The formula to calculate the lateral area of a cone is given by
where
LA=10.5pi in2
r=2.1 in
l=? slant height
substitute given values
<h2>The answer is 5 inches</h2>
F = t ⇨ df = dt
dg = sec² 2t dt ⇨ g = (1/2) tan 2t
⇔
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt
u = 2t ⇨ du = 2 dt
As integral of tan u = - ln (cos (u)), you get :
integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer
To find where on the hill the canonball lands
So 0.15x = 2 + 0.12x - 0.002x^2
Taking the LHS expression to the right and rearranging we have -0.002x^2 + 0.12x -
0.15x + 2 = 0.
So we have -0.002x^2 - 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x^2 + 0.03x -2 = 0.
This is a quadratic eqn with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25. The second solution y = 0.15 * 25 = 3.75
The equation would be -5+(-12), so the answer would be -17 :)
