Question

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Answer 1

when you have to radicals adding like that, you can treat them like variables

\[1√3+4√3=5√3\]

Answer 2

For the second one you need to factor out the second radical

96 = 2*2*2*2*2*3

pull out the groups of 2

your left with\[2*2\sqrt{2*3}=4\sqrt{6}\]

So now it's the same as the last

\[2\sqrt{6}+4\sqrt{6} = 6\sqrt{6}\]

Answer 3

For the last one you need to foil

\[64 -8\sqrt{11}+8\sqrt{11} - 11\]

= 64-11
= 53

Answer 4

What are the answer choices?

Answer 5

oh sorry you have to multiply the \[4\sqrt{6}\] by the 3 that was in the front to give you
\[2\sqrt{6} + 12\sqrt{6} = 14\sqrt{6}\]

Answer 6

Sure

Answer 7

Where it's multiplying, you can just multiply throught

14g *4g = 56g^2

so you have

\[2\sqrt{56g^2}\]

now factor out the radical

2*2*2*7*g*g

you can pull out a 2 and a g so you're left with

\[2g \sqrt{14}\]

Answer 8

For the next one you want to simplify inside the radical first

63/7 = 9

when variables are dividing, you subtract the exponents

\[\sqrt{(9x^(14))/y}\]

now factor everything out

(3*3xxxxxxxxxxxxxx)/yy

you can pull out a 3 7x's and a y

= 3x^7/y

Answer 9

For that first one, I did the same thing. Just multiply it by that 2 that was in front to give you.

\[4g \sqrt{14}\]

Answer 10

no problem. Glad to help.