The area of a circular sector of central angle α (in radians) in a circle of radius r is given by
... A = (1/2)r²×(α - sin(α))
Your area is expected to be computed as the sum of the areas of a sector with angle π/3 in a circle of radius 8 and a sector with angle π/2 in a circle of radius 6.
... A = (1/2)8²×(π/3 - sin(π/3)) + (1/2)6²×(π/2 - sin(π/2))
... A ≈ 16.07
Radii are in inches so the units of area will be in². The appropriate choice is
... 16.10 in²
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It should be noted that the geometry described is impossible. Chord CD of circle A will have length 6√2 ≈ 8.4853 inches. Chord CD of circle B will have length 8 inches. They cannot both be the same chord.
Answer:
the answer is 26
Step-by-step explanation:
you add 7 to -7 and 19. the positive and negative 7 cancels each other out. them you add 19 and 7 which is 26. that means x is 26.
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Answer:
A. -5x = -23
Step-by-step explanation:
First you have to simplify the equation - 4(x - 5) + 8x = 9x - 3
-4x + 20 + 8x = 9x -3
9x - 4x = 20 + 3
5x = 23
=> equivalent to answer A.
Observe the given data distribution table carefully.
The 5th class interval is given as,
The upper limit (UL) and lower limit (LL) of this interval are,
Thus, the upper-class limit of this 5th class is 17.4.
