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Math Help - Maths questions at GCSE level

  1. #1
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    Maths questions at GCSE level

    Can someone help answer the following questions plzz

    1)
    The area of the pentagon is 8560mmsquared

    Change 8560mmsquared top cmsquared


    2)

    Each side of another pentagon has a length of 101 mm,correct to the nearest millimetre.

    (i)Write down the least possible length of each side.


    (ii)Write down the greatest possible length of each side


    3)Jane's car uses 1 gallon of petrol for every 40 miles. A gallon of petrol costs 3.20

    (i) Work out the cost of petrol for Jane's 320 km journey.


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  2. #2
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    Quote Originally Posted by girliegal View Post
    Can someone help answer the following questions plzz

    1)
    The area of the pentagon is 8560mmsquared

    Change 8560mmsquared top cmsquared
    We know that 10 mm = 1 cm, so look at it this way:
    \frac{8560 \, mm^2}{1} = 8560 \cdot \frac{1 \, mm^2}{1} = 8560 \cdot \frac{ 1 \, mm}{1} \cdot \frac{1 \, mm}{1}

    Change each mm to cm:
    = 8560 \cdot \left ( \frac{1 \, mm}{1} \cdot \frac{1 \, cm}{10 \, mm} \right ) \left ( \frac{1 \, mm}{1} \cdot \frac{1 \, cm}{10 \, mm} \right )

     = \frac{8560}{10^2} \, cm^2 = \frac{8650}{100} \, cm^2 = 85.6 cm^2

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by girliegal View Post
    2)

    Each side of another pentagon has a length of 101 mm,correct to the nearest millimetre.

    (i)Write down the least possible length of each side.


    (ii)Write down the greatest possible length of each side
    This depends on which set of rules you use to round. My answer for i) would be 100.500...01 mm (anything greater than 100.5 mm) and my answer for ii) would be 101.5 mm.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by girliegal View Post
    3)Jane's car uses 1 gallon of petrol for every 40 miles. A gallon of petrol costs 3.20

    (i) Work out the cost of petrol for Jane's 320 km journey.
    Unfortunately I only remember one metric to English conversion: 2.54 cm = 1 in. So this is how I would do it. 320 km in terms of miles is:
    \frac{320 \, km}{1} \cdot \frac{1000 \, m}{1 \, km} \cdot \frac{100 cm}{1 m} \cdot \frac{1 \, in}{2.54 \, cm} \cdot \frac{1 \, ft}{12 in} \cdot \frac{1 \, mi}{5280 ft} = 198.83878 \, mi

    It costs 3.20 per mile. So the cost is (198.83878 mi)(3.20/mi) = 636.28

    -Dan
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    Thanks for posting.
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