// Numbas version: finer_feedback_settings {"name": "Ugur's copy of Working with standard index form", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Ugur's copy of Working with standard index form", "type": "question", "statement": "
Working with numbers that are very large or very small can be tricky.
\nStandard form allows us to simplify these numbers, using powers of 10.
\n\n
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).
$\\var{A[2]*10^2} = $ [[0]]
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", "type": "gapfill"}], "ungrouped_variables": ["A3dp", "A5dp", "small5", "A", "ran", "B", "int"], "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Convert a variety of numbers from decimal to standard index form.
"}, "preamble": {"css": "", "js": ""}, "advice": "Converting from decimal to a standard form, we are looking for $A \\times 10^n$.
\nWe need make the first number ($A$) between 1 and 10, so we put the decimal place after the first non-zero digit.
\n\n
a)
\nIn $\\var{A[2]*10^2}$, the first non-zero digit is $\\var{siground(A[2] - 0.5, 1)}$ so we get $A = \\var{A[2]}$.
\nIf 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 $n = 2$.
\n\n
b)
\nIn $\\var{A3dp*10^(-1)}$, the first non-zero digit is $\\var{siground(A3dp - 0.5, 1)}$ so we get $A = \\var{A3dp}$.
\nIf 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 $n = -1$.
\n\n
c)
\nIn $\\var{precround(A5dp*10^7,0)}$ the first non-zero digit is $\\var{siground(A5dp - 0.5, 1)}$ so we get $A = \\var{A5dp}$.
\nIf 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 $n = 7$.
\n\n
In $\\var{small5}$ the first non-zero digit is {siground({{small5}*10^5} - 0.5, 1)} so we get $A = \\var{small5*10^5}$.
\nIf 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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