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Give all answers to 2 decimal places:
", "advice": "(a)
\nThe given width = $\\var{width}$ cm and length = $\\var{length}$ cm are each rounded to the nearest 0.1cm. To find the upper and lower bounds we must add/subtract half of this value.
\nSince half of 0.1cm is 0.05cm we obtain:
\nLower bound for width $= \\var{width}-0.05=\\var{width-0.05}$
\nUpper bound for width $= \\var{width}+0.05=\\var{width+0.05}$
\nLower bound for length $= \\var{length}-0.05=\\var{length-0.05}$
\nUpper bound for length $= \\var{length}+0.05=\\var{length+0.05}$
\n\n
Since area = length $\\times$ width, we have:
\nlower bound area = lower bound length $\\times$ lower bound width $=\\var{length-0.05} \\times \\var{width-0.05} = \\var{minarea}$ cm$^2=\\var{precround(minarea,2)}$ cm$^2$ to 2d.p.
\nand
\nupper bound area = upper bound length $\\times$ upper bound width $=\\var{length+0.05} \\times\\var{width+0.05} = \\var{maxarea}$ cm$^2=\\var{precround(maxarea,2)}$ cm$^2$ to 2d.p.
\n\n(b)
\nThe given width = $\\var{width2}$ cm is rounded to the nearest 1cm. To find the upper and lower bounds we must add/subtract half of this value.
\nSince half of 1cm is 0.5cm we obtain:
\nLower bound for width $= \\var{width2}-0.5=\\var{width2-0.5}$
\nUpper bound for width $= \\var{width2}+0.5=\\var{width2+0.5}$
\n\n\nThe given area = $\\var{area}$ cm is rounded to the nearest 10cm$^2$. To find the upper and lower bounds we must add/subtract half of this value.
\nSince half of 10cm$^2$ is 5cm$^2$ we obtain:
\nLower bound for area $= \\var{area}-5=\\var{area-5}$
\nUpper bound for area $= \\var{area}+5=\\var{area+5}$
\n\nSince length = area $\\div$ width, we have:
\nlower bound length= lower bound area $\\div$ upper bound width (to get a small answer we want to be dividing a small number by a big number)
\n$=\\var{area-5} \\div \\var{width2+0.5} = \\var{minlength}$ cm$=\\var{precround(minlength,2)}$ cm to 2d.p.
\n\nand
\nupper bound length = upper bound area $\\div$ lower bound width (to get a large answer we want to be dividing a large number by a small number)
\n$=\\var{area+5} \\div \\var{width2-0.5} = \\var{maxlength}$ cm$=\\var{precround(maxlength,2)}$ cm to 2d.p.
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\n(i) the minimum possible area of the rectangle in cm$^2$ [[0]]
\n(ii) the maximum possible area of the rectangle in cm$^2$ [[1]]
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\nWhat is:
\n(i) the minimum possible length of the rectangle in cm [[0]]
\n(ii) the maximum possible length of the rectangle in cm [[1]]
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