Get insightful responses to your questions quickly and easily on IDNLearn.com. Our platform offers comprehensive and accurate responses to help you make informed decisions on any topic.

Which shows the following expression after the negative exponents have been eliminated?

[tex]\[ \frac{a^3 b^{-2}}{a b^{-4}}, \quad a \neq 0, \quad b \neq 0 \][/tex]

A. [tex]\[ \frac{a^3 b^{-4}}{a b^{-2}} \][/tex]

B. [tex]\[ \frac{a b^4}{a^3 b^2} \][/tex]

C. [tex]\[ -\frac{a^3 b^4}{a b^2} \][/tex]

D. [tex]\[ \frac{a^3 b^4}{a b^2} \][/tex]


Sagot :

To simplify the given expression [tex]\(\frac{a^3 b^{-2}}{a b^{-4}}\)[/tex] and eliminate the negative exponents, let's go step by step.

1. Start with the Original Expression:
[tex]\[ \frac{a^3 b^{-2}}{a b^{-4}} \][/tex]

2. Simplify the [tex]\(a\)[/tex] terms:
Using the property of exponents [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex], we can simplify the [tex]\(a\)[/tex] terms:
[tex]\[ \frac{a^3}{a} = a^{3-1} = a^2 \][/tex]
So the expression becomes:
[tex]\[ \frac{a^3 b^{-2}}{a b^{-4}} = \frac{a^2 b^{-2}}{b^{-4}} \][/tex]

3. Simplify the [tex]\(b\)[/tex] terms:
Using the property of exponents [tex]\(\frac{b^m}{b^n} = b^{m-n}\)[/tex], we can simplify the [tex]\(b\)[/tex] terms:
[tex]\[ \frac{b^{-2}}{b^{-4}} = b^{-2 - (-4)} = b^{-2 + 4} = b^2 \][/tex]
So the expression now becomes:
[tex]\[ a^2 b^2 \][/tex]

4. Final Simplified Expression:
Therefore, the expression with eliminated negative exponents is:
[tex]\[ a^2 b^2 \][/tex]

The correct answer from the given options is:
[tex]\[ \frac{a^3 b^{-2}}{a b^{-4}} = a^2 b^2 \][/tex]

Comparing this with the options provided:
- [tex]\(\frac{a^3 b^4}{a b^2}\)[/tex]

Clearly, the correct simplified form from the options given is:
[tex]\[ \boxed{\frac{a^3 b^4}{a b^2}} \][/tex]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.