Mature student Truncation

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    Hello, first time poster. I am preparing myself on maths as I am going back to study as a mature student. I am doing questions in my spare time. This one confuses me slightly but I have the general grasp of it.

    The value of Young’s Modulus of Elasticity (E)for a specific steel alloy is 214 854 226 000 N m-2


    Pick a suitable place to truncate this number and give the truncation error.

    I am pretty sure that I would truncate this as 21485 as you don't round up. My problem is now do I leave it at that or would I write it out as 21485 x 10^5 and would I write the truncation error as 4226000 x 10^5 or 4226000

    I also thought would they be written out as 2.1485 x 10^9 and truncation error 4.226000 x 10^4 or 0.00004226000 x 10^9





    Take the new value and give to;


    • Standard form to 3 significant figures (Which I know would be 2.15 x 10^11)
    • Engineering notation to 1 decimal place (Which I know would be 0.2 x 10^12)

    The first part stumps me as I do not know how to write out the answer and if I got the truncation error correct. Also, would I have to put NM in my answers. Sorry if this is so obvious
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    (Original post by Joel Brown)
    Hello, first time poster. I am preparing myself on maths as I am going back to study as a mature student. I am doing questions in my spare time. This one confuses me slightly but I have the general grasp of it.

    The value of Young’s Modulus of Elasticity (E)for a specific steel alloy is 214 854 226 000 N m-2


    Pick a suitable place to truncate this number and give the truncation error.

    I am pretty sure that I would truncate this as 21485 as you don't round up. My problem is now do I leave it at that or would I write it out as 21485 x 10^5 and would I write the truncation error as 4226000 x 10^5 or 4226000

    I also thought would they be written out as 2.1485 x 10^9 and truncation error 4.226000 x 10^4 or 0.00004226000 x 10^9





    Take the new value and give to;


    • Standard form to 3 significant figures (Which I know would be 2.15 x 10^11)
    • Engineering notation to 1 decimal place (Which I know would be 0.2 x 10^12)

    The first part stumps me as I do not know how to write out the answer and if I got the truncation error correct. Also, would I have to put NM in my answers. Sorry if this is so obvious
    Firstly you should always put units in your answers if they are given in the question.

    214 850 000 000 = 2.1485 * 10^11.
    The error is 4 226 000 = 4.226 * 10^6.
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    (Original post by HapaxOromenon3)
    Firstly you should always put units in your answers if they are given in the question.

    214 850 000 000 = 2.1485 * 10^11.
    The error is 4 226 000 = 4.226 * 10^6.
    Thank you I see where I went wrong. I was also working on this question and feel like I have the right answer.

    A new build estate requires a concrete path to be laid around the edge of each back garden. The first group of houses to be built have a garden 8m wide and 10m long. The path has constant width and is laid around the edge of the garden. If the area of the path is 100m2, calculate, by deriving a quadratic equation the width of the path (w) Find a quadratic expression to calculate the width of the path hence find the width using the known values. Create a spreadsheet to check your calculations and find an expression that could be used for gardens of different areas.

    Working out:

    8 x 10 = 80m^2
    8+2x multiplied by 10+2x minus 80
    (8+2x)(10+2x) - 80 = 100m^2
    80 + 16x = 20x + 4x^2 - 80 = 100m^2
    36x + 4x^2 = 100m^2

    x^2 + 9x - 25 = 0

    -b +/- √b^2 - 4ac
    ______
    2a

    -9 +/- √81-4(-25)
    ______
    -50

    -9 +/- √81+100
    ______
    -50

    -9 +/- √81-4x(x-25)

    -9 +/- √181
    ______
    2
    x= -9 1
    ____ + ____ √181 = 2.22681202 metres for the width of the path
    2 2
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    Thank you I see where I went wrong. I was also working on this question and feel like I have the right answer.

    A new build estate requires a concrete path to be laid around the edge of each back garden. The first group of houses to be built have a garden 8m wide and 10m long. The path has constant width and is laid around the edge of the garden. If the area of the path is 100m2, calculate, by deriving a quadratic equation the width of the path (w) Find a quadratic expression to calculate the width of the path hence find the width using the known values. Create a spreadsheet to check your calculations and find an expression that could be used for gardens of different areas.

    Working out:

    8 x 10 = 80m^2
    8+2x multiplied by 10+2x minus 80
    (8+2x)(10+2x) - 80 = 100m^2
    80 + 16x = 20x + 4x^2 - 80 = 100m^2
    36x + 4x^2 = 100m^2

    x^2 + 9x - 25 = 0

    -b +/- √b^2 - 4ac
    ______
    2a

    -9 +/- √81-4(-25)
    ______
    -50

    -9 +/- √81+100
    ______
    -50

    -9 +/- √81-4x(x-25)

    -9 +/- √181
    ______
    2
    x= -9 1
    ____ + ____ √181 = 2.22681202 metres for the width of the path
    2 2
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    (Original post by Joel Brown)
    Thank you I see where I went wrong. I was also working on this question and feel like I have the right answer.

    A new build estate requires a concrete path to be laid around the edge of each back garden. The first group of houses to be built have a garden 8m wide and 10m long. The path has constant width and is laid around the edge of the garden. If the area of the path is 100m2, calculate, by deriving a quadratic equation the width of the path (w) Find a quadratic expression to calculate the width of the path hence find the width using the known values. Create a spreadsheet to check your calculations and find an expression that could be used for gardens of different areas.

    Working out:

    8 x 10 = 80m^2
    8+2x multiplied by 10+2x minus 80
    (8+2x)(10+2x) - 80 = 100m^2
    80 + 16x = 20x + 4x^2 - 80 = 100m^2
    36x + 4x^2 = 100m^2

    x^2 + 9x - 25 = 0

    -b +/- √b^2 - 4ac
    ______
    2a

    -9 +/- √81-4(-25)
    ______
    -50

    -9 +/- √81+100
    ______
    -50

    -9 +/- √81-4x(x-25)

    -9 +/- √181
    ______
    2
    x= -9 1
    ____ + ____ √181 = 2.22681202 metres for the width of the path
    2 2
    It looks like your solution is correct.
 
 
 
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