For a positive integer n,if 10^{n-2} 91n,th | vedclass

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(C) We test values for $n$ to check the inequality $10^{n-2} > 91n$: For $n=1$: $10^{-1} = 0.1$,$91(1) = 91$. $0.1 > 91$ is false. For $n=2$: $10^{0}

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$\left\{ 5, 6, 7, 8, \dots \right\}$

Vedclass · Answer

(C) We test values for $n$ to check the inequality $10^{n-2} > 91n$: For $n=1$: $10^{-1} = 0.1$,$91(1) = 91$. $0.1 > 91$ is false. For $n=2$: $10^{0} = 1$,$91(2) = 182$. $1 > 182$ is false. For $n=3$: $10^{1} = 10$,$91(3) = 273$. $10 > 273$ is false. For $n=4$: $10^{2} = 100$,$91(4) = 364$. $100 > 364$ is false. For $n=5$: $10^{3} = 1000$,$91(5) = 455$. $1000 > 455$ is true. For $n=6$: $10^{4} = 10000$,$91(6) = 546$. $10000 > 546$ is true. Since the exponential function $10^{n-2}$ grows much faster than the linear function $91n$,the inequality will hold for all $n \ge 5$. Thus,the set of values is $\left\{ 5, 6, 7, 8, \dots \right\}$.

For a positive integer $n$,if $10^{n-2} > 91n$,then the complete set of values of $n$ is:

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