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When $n$ is positive, we multiply $10$ by itself $n$ times,

\\[\\text{e.g. } 10^3 = 10 \\times 10 \\times 10 = 1000 \\text{ .}\\]

When $n$ is negative, we can think of $10^{-n}$ as $\\frac{1}{10^{n}}$,

\\[\\text{e.g. } 10^{-3} = \\frac{1}{10^3} = \\frac{1}{1000} = 0.001\\text{ .}\\]

When $n = 0$:

\\[10^{0} = 1 \\text{ .}\\]

Generally, we can think of $10^n$ as a number in standard form $1 \\times 10^n$. Then $n$ always tells us the number of decimal places to move the decimal point in $1.0$, for example

\\[10^{-3} = 1.0 \\times 10^{-3} \\text{ and since } n = - 3 \\text{, we go } 3 \\text{ places back as follows: } 1.0 ⇒ 0.1 ⇒ 0.01 ⇒ 0.001 \\text{ .}\\]

A complete table of powers of ten for $n$ from $-6$ to $6$ is:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$n$	$10^n$
$-6$	$0.000001$
$-5$	$0.00001$
$-4$	$0.0001$
$-3$	$0.001$
$-2$	$0.01$
$-1$	$0.1$
$0$	$1$
$1$	$10$
$2$	$100$
$3$	$1000$
$4$	$10000$
$5$	$100000$
$6$	$1000000$

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Fill in a table of powers of 10.

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Use the table below to find $f'(\\var{a})$ and $w'(\\var{b})$ if $f(x)=g(h(x))$ and $w(x)=h(g(x))$.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$x$	$g(x)$	$g'(x)$	$h(x)$	$h'(x)$
1	3	$\\var{gp[0]}$	3	$\\var{hp[0]}$
2	1	$\\var{gp[1]}$	2	$\\var{hp[1]}$
3	2	$\\var{gp[2]}$	1	$\\var{hp[2]}$

$f'(\\var{a})=$[[0]]

$w'(\\var{b})=$[[1]]

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Answer the following:

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