Mathematics, 26.05.2021 14:00 dmccray3357

Find the area of the shaded segment VB and length of VB

Answers: 2

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Answer from: Maryhejsbsksb

Area of length Vb = 25 square units

Area of segment Vb = 26 square units

Step-by-step explanation:

Area of length Vb = theta/360 * 2πr

Area of length Vb = 120/360 * 2*3.14 * 12

Area of length Vb = 1/3 * 75.36

Area of length Vb = 25 square units

Ares of length Vb is 25.1 sq. units

Area of shaded portion = Area of sector VUB - Area of triangle VUB

Area of sector VUB = theta/360 * πr²

Area of sector VUB = 120/360 * 3.14*144

Area of sector VUB = 1/3 * 452.16

Area of sector VUB =150.72

Area of triangle VUB = r²sintheta

Area of sector VUB = = 12²sin 120

Area of sector VUB = = 144(0.8660)

Area of sector VUB = 124.71

Area of the shaded segment = 150.72 - 124.71

Area of the shaded segment = 26.01

Area of the shaded segment = 26 sq. units

Area of length Vb = 25 square units

Area of segment Vb = 26 square units

Step-by-step explanation:

Area of length Vb = theta/360 * 2πr

Area of length Vb = 120/360 * 2*3.14 * 12

Area of length Vb = 1/3 * 75.36

Area of length Vb = 25 square units

Ares of length Vb is 25.1 sq. units

Area of shaded portion = Area of sector VUB - Area of triangle VUB

Area of sector VUB = theta/360 * πr²

Area of sector VUB = 120/360 * 3.14*144

Area of sector VUB = 1/3 * 452.16

Area of sector VUB =150.72

Area of triangle VUB = r²sintheta

Area of sector VUB = = 12²sin 120

Area of sector VUB = = 144(0.8660)

Area of sector VUB = 124.71

Area of the shaded segment = 150.72 - 124.71

Area of the shaded segment = 26.01

Area of the shaded segment = 26 sq. units

Answer from: Quest

the answer on e2020 is

4/5

3/5

4/3

step-by-step explanation:

Answer from: Quest

the dimensions are: $(2b+-+1)$

step-by-step explanation:

we want to find the dimensions of a rectangular box with volume of

$v(b)=50b^3+75b^2-2b-3$

we factor to obtain;

$v(b)=(2b+3)(25b^2-1)$

we rewrite the rightmost factor to obtain;

$v(b)=(2b+)^2-1^2)$

recall that:

$a^2-b^2=(a+b)(a-b)$

we apply difference of two squares formula to obtain;

$v(b)=(2b+3)(5b-1)(5b+1$

hence the dimensions are

$(2b+-+1)$

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Find the area of the shaded segment VB and length of VB
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