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# Math

+1
152
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4(3x^2 y^4)^3/(2x^3y^5)^4

Here is what it looks like

https://prnt.sc/pnu56m

Oct 25, 2019

#1
+9149
+2

Simplify   4(3x^2 y^4)^3/(2x^3y^5)^4

Hello Guest!

$$4(3x^2 y^4)^3/(2x^3y^5)^4$$

$$=\large \frac{4(27x^6y^{12})}{16x^{12}y^{20}}$$

$$=\large \frac{27}{4x^6y^8}$$

!

Oct 25, 2019
#2
+1

Sorry I don't understand what you did here

Like what are some fo the rules I was supposed to do while solving?

Oct 25, 2019
#3
+108732
+1

4(3x^2 y^4)^3/(2x^3y^5)^4

$$\frac{4(3x^2 y^4)^3}{(2x^3y^5)^4}\\~\\$$

$$4*(3x^3y^4)^3\\ =4*(3*x^3*y^4)^3\\ =4*3^3*(x^3)^3*(y^4)^3\\ =4*27*x^{3*3}*y^{4*3}\\ =4*27*x^{9}*y^{12}\\$$

Now I have done what is necessary to simplify the top.

Now it is your turn to simplify the bottom.

After that you can put it back to one problem and  cancel.

Oct 26, 2019