Exercise 21 Let x represents a length in cm where x > 1.5.
AB = 2x - 3, BC = 4x – 6,
AD = 2x + 3 and BE = 4x +1
• ABC is right triangle of area Sı.
• CBFG is a rectangle of area S2.
ADEB is a right trapezoid of area S3.
1) Express S1, S2 and S3 in terms of x.
2) Let S = Si + S2 - S3.
a) Show that S = (2x - 3)(5x - 9).
b) For what values of x, the areas of AFGC and ADEB are equal?
c) Expand and reduce S.
d) What is the value of x, so that the area of AFGC exceed the area of ADEB by
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4.08%

step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.

d no solution

step-by-step explanation: