46.3k views
9 votes
Multiplicative inverse of (7/3)^-3

2 Answers

3 votes

Hi there! I am EnjoyingLife, and I hope you find my answer helpful! :)

STEPS:

First we have a negative power so we flop the number over

Then since we have a fraction to a power we raise both the top & bottom to that power.

Then, we flop the number over once more to find the multiplicative inverse.

Hope it helps!

User OpenBSDNinja
by
7.5k points
7 votes

Hello.

First, let's calculate


\tt{\displaystyle((7)/(3)) ^(-3)

Remember the Properties of Exponents:

If we have a number to a negative power, we flop the number over:


\tt{a^(-b)=\displaystyle(1)/(a^b)

Now, let's use this property to simplify our expression.

Flop the number over:


\tt{\displaystyle((3)/(7) )^3

Now, we should recall another property of exponents:


\tt{\displaystyle((a)/(b) )^m=(a^m)/(b^m)

Since we have a fraction to a power, we should raise both the numerator and the denominator to the power:


\tt{\displaystyle(3^3)/(7^3)

3 cubed is 9, and 7 cubed is 343:


\tt{\displaystyle(9)/(343)

Now, the multiplicative inverse of that number is simply that number flopped over:


\Large\boxed{\tt{\displaystyle(343)/(9) }}

I hope it helps.

Have a great day.


\boxed{imperturbability}

User Henriale
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories