The expression that represents the volume, in cubic units, of the shaded region of the composite figure is: A. One-half(14)(10)(8) – π(2.52)(8).
<h3>Expression</h3>
Given:
Base side length=14 and 10 units
Height=8 units
Diameter=5 units
Hence:
Expression= 1-( half 14)(10)(8) - π(2.52)(8)
Expression=1-(7) (10)(8) - π(2.52)(8)
Expression=1- (560)- 63.33
Expression=-559-63.33
Expression=-622.33
Therefore the expression that represents the volume, in cubic units, of the shaded region of the composite figure is: A. One-half(14)(10)(8) – π(2.52)(8).
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Answer:
16) 62.8 m²
17) 19.4 m²
18) 103.5 m²
Step-by-step explanation:
The formula to find area of a circle is:
. To find radius, find the half of the diameter.
To solve number 16, first find the area of the unshaded circle using the formula:
. This is also 3.14*16. Multiply to get 50.24 m². Now find the area of the larger circle using the formula:
. This is also 3.14*36. Multiply to get 113.04 m². Now subtract 113.04 - 50.24 to get the shaded area or 62.8 m².
To solve number 17, first find the area of the square using the formula: side x side. In this case multiply: 5.25 x 5.25 to get 27.5625. Round to the nearest tenth to get 27.6 m². Now find the area of the circle using the formula:
. This is also 3.14*2.625. Multiply to get 8.2425. Round to the nearest tenth to get 8.2. Subtract 27.6 - 8.2 to get 19.4 m².
To solve number 18, find the area of one unshaded circle using the formula:
. This is also 3.14*3.0625. Multiply to get 9.61625. Round to the nearest tenth to get 9.6 m². Add 9.6 + 9.6 to find the area of both unshaded circles. You get 19.2 m². Now find the area of the shaded circle using the formula:
. This is also 3.14*39.0625. Multiply to get 122.65625. Round to the nearest tenth: 122.7. Subtract 122.7 - 19.2 to get 103.5 m².
Hope it helps and is correct!
Answer:
r = 
Step-by-step explanation:
We simply are rearranging C= 2πr in terms of <em>r</em>. We just divide 2π on both sides.
Answer:
0.44444444444
Step-by-step explanation:
Answer:
3(x-6)
Step-by-step explanation: