please help me!!!

Correct, otherwise it would be:

(a5b0)3

---------

(2ba3)

multiplication by 0 in the numerator would make it:

0

----

6ba

and 0 divided by any number (other than 0) is 0:

0

please help me!!!

Tutors, please sign in to answer this question.

Placentia, CA

Correct, otherwise it would be:

(a5b0)3

---------

(2ba3)

multiplication by 0 in the numerator would make it:

0

----

6ba

and 0 divided by any number (other than 0) is 0:

0

Niceville, FL

I'm going to assume the equation you wrote is this:

(a^{5}•b^{0})^{3}

--------- = ?

(2b•a^{3})

The "cube" on the numerator's parentheses means

a5•b0 • a5•b0 • a5•b0

Since multiplying numbers means adding the exponents of the like terms, then

a^{(5+5+5)}•b^{(0+0+0)} = a^{15}•b^{0}

Thus you have

a^{15}•b^{0}

-------- = ?

2b•a^{3}

The zero exponent means "1", therefore

a^{15}•1

-------- = ?

2b•a^{3}

-------- = ?

2b•a

Multiplication is commutative, so you can rearrange the denominator

a^{15}•1

-------- = ?

a^{3}•2b

-------- = ?

a

And you can combine like terms into separate fractions, like this

a^{15} 1

---- • ---- = ?

a^{3} 2b

---- • ---- = ?

a

Dividing numbers means subtracting the exponents of the denominator from the exponents of the numerator, so

a^{(15-3)} = a^{12}

and

a^{12} 1

---- • ---- = ?

1 2b

---- • ---- = ?

1 2b

Recombining results in

a^{12}

------ = ?

2b

2b

So the answer is

(a^{5}•b^{0})^{3 } a^{12}

--------- = -----

(2b•a^{3}) 2b

--------- = -----

(2b•a

Of course, this is all assuming my interpretation of your problem is correct. Even if it's not, the principles shown here still hold.

Hope this helps!

Anna L.

Affordable Columbia Educated Engineer

New York, NY

4.9
(30 ratings)

Kevin S.

Personalized Tutoring Services

Brooklyn, NY

5.0
(210 ratings)

Jeffrey G.

Former Med-Student turned Professional Science Tutor

Brooklyn, NY

5.0
(364 ratings)

- Math Help 4691
- Math 8195
- Math Problem 892
- Math Question 709
- Math Word Problem 3585
- Math Help For College 1275
- Mathematics 474
- Math Answers 374
- Maths 584
- Precalculus 1375