Question

Help with a few problems for a medal and a fan?

Answer 1

\[\frac{ 1 }{ p ^{2}-q ^{-4}r ^{0} }\]

Answer 2

??

Answer 3

latex is not working, you'll have to write it out :(

Answer 4

I can see it. Is this it?

\(\Large \dfrac{1}{p^2 - q^{-4}r^0} \)

Answer 5

lol its just a bunch of parenthesis xD

Answer 6

Yes

Answer 7

The first thing to do is to get rid of the negative exponent in the denominator.
Write it as a fraction.

Answer 8

\(\Large \dfrac{1}{p^2 - q^{-4}r^0}\)

\(\Large= \dfrac{1}{p^2 - \frac{r^0}{q^4} }\)

What is \(r^0\) ?

Answer 9

0?

Answer 10

No. In general a number raised to the zero power is equal to 1. The only exception is 0^0.

Answer 11

That means, assuming r is not 0, r^0 = 1.

Answer 12

\(\Large= \dfrac{1}{p^2 - \frac{1}{q^4} }\)

Since the denominator has a denominator, multiply the numerator and denominator of the main fraction by q^4.

\(\Large= \dfrac{q^4}{q^4} \cdot \dfrac{1}{p^2 - \frac{1}{q^4} }\)

Answer 13

Okay so now we have \[\frac{ 1 }{ p ^{2}-q ^{4} } \] ?

Answer 14

Wait what

Answer 15

Just because r^0 is 1, it doesn't mean the small fraction disappears. \( \dfrac{r^0}{q^4}\) became \( \dfrac{1}{q^4} \). It is still a fraction.

Answer 16

Then you multiply the fraction by \(\dfrac{q^4}{q^4}\).

\(\Large= \dfrac{q^4}{q^4} \cdot \dfrac{1}{p^2 - \frac{1}{q^4} }\)

\( \Large= \dfrac{q^4}{p^2q^4 - \frac{q^4}{q^4} } \)

\(\Large= \dfrac{q^4}{p^2q^4 - 1 } \)

Now you no longer have a negative exponent or a complex fraction.