// Numbas version: finer_feedback_settings {"name": "31. Converting standard form", "metadata": {"description": "
Conversions between standard form and non-standard form.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following number in scientific notation.
", "advice": "Suppose we have the number $\\var{q2}$. In scientific notation, this number would start with $\\var{dec2}$ since we only want one digit in front of the decimal point. The decimal point is currently to the right of the last digit in $\\var{q2}$ and needs to be between the first and second digits, i.e $\\var{dec2}$. Count the places that the digits must move and you get $\\var{pow2}$ places. That is,
\n\n\\[\\var{q2}=\\var{dec2}\\times 10^{\\var{pow2}}\\]
\n\nWe have a positive $\\var{pow2}$ as the power because we need to make the number $\\var{dec2}$ bigger to get to $\\var{q2}$.
\n\nUse this link to find some resources which will help you revise this topic.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following number in scientific notation.
", "advice": "Suppose we have the number $\\var{q2}$. In scientific notation, this number would start with $\\var{dec2}$ since we only want one digit in front of the decimal point. The decimal point is currently to the right of the last digit in $\\var{q2}$ and needs to be between the first and second digits, i.e $\\var{dec2}$. Count the places that the digits must move and you get $\\var{pow2}$ places. That is,
\n\n\\[\\var{q2}=\\var{dec2}\\times 10^{\\var{pow2}}\\]
\n\nWe have a positive $\\var{pow2}$ as the power because we need to make the number $\\var{dec2}$ bigger to get to $\\var{q2}$.
\n\nUse this link to find some resources which will help you revise this topic.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following as an ordinary number.
", "advice": "To write convert a number in standard form back to an ordinary number we need to move the decimal point to the right if it's a positive power and left if it's a negative power. Since $\\var{dec2}\\times 10^\\var{pow2}$ has a positive power we are going to move the decimal point right $\\var{pow2}$ places (i.e. we need to make $\\var{dec2}$ bigger). That is,
\n\\[\\var{dec2}\\times 10^{\\var{pow2}}=\\var{q2}.\\]
\nUse this link to find some resources which will help you revise this topic.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following as an ordinary number.
", "advice": "To write convert a number in standard form back to an ordinary number we need to move the decimal point to the right if it's a positive power and left if it's a negative power. Since $\\var{dec2}\\times 10^\\var{pow2}$ has a positive power we are going to move the decimal point right $\\var{pow2}$ places (i.e. we need to make $\\var{dec2}$ bigger). That is,
\n\\[\\var{dec2}\\times 10^{\\var{pow2}}=\\var{q2}.\\]
\nUse this link to find some resources which will help you revise this topic.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following number in scientific notation.
", "advice": "Suppose we have the number $\\var{q2}$. In scientific notation, this number would start with $\\var{dec2}$ since we only want one digit in front of the decimal point. Count the places that the digits must move and you get $\\var{-pow2}$ places to the right. That is,
\n\\[\\var{q2}=\\var{dec2}\\times 10^{\\var{pow2}}\\]
\n\nWe have a negative $\\var{-pow2}$ as the power because we need to make the number $\\var{dec2}$ smaller to get to $\\var{q2}$.
\n\nUse this link to find some resources which will help you revise this topic.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following number in scientific notation.
", "advice": "Suppose we have the number $\\var{q2}$. In scientific notation, this number would start with $\\var{dec2}$ since we only want one digit in front of the decimal point. Count the places that the digits must move and you get $\\var{-pow2}$ places to the right. That is,
\n\\[\\var{q2}=\\var{dec2}\\times 10^{\\var{pow2}}\\]
\n\nWe have a negative $\\var{-pow2}$ as the power because we need to make the number $\\var{dec2}$ smaller to get to $\\var{q2}$.
\n\nUse this link to find some resources which will help you revise this topic.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following as an ordinary number.
", "advice": "To write convert a number in standard form back to an ordinary number we need to move the decimal point to the right if it's a positive power and left if it's a negative power. Since $\\var{dec2}\\times 10^\\var{pow2}$ has a negative power we are going to move the decimal point left $\\var{-pow2}$ places (i.e. we need to make $\\var{dec2}$ smaller). That is,
\n\\[\\var{dec2}\\times 10^{\\var{pow2}}=\\var{q2}.\\]
\nUse this link to find some resources which will help you revise this topic.
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", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{q2}", "maxValue": "{q2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "31.h Standard form to small number", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": ["converting", "scientific notation", "standard form"], "metadata": {"description": "Convert numbers between 0 and 1 intro standard form/scientific notation.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following as an ordinary number.
", "advice": "To write convert a number in standard form back to an ordinary number we need to move the decimal point to the right if it's a positive power and left if it's a negative power. Since $\\var{dec2}\\times 10^\\var{pow2}$ has a negative power we are going to move the decimal point left $\\var{-pow2}$ places (i.e. we need to make $\\var{dec2}$ smaller). That is,
\n\\[\\var{dec2}\\times 10^{\\var{pow2}}=\\var{q2}.\\]
\nUse this link to find some resources which will help you revise this topic.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following number in standard form.
", "advice": "First notice that $\\var{Num1}\\times 10^\\var{Pow1}$ is not in standard form since $\\var{Num1}$ is not between 1 and 10. There are two options for how to put this in standard form. We could first write $\\var{Num1}\\times 10^\\var{Pow1}$ as an ordinary number which would give us
\n\\[\\var{Num1}\\times 10^\\var{Pow1}=\\var{Num1*10^Pow1}\\]
\nwhich we can then write in standard form as
\n\\[\\var{Num1*10^Pow1}=\\var{dec2}\\times 10^\\var{pow2}.\\]
\nAlternatively, we can write $\\var{Num1}$ in standard form which would give us
\n\\[\\var{Num1}\\times 10^\\var{Pow1}=\\var{dec2}\\times10^\\var{DecOffset}\\times 10^\\var{Pow1}\\]
\nand then we simply use the rules of exponents to multiply the two powers of 10 together to get
\n\\[\\var{dec2}\\times10^\\var{DecOffset}\\times 10^\\var{Pow1}=\\var{dec2}\\times 10^\\var{pow2}.\\]
\nUse this link to find some resources which will help you revise this topic.
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