Sure, let's tackle each part of the question one by one with detailed explanations.
1. Fill in the blanks:
(a) Among [tex]\(\frac{-6}{11}, \frac{-6}{13}, \frac{-6}{7}\)[/tex] the greatest rational number is:
- To compare these rational numbers, we need to understand that larger negative denominators result in larger negative values (closer to zero).
[tex]\[
\frac{-6}{11} \approx -0.545, \quad \frac{-6}{13} \approx -0.462, \quad \frac{-6}{7} \approx -0.857
\][/tex]
Comparatively, [tex]\(\frac{-6}{13}\)[/tex] is the greatest (least negative) rational number among the three given.
- The answer is [tex]\(\frac{-6}{13}\)[/tex].
(b) A negative number raised to the power of zero is equal to:
- Any non-zero number raised to the power of zero is always equal to 1.
- Hence, the result is [tex]\(1\)[/tex].
(c) [tex]\(1+3+5+7+\ldots+17\)[/tex]:
- This is the sum of the first nine odd numbers. The formula for the sum of the first [tex]\(n\)[/tex] odd numbers is [tex]\(\sum_{i=1}^{n} (2i-1) = n^2\)[/tex].
- Here, [tex]\(n = 9\)[/tex] (because there are 9 terms), so:
[tex]\[
9^2 = 81
\][/tex]
- The answer is [tex]\(81\)[/tex].
(d) The smallest number after 1 that is a perfect square as well as a perfect cube:
- A number that is both a perfect square and a perfect cube must be of the form [tex]\(n^6\)[/tex] (since [tex]\((n^2)^3 = (n^3)^2 = n^6\)[/tex]).
- The smallest number greater than 1 that fits this criterion is [tex]\(2^6 = 64\)[/tex].
- The answer is [tex]\(64\)[/tex].
(e) If [tex]\(6^{x-2}=1\)[/tex], then [tex]\(x\)[/tex] equals:
- Any number raised to the power of zero is 1. Therefore, [tex]\(6^{0} = 1\)[/tex].
- Equating the exponents:
[tex]\[
x - 2 = 0 \implies x = 2
\][/tex]
- The answer is [tex]\(2\)[/tex].
So, the filled blanks are:
(a) [tex]\(\frac{-6}{13}\)[/tex]
(b) [tex]\(1\)[/tex]
(c) [tex]\(81\)[/tex]
(d) [tex]\(64\)[/tex]
(e) [tex]\(2\)[/tex]
