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Answer the following:

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Fill in a table of powers of 10.

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

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

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

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

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When $n = 0$:

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

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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

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\\[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{ .}\\]

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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$
\n

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

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\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)$
13$\\var{gp[0]}$3$\\var{hp[0]}$
21$\\var{gp[1]}$2$\\var{hp[1]}$
32$\\var{gp[2]}$1$\\var{hp[2]}$
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$w'(\\var{b})=$[[0]]

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