How do you simplify #zr \cdot 2z ^ { - 4} r ^ { - 3}#?

Question

1626 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-23T01:35:36

#2/(z^3r^2)#

Add exponents when multiplying:
#2z^(-3)r^(-2)=#

#2/(z^3r^2)#

Answer 2 · 2018-07-23T01:46:17

#zr*2z^(-4)r^(-3)=2/(z^3r^2)#

Simplify:

#zr*2z^(-4)r^(-3)#

Take out the constant #2#.

#2zrz^(-4)r^(-3)#

Apply product rule: #a^ma^n=a^(m+n)#

#2z^(1+(-4))r^(1+(-3))#

Simplify.

#2z^(-3)r^(-2)#

Apply negative exponent rule: #a^(-m)=1/a^m#

#2/(z^3r^2)#

Answer 3 · 2018-07-23T04:16:19

#2/(z^3r^2)#

We can rewrite this with the constant out front as

#2z*z^(-4)*r*r^(-3)#

When we multiply exponents, we add the powers. We now have

#2z^(-3)r^(-2)#

We can make the negative exponents positive by bringing them to the denominator. We get

#2/(z^3r^2)#

Hope this helps!