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Working with numbers that are very large or very small can be tricky.

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Standard form allows us to simplify these numbers, using powers of 10.

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The standard index form can be defined as
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\\[A \\times 10^n,\\]
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where $1 ≤ A < 10$ and $n$ is an integer, e.g. $2.26 \\times 10^5$ is a standard form of a number 226000.
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Write the following in standard index form (for example, for $2.01\\times 10^5$ we would write 2.01*10^5 in the gap).

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$\\var{A[2]*10^2} = $  [[0]]

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$\\var{A3dp*10^(-1)} = $  [[0]]

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$\\var{precround(A5dp*10^7,0)} =$  [[0]]

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$\\var{small5} = $  [[0]]

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Convert a variety of numbers from decimal to standard index form.

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Converting from decimal to a standard form, we are looking for $A \\times 10^n$.

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We need make the first number ($A$) between 1 and 10, so we put the decimal place after the first non-zero digit.

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a)

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In $\\var{A[2]*10^2}$, the first non-zero digit is $\\var{siground(A[2] - 0.5, 1)}$ so we get $A = \\var{A[2]}$.

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If we moved the decimal place in $\\var{A[2]}$ so it matches our original number $\\var{A[2]*10^2}$, we would go 2 places to the right, so $n = 2$.

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b)

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In $\\var{A3dp*10^(-1)}$, the first non-zero digit is $\\var{siground(A3dp - 0.5, 1)}$ so we get $A = \\var{A3dp}$.

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If we moved the decimal place in $\\var{A3dp}$ so it matches our original number $\\var{A3dp*10^(-1)}$, we would go 1 place to the left, so $n = -1$.

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c)

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In $\\var{precround(A5dp*10^7,0)}$ the first non-zero digit is $\\var{siground(A5dp - 0.5, 1)}$ so we get $A = \\var{A5dp}$.

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If we moved the decimal place in $\\var{A5dp}$ so it matches our original number $\\var{precround(A5dp*10^7,0)}$, we would go 7 places to the right, so $n = 7$.

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d)

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In $\\var{small5}$ the first non-zero digit is {siground({{small5}*10^5} - 0.5, 1)} so we get $A = \\var{small5*10^5}$.

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If we moved the decimal place in $\\var{small5*10^5}$ so it matches our original number $\\var{small5}$, we would go 5 places to the left, so $n = -5$.

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