Question

what is the simplest form of the expression?
sqr root 6x^8y^9/ sqr root 5x^2y^4

a. https://study.ashworthcollege.edu/access/content/group/59841a3a-ae83-40e0-9ad2-cdbb53b336a0/Algebra%20II%20Part%201/Exam%205%20NEW/image016.png
b. https://study.ashworthcollege.edu/access/content/group/59841a3a-ae83-40e0-9ad2-cdbb53b336a0/Algebra%20II%20Part%201/Exam%205%20NEW/image018.png
c.https://study.ashworthcollege.edu/access/content/group/59841a3a-ae83-40e0-9ad2-cdbb53b336a0/Algebra%20II%20Part%201/Exam%205%20NEW/image017.png

Answer 1

@yttrium

Answer 2

well the problem can be rewritten as

\[\sqrt{\frac{6x^8y^9}{5x^2y^4}}\]

the rule for dividing the same base is subtract the powers

\[\frac{x^a}{x^b} = x^{a - b}\]

hope this helps

Answer 3

Just rationalise the denominator by multiplying the numerator and the denominator by sqrt(5x^2y^4) So, you know what it will be look like?

Answer 4

I think campbell's approach is much easier. If you understand what he is trying to say.

Answer 5

i dont understand..

Answer 6

what you can apply is

\[\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\]

Answer 7

Well. The expression \[\frac{ \sqrt{6x^8y^9} }{ \sqrt{5x^2y^4} }\] is same as \[\sqrt{\frac{ 6x^8y^9 }{ 5x^2y^4 }}\]

Answer 8

And since they have common terms xy. You can just subtract the powers of each variable x and y. and divide their coefficients.

Answer 9

You understand now?

Answer 10

yes

Answer 11

so what you think is the answer, then?

Answer 12

im working it out now

Answer 13

b?

Answer 14

No.

Answer 15

c

Answer 16

so can you post what you worked out..?

Answer 17

it just makes it easy to see what you do and don't understand

Answer 18

it's actually a