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Math Help - Coordinate geometry

  1. #1
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    Coordinate geometry

    Hello!







    Please help me with these two problems


    Problem 1
    Up to a certain height the temperature above sea level can be modelled by a linear function of height h in meters. At the height 500m the temperature is 15 degrees celsius, at a height of 10 000m the temperature is -80 degrees. Determine an equation for the relation between height and temperature.




    I tried calculating the gradient, but I always got the wrong answer.



    Problem 2



    The width of the canal is 10.0m and the greatest depth is 5m. The cross-section of the canal has the shape of a parabola.
    How far from the bank is the depth 2.5m?



    I have tried everything, from trying to find the zero points to finding the axis of symmetry and nothing works.


    Please help me, I am grateful for any help, no matter how little.
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  2. #2
    Moo
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    Hello

    Problem 1
    Up to a certain height the temperature above sea level can be modelled by a linear function of height h in meters. At the height 500m the temperature is 15 degrees celsius, at a height of 10 000m the temperature is -80 degrees. Determine an equation for the relation between height and temperature.
    Let x be the height and y the temperature.

    y=f(x)=ax+b, with a and b to determine.

    You know the gradient, which may be \frac{y_B-y_A}{x_B-x_A}=\frac{-80-15}{10000-500}=-0.01

    Is that what you want ?

    Now, replace in y=ax+b to get b
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  3. #3
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    Quote Originally Posted by Coach View Post

    Problem 1
    Up to a certain height the temperature above sea level can be modelled by a linear function of height h in meters. At the height 500m the temperature is 15 degrees celsius, at a height of 10 000m the temperature is -80 degrees. Determine an equation for the relation between height and temperature.

    Problem 2

    The width of the canal is 10.0m and the greatest depth is 5m. The cross-section of the canal has the shape of a parabola.
    How far from the bank is the depth 2.5m?
    ...
    to #1: Calculate the equation of a straight line passing through the points (500, 15) and B(10000, -80)

    You should get: y = -\frac1{100} x + 20

    to #2: Use coordinates: The vertex of the parabola is V(0, -5) and the shores of the canal are at P(-5, 0) and Q(5, 0). With these three points it is possible to calculate the equation of the parabola.

    You should get: y = \frac15 x^2 -5
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  4. #4
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    Hi, Moo!



    yes that's how far I got too, but the sample answer says t is about -0.0068h-11.6
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  5. #5
    Moo
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    This is strange o.O

    Perhaps they transformed degrees into Kelvin degrees or Fahrenheit ?
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