### Solutions to Episode Five

a) f(x)= 4x-x^2 and g(x)=5-2x

in this case 4x-x^2 lagrer than or equal to 5-2x. The graph interest where 4x-x^2=5-2x

solving this equation,

4x-x^2=5-2x

x^2-6x+5=0

(x-1)(x-5)=0

x=1 or x=5

we substitute in either formula and find that graphs intersect at (1,3) and (5,-5)

then,

Area(x) =

in this case 4x-x^2 lagrer than or equal to 5-2x. The graph interest where 4x-x^2=5-2x

solving this equation,

4x-x^2=5-2x

x^2-6x+5=0

(x-1)(x-5)=0

x=1 or x=5

we substitute in either formula and find that graphs intersect at (1,3) and (5,-5)

then,

Area(x) =

**S***(from 1 to 5)*((4x-x^2) - (5-2x)) dx=(3x^2- (1/3 )x^3- 5x)

*(from 1 to 5)*=75-125/3-25)-(3-1/3-5)

=32/3

*b)*1/14**S***(from 6 to 14)*-(80-10 cos (pai t/12) dt=-1/8(697.2957795)

=-87 F

c) -(80 -10 cos (pai t/12) >/ -78

80 - 10 cos (pai t/12) /<>

-5.230>/t>/-18.769

between this time it will have the storm

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