Ok so assume that if you have
xy=0 then x and y=0
-1/4(x^2-4x+3)>0
multiply both sides by -4 to clear fraction (-4/-4=1)
flip sign
x^2-4x+3<0
factor
(x-3)(x-1)<0
we know that (+) times (-)=(-) and
(-) times (-)=(+) so we don't want them to be both negative, we want different sign
we cannnot have 3 since it would be (0)(2)<0 which is false
we cannot have 1 either since (-2)(0)<0 is also false
lets see if the solution is in betwen 3 and 1 or outside of 3 and 1
ltry 2
2^3-4(2)+3<0
8-8+3<0
3<0
false
therefor it is outside ie
x>3 and x<1
Answer:
Volume = 16 unit^3
Step-by-step explanation:
Given:
- Solid lies between planes x = 0 and x = 4.
- The diagonals rum from curves y = sqrt(x) to y = -sqrt(x)
Find:
Determine the Volume bounded.
Solution:
- First we will find the projected area of the solid on the x = 0 plane.
A(x) = 0.5*(diagonal)^2
- Since the diagonal run from y = sqrt(x) to y = -sqrt(x). We have,
A(x) = 0.5*(sqrt(x) + sqrt(x) )^2
A(x) = 0.5*(4x) = 2x
- Using the Area we will integrate int the direction of x from 0 to 4 too get the volume of the solid:
V = integral(A(x)).dx
V = integral(2*x).dx
V = x^2
- Evaluate limits 0 < x < 4:
V= 16 - 0 = 16 unit^3
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What area of math is this,I may be able to help?
