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Multiply two numbers in standard form, then divide two numbers in standard form.

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Needs marking algorithm to allow equal values in standard form to gain equal marks

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Simplify, leaving the result in standard index form.

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Write you answer in the form n*10^x

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We use the rules for multiplying and dividing powers.

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To multiply powers we add, for example $10^2 \\times 10^6 = 10^{(2+6)} = 10^{8}$.

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To divide powers we subtract, for example $10^2 \\div 10^6 = 10^{(2-6)} = 10^{-4}$.

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a)

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\\[ \\begin{align} (\\var{int} \\times 10^\\var{ran - 3} ) \\times (\\var{int - 11} \\times 10^\\var{ran})  &= \\var{int} \\times (\\var{int-11}) \\times 10^{(\\var{ran - 3} + \\var{ran})} \\\\&= (\\var{int*(int-11)}) \\times 10^{\\var{ran - 3 + ran}} \\end{align} \\]

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We want to write this number in standard form so we move decimal place one place to the left to get

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\\[(\\var{int*(int-11)/10}) \\times 10^{(\\var{ran - 3 + ran} + 1)} = (\\var{int*(int-11)/10}) \\times 10^{\\var{ran - 2 + ran}} \\text{.}\\]

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b)

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Similarly,

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\\[ \\begin{align} (\\var{int*int} \\times 10^\\var{ran -4} ) \\div (\\var{int} \\times 10^\\var{ran - 2})  &= \\var{int*int} \\div \\var{int} \\times 10^{(\\var{ran - 4} - \\var{ran -2})} \\\\&= \\var{int} \\times 10^{-2} \\text{.}\\end{align} \\]

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$(\\var{int} \\times 10^\\var{ran - 3} ) \\times (\\var{int - 11} \\times 10^\\var{ran})  =$  [[0]]

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$(\\var{int*int} \\times 10^\\var{ran -4} ) \\div (\\var{int} \\times 10^\\var{ran - 2})  =$  [[0]]

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