1. ## Coordinate geometry

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.

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.

2. 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

3. Originally Posted by Coach

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: $y = \frac15 x^2 -5$

4. Hi, Moo!

yes that's how far I got too, but the sample answer says t is about -0.0068h-11.6

5. This is strange o.O

Perhaps they transformed degrees into Kelvin degrees or Fahrenheit ?