{-4}=\frac{1}{y^4}$

Therefore, the given expression is simplified as:

$\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}$