Our aim is to calculate the Radius so that to use the formula related to the area of a segment of a circle, that is: Aire of segment = Ф.R²/2
Let o be the center of the circle, AB the chord of 8 in subtending the arc f120°
Let OH be the altitude of triangle AOB. We know that a chord perpendicular to a radius bisects the chord in the middle. Hence AH = HB = 4 in
The triangle HOB is a semi equilateral triangle, so OH (facing 30°)=1/2 R. Now Pythagoras: OB² = OH² + 4²==> R² = (R/2)² + 16
R² = R²/4 +16. Solve for R ==> R =8/√3
OB² = OH² +
<u>Graph 1</u>
coordinates : (0,-5) (-6,-2)
m = (y2-y1)/(x2-x1)
m = (-2-(-5))/(-6-0)
m = 3/-6 = -1/2
Substitute the values,
y = mx + c
-2 = (-1/2)(-6) + c
-2 = 3 + c
c = -5
Thus, equation is y = -1/2x -5
<u>Graph </u><u>2</u>
coordinates : ( 4,-2) (-5,3)
m = (y2-y1)/(x2-x1)
m = (3-(-2))/(-5-4)
m = 5/-9
Substitute the values,
y = mx + c
-2 = (-5/9)(4) + c
-2 = -20/9 + c
c = 2/9
Thus, the equation is y = -5/9 + (2/9)
Hope it helps :)
30inc=76.2cm
76.2*pi=76.2pi
1cm=0.0328084 foot
1 mile=<span>5280 ft
</span>5280/0.0328084=160934.4
1 mile=160934.4 cm
160934.4 /76.2pi=672.27
672 revolutions i think
The cross section refers to the area where we cut in. So id say its B.Trapezoid, bc its cutting into the cube which sits on top of the rectangle
Yes your answer is correct
Just plug in A - 2B + C
5 - 2(4) + 2 = -1
2 - 3(2) + 8 = 4
3 - 2(-1) + 0 = 5
0 - 2(6) + 5 = -5
so
[-1 4]
[5 -7]
