Simplify the expression and rationalize the denomintor of 5 squareroof of 2x^2y^7 over 81x?

Question

precal prob

Answer 1

first you need to do is clear your radical. the square root of a number is that number raised to the one half power
.._______
|/ aaaaaaa is the same as aaaaaa ^1/2

now you simply follow your rules of exponents when it comes to multiplying and dividing exponents in this case.

5* ( (2x^2* y^7)^1/2) / 81x

5* [ (2x^2)^1/2 * (y^7)^1/2 ] / 81x

5*[ { 2x^ (2* 1/2) } * { y^(7*1/2)} ] / 81x

5*[ { 2x^(2/2)} * {y^ (7/2)} ] / 81x

didn't forget when dividing exponents simply subtract exponents from each other or used additive inverse and when multipying exponents simply add

And if the exponent has friction in it you do not need a common denimontor when mutliplying and divided the exponent. If dividing you simply take the additive inverse to clear the fraction subtracting the bottom number from the top number of the friction.

5 * [ 2x * y^5] / 81x

[10x * 5y^5] / 81x

[10*5y^5] / 81

50y^5 / 81

Answer 2

5 sqrt[(2x^2 y^7)/81x]
(5/9)x y^3 sqrt[(2 y/x] =
(5/9)xy^3 sqrt[2 y x/x^2] =
(5/9)y^3 sqrt(2 x y )