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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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is the second one C?
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Question 7: A \[(8+\sqrt{2})(4-3\sqrt{2}) \rightarrow 32 -24\sqrt{2}+4\sqrt{2}-3\sqrt{2}\sqrt{2}\]\[32-20\sqrt{2}-3(2) \rightarrow 32-6-20\sqrt{2} \rightarrow 26-20\sqrt{2}\]
thanks guys!
and with the first one, just multiply the two fractions. 2/3 x 4/5 = 8/15
what about the #6 ?
Which of the following is a solution to √x-4+9=6
x = 17 x = 15 x = 5 x = 3
is that sqrt(x - 4) + 9 = 6?
x = –3 x = –7 x = 1 x = 13 I put the wrong answer choices
yes
ok, so subtract 9 first
√x-4-3 ?
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
i have one more. I gonna see if I can do it
6√x+1=24
I got it wrong. i got 88
x = 17 x = 15 x = 5 x = 3
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
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