Question

A rectangle has a length of the fifth root of 16 inches and a width of 2 to the 1 over 5 power inches. Find the area of the rectangle.

A.) 2 to the 3 over 5 power inches squared
B.) 2 to the 4 over 5 power inches squared
C.) 2 inches squared
D.) 2 to the 2 over 5 power inches squared

Answer 1

So the length is a little tricky huh?\[\large\rm \ell=\sqrt[5]{16}\]Do you remember how to rewrite this with a `rational exponent`?

Answer 2

@zepdrix not sure i haven't done this since April ehh lol

Answer 3

Example:\[\large\rm \sqrt[a]{x^b}\quad=x^{b/a}\]The exponent I've created in this exponent has two parts to it.
The numerator is the degree of the "power" that you're used to seeing.
The denominator is the degree of the root.

So another example:\[\large\rm \sqrt[7]{x}\]Well, notice that there is no power labeled on x, so it's implicitly a 1.\[\large\rm \sqrt[7]{x^1}\quad= x^{1/7}\]

Answer 4

Oh oh oh, they gave us a 16...
and a 2 in the other dimension.
So we have a little bit more work to do :) my bad.

Answer 5

I see now... No I never learned that

Answer 6

We would like to rewrite this 16 as a power of 2.

Answer 7

\(\large\rm 16=2^?\)

Answer 8

How many 2's do you have to multiply together to get to 16?

Answer 9

8

Answer 10

No that's way too many :) hehe

Answer 11

If I multiply eight 2's together,\[\large\rm 2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2=256\]

Answer 12

Oh sorry 4 right?

Answer 13

2*2*2*2=16, ok that looks better :)

Answer 14

\[\large\rm \ell=\sqrt[5]{16}\]\[\large\rm \ell=\sqrt[5]{2^4}\]\[\large\rm \ell=2^{4/5}\]And we also have our width,\[\large\rm w=2^{1/5}\]

Answer 15

Remember how to find area of a rectangle?

Answer 16

@zepdrix ha um no i dont think so.. (lol its not that i didn't listen in math class, my teacher never thought us much)

Answer 17

To find the area, or "space", inside the rectangle, we simply multiply length times width.