Question

What is the exact value of 3q^8 over sqrt q^19

Answer 1

\[\frac{3Q^{8}}{\sqrt{Q^{19}}}\]
Simplify the denominator.

Answer 2

How?

Answer 3

Rewrite it using these two general formulas:
\[\sqrt{AB} = \sqrt{A} \sqrt{B}\]
\[A^{X+Y} = A^{X}*A^{Y}\]

Answer 4

Which is ??? I'm really bad at this..

Answer 5

\[A^{T} = A^{X+Y}\]

Answer 6

whats substituted into the equation?

Answer 7

\[Q^{19} = Q^{X+Y} = Q^{X}*Q^{Y}\]
\[\sqrt{Q^{19}} = \sqrt{Q^{X+Y}} = \sqrt{Q^{X}*Q^{Y}} = \sqrt{Q^{X}}*\sqrt{Q^{Y}}\]

Answer 8

Of course you need to figure out X and Y. You can always reduce it bit by bit, until you can't reduce it anymore.

Answer 9

Example:
\[\sqrt{Q^{5}} = \sqrt{Q^{2+3}} = \sqrt{Q^{2}*Q^{3}} = \sqrt{Q^{2}}*\sqrt{Q^{3}} = Q*\sqrt{Q^{3}}\]
\[Q*\sqrt{Q^{3}} = Q*\sqrt{Q^{2+1}} = Q*\sqrt{Q^{2}*Q^{1}} = Q*\sqrt{Q^{2}}*\sqrt{Q^{1}} = Q*Q*\sqrt{Q^{1}} = Q^{2}\sqrt{Q}\]
The same thing can be done quicker as:
\[\sqrt{Q^{5}} = \sqrt{Q^{4+1}} = \sqrt{Q^{4}*Q^{1}} = \sqrt{Q^{4}}*\sqrt{Q^{1}} = Q^{2}*\sqrt{Q^{1}} = Q^{2}\sqrt{Q}\]

Answer 10

so the answer is ?? lol

Answer 11

Do you just want the answer, or do you want to know how to solve it for yourself?

Answer 12

The answer please, it's a final test.