X(a+b)-y(a+b)
*factor out (a+b)*
=(a+b)(x-y)
Answer:
Yes :)
Step-by-step explanation:
The solution would be like
this for this specific problem:
<span>V = ∫ dV </span><span>
<span>= ∫0→2 ∫
0→π/2 ∫ 0→ 2·r·sin(φ) [ r ] dzdφdr </span>
<span>= ∫0→2 ∫
0→π/2 [ r·2·r·sin(φ) - r·0 ] dφdr </span>
<span>= ∫0→2 ∫
0→π/2 [ 2·r²·sin(φ) ] dφdr </span>
<span>= ∫0→2 [
-2·r²·cos(π/2) + 2·r²·cos(0) ] dr </span>
<span>= ∫0→2 [
2·r² ] dr </span>
<span>=
(2/3)·2³ - (2/3)·0³ </span>
<span>= 16/3 </span></span>
So the volume of the
given solid is 16/3. I am hoping that these answers have satisfied
your query and it will be able to help you in your endeavors, and if you would
like, feel free to ask another question.
Here is a reference to the Inscribed Quadrilateral Conjecture it says that opposite angles of an inscribed quadrilateral are supplemental.
Explanation:
The conjecture, #angleA and angleC# allows us to write the following equation:
#angleA + angleC=180^@#
Substitute the equivalent expressions in terms of x:
#x+2+ x-2 = 180^@#
#2x = 180^@#
#x = 90^@#
From this we can compute the measures of all of the angles.
#angleA=92^@#
#angleB=100^@#
#angleC=88^@#
<span>#angleD= 80^@#</span>
since the polygon has 8 chambers the angle of one chamber = 360/8 = 45°
so the angle WOP = 45°
now drawing a imaginary line which seperates the triangleWOP into half
so the angle of imaginary line is 45/2 = 22.5
sin(theta) = opp/hyp
since WP (opp) is unknown. take as x
sin(22.5) = x/44
0.383×4 = x
x = 1.532 (app) = 1.5 cm
perimeter = 1.5 × 8
= 12 cm
area of triangle WOP = 2×1/2 ×bh = bh
= 4×1.5
= 6 cm²
so the area of regular polygon = 8× area of a triangle
= 48 cm²
