Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Let's simplify the expression [tex]\(\frac{x^0 y^{-3}}{x^2 y^{-1}}\)[/tex] step-by-step.
1. Simplify the numerator and the denominator separately:
- The numerator is [tex]\(x^0 y^{-3}\)[/tex]:
- [tex]\(x^0 = 1\)[/tex] (Any number raised to the power of 0 is 1).
- So, the numerator becomes [tex]\(1 \cdot y^{-3} = y^{-3}\)[/tex].
- The denominator is [tex]\(x^2 y^{-1}\)[/tex]:
- It remains [tex]\(x^2 y^{-1}\)[/tex].
The expression now looks like this:
[tex]\[ \frac{y^{-3}}{x^2 y^{-1}}. \][/tex]
2. Rewriting the expression using properties of exponents:
We can split the fraction into two separate fractions:
[tex]\[ \frac{y^{-3}}{x^2 y^{-1}} = \frac{y^{-3}}{x^2} \cdot \frac{1}{y^{-1}}. \][/tex]
3. Simplify each part of the expression:
- The first part is [tex]\(\frac{y^{-3}}{x^2}\)[/tex]:
[tex]\[ \frac{y^{-3}}{x^2}. \][/tex]
- The second part is [tex]\(\frac{1}{y^{-1}}\)[/tex]:
- We can rewrite [tex]\(y^{-1}\)[/tex] as [tex]\(\frac{1}{y^{-1}} = y^1 = y\)[/tex].
Now, we multiply these simplified parts together:
[tex]\[ \frac{y^{-3}}{x^2} \cdot y = \frac{y^{-3} \cdot y^1}{x^2}. \][/tex]
4. Combine the exponents:
Using the properties of exponents, when you multiply terms with the same base, you add the exponents:
[tex]\[ y^{-3} \cdot y^1 = y^{-3+1} = y^{-2}. \][/tex]
Thus, the expression simplifies to:
[tex]\[ \frac{y^{-2}}{x^2}. \][/tex]
5. Express with positive exponents:
Since [tex]\(y^{-2}\)[/tex] can be rewritten with a positive exponent as [tex]\(\frac{1}{y^2}\)[/tex]:
[tex]\[ \frac{y^{-2}}{x^2} = \frac{1}{x^2 y^2}. \][/tex]
Therefore, the simplified form of the given expression [tex]\(\frac{x^0 y^{-3}}{x^2 y^{-1}}\)[/tex] is [tex]\(\frac{1}{x^2 y^2}\)[/tex].
The correct answer is (B) [tex]\(\frac{1}{x^2 y^2}\)[/tex].
1. Simplify the numerator and the denominator separately:
- The numerator is [tex]\(x^0 y^{-3}\)[/tex]:
- [tex]\(x^0 = 1\)[/tex] (Any number raised to the power of 0 is 1).
- So, the numerator becomes [tex]\(1 \cdot y^{-3} = y^{-3}\)[/tex].
- The denominator is [tex]\(x^2 y^{-1}\)[/tex]:
- It remains [tex]\(x^2 y^{-1}\)[/tex].
The expression now looks like this:
[tex]\[ \frac{y^{-3}}{x^2 y^{-1}}. \][/tex]
2. Rewriting the expression using properties of exponents:
We can split the fraction into two separate fractions:
[tex]\[ \frac{y^{-3}}{x^2 y^{-1}} = \frac{y^{-3}}{x^2} \cdot \frac{1}{y^{-1}}. \][/tex]
3. Simplify each part of the expression:
- The first part is [tex]\(\frac{y^{-3}}{x^2}\)[/tex]:
[tex]\[ \frac{y^{-3}}{x^2}. \][/tex]
- The second part is [tex]\(\frac{1}{y^{-1}}\)[/tex]:
- We can rewrite [tex]\(y^{-1}\)[/tex] as [tex]\(\frac{1}{y^{-1}} = y^1 = y\)[/tex].
Now, we multiply these simplified parts together:
[tex]\[ \frac{y^{-3}}{x^2} \cdot y = \frac{y^{-3} \cdot y^1}{x^2}. \][/tex]
4. Combine the exponents:
Using the properties of exponents, when you multiply terms with the same base, you add the exponents:
[tex]\[ y^{-3} \cdot y^1 = y^{-3+1} = y^{-2}. \][/tex]
Thus, the expression simplifies to:
[tex]\[ \frac{y^{-2}}{x^2}. \][/tex]
5. Express with positive exponents:
Since [tex]\(y^{-2}\)[/tex] can be rewritten with a positive exponent as [tex]\(\frac{1}{y^2}\)[/tex]:
[tex]\[ \frac{y^{-2}}{x^2} = \frac{1}{x^2 y^2}. \][/tex]
Therefore, the simplified form of the given expression [tex]\(\frac{x^0 y^{-3}}{x^2 y^{-1}}\)[/tex] is [tex]\(\frac{1}{x^2 y^2}\)[/tex].
The correct answer is (B) [tex]\(\frac{1}{x^2 y^2}\)[/tex].
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.