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correction - mathhelper
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correction - mathhelper
Answers (4)
that should have been a = -6 and y = -6(x - 2)^2 + 1
sorry, I found the area by mistake, you wanted the volume vol = ∫ π y^2 dx from 0 to 2 = π ∫ 1/(x+1) dx from 0 to 2 = π [ ln(x1) ] from 0 to 2 = π(ln3 - ln1] = π ln3
Oooops .... My ratio should have been: x/2 = 5/√29 x = 10/√29 x =1.8569 We could have used trig from the slope of 2.5/1 tan P = 2.5/1 angle P = 68.1986° now you want the base side cos 68.1986 = x/5 x = 5cos68.1986 = appr 1.8569
for a) should say: we could have WWWW = (1/3)^4 = 1/81 ... prob(no draws) = 1/81 + 1/6 + 1/6 + 2/27 = 34/81
