Find an equation for the plane containing the point (1, 2, 3) and parallel to the plane x 2y 3z
1 answer:
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Alrighty
find where they intersect
9x²ln(x)=36ln(x)
divide both sides by 9
x²ln(x)=4ln(x)
so
so x=1 and and 2 (x can't be 0 or -2 because ln(0) and ln(-2) don't exist)
so intersect at x=1 and x=2
which is on top?
9(1.5)²ln(1.5)=20.25ln(1.5)
36ln(1.5)=36ln(1.5)
36ln(1.5) is on top
so
that will be
the area is
![[36x(ln(x)-1)-x^3(3ln(x)-1)]^2_1=](https://tex.z-dn.net/?f=%20%5B36x%28ln%28x%29-1%29-x%5E3%283ln%28x%29-1%29%5D%5E2_1%3D)
the area between the curves is 48ln(2)-29
find where they intersect
9x²ln(x)=36ln(x)
divide both sides by 9
x²ln(x)=4ln(x)
so
so x=1 and and 2 (x can't be 0 or -2 because ln(0) and ln(-2) don't exist)
so intersect at x=1 and x=2
which is on top?
9(1.5)²ln(1.5)=20.25ln(1.5)
36ln(1.5)=36ln(1.5)
36ln(1.5) is on top
so
that will be
the area is
the area between the curves is 48ln(2)-29
Hope this helps!! Lol
