Question

√20x^13y^5
------------
5xy^7

Answer 1

(he √ extends over the y^5, too

Answer 2

@twicker ?

Answer 3

√(20/5) = 2


√x^(13-1) = √x^12= √x^4.√x^3 = (x^2)(√x^3)

√y^(5--7) = √y^12= √y^4.√y^3 = (y^2)(√y^3)

= 2(xy)^2.√((xy)^3))

Answer 4

\(\Huge\color{blue}{ \sf \sqrt{ \frac{ 20x^{13}y^5 }{ 5xy^7 } } }\)

try to cancel some exponents first.

Answer 5

so then it becomes √20x^12/√5y^2?

Answer 6

@SolomonZelman

Answer 7

i dont get it @twicker

Answer 8

First step is to split the equation. to values that are possible to find the squareroot easily
So, the first part if have taken 20/5 = 4.

Then √4 is 2. Set that aside and go to the next part which is the x
√x^(13-1) = √x^12= √x^4.√x^3 = (x^2)(√x^3)

The last part
√y^(5--7) = √y^12= √y^4.√y^3 = (y^2)(√y^3)

Combine all now: 2.(x^2)(√x^3).(y^2)(√y^3)

=2(xy)^2.√((xy)^3)) read: 2 multiplies by the square of xy multiplied by the squareroot of the cube of xy

Answer 9

can you draw it out?

Answer 10

Is this the answer correct because if it is not, there's no point repeating the wrong answer. But if it right, I'll try drawing it out

Answer 11

i just cant picture the 2(xy)^2.√((xy))^3

Answer 12

@twicker

Answer 13

a^x . b^x = (ab)^x
eg: 2^2 * 3^2 = 4*9= 36=
(2*3)^2= 36