Question

Simplify these expressions. Use only positive exponents.


p^-5

1/7t^-3

(6m^-2n^3)^0

Answer 1

@jdoe0001

Answer 2

well... just the first one is almost a given one
recall that \(\bf a^{-{\color{red} n}} \implies \cfrac{1}{a^{\color{red} n}}\qquad \qquad

\cfrac{1}{a^{\color{red} n}}\implies a^{-{\color{red} n}}
\\ \quad \\
% negative exponential denominator
a^{{\color{red} n}} \implies \cfrac{1}{a^{-\color{red} n}}
\qquad \qquad

\cfrac{1}{a^{-\color{red} n}}\implies \cfrac{1}{\frac{1}{a^{\color{red} n}}}\implies a^{{\color{red} n}}
\\ \quad \\

% rational with exponential terms
\cfrac{a^{\color{red}{ m}}}{a^{\color{purple}{ n}}}\implies \cfrac{a^{\color{red}{ m}}}{1}\cdot \cfrac{1}{a^{\color{purple}{ n}}}\implies a^{\color{red}{ m}}\cdot a^{-\color{purple}{ n}}\implies a^{{\color{red}{ m}}{-\color{purple}{ n}}}\)

Answer 3

It just all looks like gibberish....

Answer 4

hheheh w0t?

Answer 5

are you hmm using Internet Explorer by any chance? or you can see the red numbers?

Answer 6

I can see them. I just don't understand them.

Answer 7

the "a" is any value pretty much
raised to some "n" exponent
if the exponent is negative, you flip it over the fraction, to get a positive exponent :)

Answer 8

so \(\bf p^{-5}\implies \cfrac{1}{p^{\color{red}{ 5}}}\)

Answer 9

\(\bf \cfrac{1}{7t^{-3}}\implies \cfrac{1}{7}\cdot \cfrac{1}{t^{-3}}\implies \cfrac{1}{7}\cdot t^{\color{red}{ 3}}\implies \cfrac{t^3}{7}\)

Answer 10

the last one is raised to the "0" exponent, and well.. anything raised to 0 (except 0 itself), is 1 :)

Answer 11

@zepdrix

Answer 12

Do you think you could help me with the last one please?

Answer 13

hi book c:

Answer 14

hi! :)

Answer 15

When you raise something to the 1st power,
you leave it alone.
\(\rm (stuff)^1=stuff\)

When you raise something to the 2nd power,
you multiply it by a copy of itself.
\(\rm (stuff)^2=(stuff)(stuff)\)

So when you raise something to the 0th power,
you're actually dividing the only copy out of itself.
\(\rm (stuff)^0=(stuff)\div(stuff)=1\)

Answer 16

Anything* divided by itself, gives us 1, ya?
Because it divides evenly into itself one time.

Answer 17

yeah, but i'm not sure I see where this is going...

Answer 18

(sorry i'm just about clueless when it comes to math.)

Answer 19

The point is that,\[\large\rm (6m^{-2}n^3)^0=(6m^{-2}n^3)\div(6m^{-2}n^3)\]\[\large\rm (6m^{-2}n^3)^0=1\]you don't actually have to distribute the 0 to each item,
the 0 power always give us 1 as a result :)

Answer 20

So it looks long (maybe a bit complicated) and it only ends up as a 1 because of the 0th power?

Answer 21

Yes :)\[\large\rm 4^0=1\]\[\large\rm (7x)^0=1\]\[\large\rm (12x^2y^3)^0=1\]\[\large\rm (75a^{-4}b^{3}c^{54}d^{12})^0=1\]

Answer 22

Keep in mind that ONLY the part with the 0th power on it turns into a 1 though,
Example:\[\large\rm 7\color{orangered}{(12x^2)^0}=7\cdot \color{orangered}{1}\]

Answer 23

Good to know! Thank you so much! I think that because of your help (and the time you took to explain things so I understood.) I'm going to have to knowledge to pass my test next week. I really don't think I can thank you enough!

Answer 24

lol aw c:

Answer 25

Keep up the hard work, you'll do great.
Be blessed! c: