Question

Can someone help me?

Answer 1

:c please help

Answer 2

would it be\[4\sqrt{y^5/2}\]?

Answer 3

I will help you! We will take this one step at a time, and I tell that to my Algebra students every day. One thing at a time. For the numerator, simplify as much as possible before taking the square root of it. Find what numbers go into 64, trying to find one of the numbers that is a perfect square. Simplifying the numbers part is this\[\sqrt{8\times8}\]That shows us that 64 is itself a perfect square, because the definition of a perfect square is a number multiplied by itself. The "x" term is singular and nothing can be done with it. However, the y^5 can be simplified. Divide this into two y's, one being the largest even number under 5, (4), and the other is y^1. Here...\[\sqrt{y ^{4}y ^{1}}\]To find the perfect square in here, divide the even number by 2 to be able to pull out a y^2. Heres the whole thing under one radical. And this is only the numerator.\[\sqrt{8\times8 y ^{4}y ^{1}}\]You can pull out an 8 and a y^2 now, leaving the single x and the single y under the radical. I see I forgot to put the x there; sorry!\[8y ^{2}\sqrt{xy}\]Now for the denominator. The 8 can be broken up into 2 numbers multiplied by one another, and one of them is a perfect square.\[\sqrt{4\times2 x}\]The 4 is a perfect square, but that's the only thing that is, so here's what we can pull out\[2\sqrt{2x}\]Now we need to put them back together.\[\frac{ 8y ^{2} \sqrt{xy}}{ 2\sqrt{2x} }\]The 8 outside the radical in the numerator can cancel with the 2 in the denominator and you're done!

Answer 4

@IMStuck ohhh thank you!!! don't you simplify the x as well?

Answer 5

@IMStuck