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how do you work out gradient?

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    if you have got a agra[h with a sytraight line on it, how do you work out teh gradient, i think know the formula m=(y1-y2) / (x1- x2) but am struggling to use it, can someone give me an example please!?

    thanks!
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    Two points on the straight line: say, (6,4) and (-3,2)

    We'll assume that (6,4) is (x1,y1) and (-3,2) is (x2,y2)

    Use the formula: m=y2-y1/x2-x1
    =2-4/-3-6
    =-2/-9

    So the gradient of the straight line is 2/9 (two negatives equal a positive).
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    (Original post by Ramadulla)
    Two points on the straight line: say, (6,4) and (-3,2)

    We'll assume that (6,4) is (x1,y1) and (-3,2) is (x2,y2)

    Use the formula: m=y2-y1/x2-x1
    =2-4/-3-6
    =-2/-9

    So the gradient of the straight line is 2/9 (two negatives equal a positive).
    i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

    thank you soo much!
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    (Original post by Ramadulla)
    Two points on the straight line: say, (6,4) and (-3,2)

    We'll assume that (6,4) is (x1,y1) and (-3,2) is (x2,y2)

    Use the formula: m=y2-y1/x2-x1
    =2-4/-3-6
    =-2/-9

    So the gradient of the straight line is 2/9 (two negatives equal a positive).
    Couldn't of explained it better, myself!
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    (Original post by 786_786)
    i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

    thank you soo much!
    He... just did.
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    (Original post by nuodai)
    He... just did.
    yeh but the answers there so i dont know if id be able to work it out in the exam myself...ah never mind ill find an example on th net somewhere
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    (Original post by 786_786)
    i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

    thank you soo much!
    Work out the gradient of a line which passes through the x-axis at 4, and which also passes through the point (14,9)
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    (Original post by 786_786)
    i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

    thank you soo much!
    Example:
    Work out the gradient of line A and line B where A(10,9) and B(6,-7). Make an equation in the form of y=mx+c

    Working:

    Okay, so we have coordinates of A and B, so we use the formula: m=(y2-y1)/(x2-x2). So substitute the coordinates of both A and B into the formula: m=(-7-9)/(6-10), which then equals: m=-16/-4 which simplifies to +4.

    We then substitute into the formula, and can use any of the points above for either A and B. So we substitute our gradient (m) into the (m) of the formula and any point, we will pick point A (easier as no negative or positive signs-less likely to have a mistake).

    So: y-9=4(x-10)
    y-9=4x+40
    y=4x+49

    That's and example! :yep:
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    (Original post by sohanshah)
    Work out the gradient of a line which passes through the x-axis at 4, and which also passes through the point (14,9)
    is it 0.8? :o:
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    9/10

    remembering the first co-ordinate is (4,0)
  11. Offline

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    (Original post by sohanshah)
    9/10

    remembering the first co-ordinate is (4,0)
    forget it ill just miss the gradient question out!
  12. Offline

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    (Original post by 786_786)
    forget it ill just miss the gradient question out!
    what exam is this for?
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    It's worth pursuing this as it's money for old rope once you've got it.
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    if you have A(x_1,y_1), B(x_2,y_2)

    then the gradient between the two points is \frac{y_2 - y_1}{x_2 - x_1}
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    (Original post by 786_786)
    i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

    thank you soo much!
    Level A = 16.050
    Level B = 16.042

    Distance between = 2.4m
    change in level = 0.008

    0.008/2.4 = Gradient
    2.4/0.008 = 1 in 300

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