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miah
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miah
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is the second one C?

AriPotta
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Lynncake
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Question 7: A \[(8+\sqrt{2})(43\sqrt{2}) \rightarrow 32 24\sqrt{2}+4\sqrt{2}3\sqrt{2}\sqrt{2}\]\[3220\sqrt{2}3(2) \rightarrow 32620\sqrt{2} \rightarrow 2620\sqrt{2}\]

miah
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thanks guys!

AriPotta
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and with the first one, just multiply the two fractions. 2/3 x 4/5 = 8/15

miah
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what about the #6 ?

miah
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Which of the following is a solution to √x4+9=6

miah
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x = 17
x = 15
x = 5
x = 3

miah
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@AriPotta

AriPotta
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is that sqrt(x  4) + 9 = 6?

miah
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x = –3
x = –7
x = 1
x = 13
I put the wrong answer choices

miah
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yes

AriPotta
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ok, so subtract 9 first

miah
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√x43 ?

AriPotta
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sqrt(x  4) + 9 = 6
sqrt(x  4) = 3
then square it
[sqrt(x  4)]^2 = 3^2
x  4 = 9
add 4
x = 13

miah
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i have one more. I gonna see if I can do it

miah
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6√x+1=24

miah
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I got it wrong.
i got 88

miah
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x = 17
x = 15
x = 5
x = 3

AriPotta
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divide by 6
6sqrt(x + 1) = 24
sqrt(x + 1) = 4
square it
[sqrt(x + 1)]^2 = 4^2
x + 1 = 16
subtract 1
x + 1 = 16
x = 15

miah
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