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Q11: \[\sqrt{16*5}-3\sqrt{9*7}-\sqrt{25*5}-\sqrt{25*7}=4\sqrt{5}-9\sqrt{7}-5\sqrt{5}-5\sqrt{7}\]\[-\sqrt{5}-14\sqrt{7}\]The answer to Q11: A

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sorry, i can't really help. i'm trying to get my own work done lol. best of luck :]
Thanks
Q12: \[\frac{ 4 }{ 6-\sqrt{7} }*\frac{ 6+\sqrt{7} }{ 6+\sqrt{7} }=\frac{ 4(6+\sqrt{7}) }{ 36-7 }\]\[\frac{ 24+4\sqrt{7} }{ 29 }\]The answer to Q12: B
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I can't give direct answers, but I can work with you through the answer.
ohkay
In this one \[\sqrt[8]{x^3}\]from the basic analysis that \[\sqrt[]{x}=\sqrt[2]{x}=x^{\frac{ 1 }{ 2 }}\]x is raised to the power of 1 (make it the numerator) and the 2 (make it the denominator) Do you follow?
yeah
So what's the answer?

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