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Pre calculus with integration and differentiation

Mathematics
Tutor: None Selected Time limit: 1 Day

find area under curve with 90 m length and 18.5 m right in middle. first have to find equation of the curve

Apr 23rd, 2015

It sounds like you are describing an inverted parabola with zeros at +45 and -45, with vertex at (0,18.5)

In which case F(x) = -ax^2 + 18.5

when F(x) = 0, x = +/- 45

-a(45)^2 + 18.5 = 0

a = 18.5/45^2

a = 0.0089

F(x) = -0.0089x^2 +18.5

the integral of that is -0.0267x^3 + 18.5x taken from 0 to 45

calculate that, and then multiply by 2 because the area is symmetrical about the y -axis

Apr 23rd, 2015

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Apr 23rd, 2015
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Apr 23rd, 2015
May 25th, 2017
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