Join the IDNLearn.com community and start finding the answers you need today. Our Q&A platform is designed to provide quick and accurate answers to any questions you may have.
Sagot :
To find the value of the given expression [tex]\(\frac{j^3 k}{h^0}\)[/tex] where [tex]\(h = 8\)[/tex], [tex]\(j = -1\)[/tex], and [tex]\(k = -12\)[/tex], we follow these steps:
1. Evaluate [tex]\(h^0\)[/tex]:
- Any number raised to the power of 0 is 1. Hence, [tex]\(h^0 = 1\)[/tex].
2. Simplify the expression using the evaluated value:
- The expression simplifies to [tex]\(\frac{j^3 k}{1}\)[/tex] because [tex]\(h^0 = 1\)[/tex].
3. Calculate [tex]\(j^3\)[/tex]:
- [tex]\(j = -1\)[/tex]
- [tex]\((-1)^3 = -1\)[/tex]
4. Multiply [tex]\(j^3\)[/tex] by [tex]\(k\)[/tex]:
- [tex]\(k = -12\)[/tex]
- So, [tex]\((-1)^3 \cdot (-12) = -1 \cdot -12 = 12\)[/tex]
5. Simplify the final expression:
- The final expression is [tex]\(\frac{12}{1} = 12\)[/tex]
Thus, the value of the expression [tex]\(\frac{j^3 k}{h^0}\)[/tex] is [tex]\(\boxed{12}\)[/tex].
Therefore, the correct answer is:
A. 12
1. Evaluate [tex]\(h^0\)[/tex]:
- Any number raised to the power of 0 is 1. Hence, [tex]\(h^0 = 1\)[/tex].
2. Simplify the expression using the evaluated value:
- The expression simplifies to [tex]\(\frac{j^3 k}{1}\)[/tex] because [tex]\(h^0 = 1\)[/tex].
3. Calculate [tex]\(j^3\)[/tex]:
- [tex]\(j = -1\)[/tex]
- [tex]\((-1)^3 = -1\)[/tex]
4. Multiply [tex]\(j^3\)[/tex] by [tex]\(k\)[/tex]:
- [tex]\(k = -12\)[/tex]
- So, [tex]\((-1)^3 \cdot (-12) = -1 \cdot -12 = 12\)[/tex]
5. Simplify the final expression:
- The final expression is [tex]\(\frac{12}{1} = 12\)[/tex]
Thus, the value of the expression [tex]\(\frac{j^3 k}{h^0}\)[/tex] is [tex]\(\boxed{12}\)[/tex].
Therefore, the correct answer is:
A. 12
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.