Question

\left( sqrt{15x^5y^7} \right)\left( sqrt{5x^3y^5} \right)

Answer 1

Maybe you should try writing the equation here...

Answer 2

\(\left( \sqrt{15x^5y^7} \right)\left( \sqrt{5x^3y^5} \right)\)

Answer 3

\[\left( \sqrt{15x^5y^7} \right)\left( \sqrt{5x^3y^5} \right)\]

I know you multiple them across getting

\[\sqrt{75x^7y^\left( 13 \right)}\]

I dont know where to go from here

Answer 4

Pull out the perfect square factors

Answer 5

?

Answer 6

factors of 75 are 25 and 3.
25 is a perfect square, so...
\(\sqrt{75} = 5\sqrt{3}\)
Do the same with the variables.

Answer 7

By the way, you multiplied the variables incorrectly. When multiplying you add the exponents.
\(x^5 \times x^3 = x^{5+3} = x^8\)

Answer 8

ok yes I remeber how to do it with real numbers, what about variables though how can you pull out the squares of variables.

Answer 9

yes i ment to the power of 8 but the text is small on here so thought the 3 was a 2 is why i got the power of 7, power of 8 is correct

Answer 10

\(x \times x = x^2\)
\(\sqrt{x^2} = x\)

Answer 11

\(\Large{\left( \sqrt{15x^5y^7} \right)\left( \sqrt{5x^3y^5} \right)}\)
`\(\Large{\left( \sqrt{15x^5y^7} \right)\left( \sqrt{5x^3y^5} \right)}\)`

Answer 12

Any variable squared is a perfect square.

Answer 13

ok so x would be x to the 4th, how about y since its to the power of 13

Answer 14

nvm y to the 12th mistype got it thx

Answer 15

You got it!