Question

In need of help again. stuck on this problem and I've already had 4 people try to help me. 11√49a^5/4a^3

Answer 1

put your problem in at http://www.wolframalpha.com/?fp=right

Answer 2

it didnt work

Answer 3

quick question is it

1.

\[\sqrt{\frac{49a^5}{4a^3}}\]

or

2.

\[\frac{\sqrt{49a^5}}{4a^3}\]

Answer 4

(77/4) a^2 ? hard to know where the square roots stop

Answer 5

\[11\sqrt{\frac{ 49a^5 }{ 4a^3 }}\]

Answer 6

the answers gonna come out to be \[\frac{ 77a }{ 2}\] I just don't know how haha

Answer 7

ok... so look at it this way... cancel the a's first

so its

\[\sqrt{\frac{49a^2 \times a^3}{4 \times a^3}}\]

what cancels...from the numerator and denominator..?

Answer 8

oops forgot the 11 outside.. pick that up in a min

Answer 9

uh a^3?

Answer 10

ok... thats great so you have

and applying a fule about radicals it can be split to the square root of the factors..

\[11 \sqrt{\frac{49a^2}{4}} = \frac{11 \sqrt{49a^2}}{4}\]


to take the square root of the numerator and denominator which gives

\[\frac{11 \times 7a}{2} \]

and hence your answer

Answer 11

Thank you so much!!