the height is at 40 when
t^2-14t+40 = 0
(t-4)(t-10) = 0
SO, it is above 40 for 6 seconds (between t=4 and t=10)
x(x+7)/2 = 60
x^2+7x-120 = 0
(x+15)(x-8)=0
The triangle is the familiar 8-15-17 right triangle.
1.A projectile's height (in metres) is given by the expression t(14-t), where time t is in seconds. How long is the projectile above a height of 40m?
2. The area of a right-angled triangle is 60 square units and the lengths of the two shorter sides differ by 7 units. Find the length of the hypotenuse.
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