Please everyone I need help?

Question

1) ^3 square root of - 8/27

2)square root of 13 times the square root of 5

3) -9 - square root of 108/3

4) 6 - the square root of 20/2

Please I need help with these four questions

Answer 1

1. 3rdrt(-8)=-2
3rdrt(27)=3
2/3

2. You can multiply in the radical, ie sqrt(3)*sqrt(2)=sqrt(6)
so 2 is sqrt(65)

3. We'll ignore the -9 for now
sqrt(108)/sqrt(3)
You cannot have a radical in the denominator, multiply by sqrt(3)
sqrt(108*3)=sqrt(324)
Sqrt(324)=18
Now we have 18 over the denominator of sqrt(3) sqared, which equals 3
18/3=6
(Here's the -9 again) -9-6=
-15

4. We'll ignore the 6 for now
Sqrt(20)/sqrt(2)
Same as the last problem, you can't have a radical in the denominator, so we multiply the numerator and denominator by sqrt(2), giving us sqrt(40)/2
You can pull out a perfect sqare, 4, from sqrt(40), giving us a 2sqrt(10)/2
The twos cancel eachother out, so now we have sqrt(10)
Bringing the 6 back into the expression, we can't simplify it anymore without a calculator
6-sqrt(10)

Answer 2

1. I hope you mean the cube root, cuz then it's easy. What cubed is 8? (2) what cubed is 27? (3) Cube root of 8/27 is 2/3 BECAUSE (2/3)^3 = 8/27

2. You can simplify neither one. When you have two square roots multiplied, you can multiply them under one radical. 13*5 = 65. You can't really do anything with 65 (i.e., take out 4, 9, 16, etc.), so just leave it.

3 & 4. What is the "9" and the "6" there for? Do you just mean square root of 108/3? PLEASE EDIT

Answer 3

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Answer 4

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