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Paladinen [302]
3 years ago
13

I need to find the area of the shaded region

Mathematics
1 answer:
spin [16.1K]3 years ago
4 0
Well first find the area of the semi-circle. 

If the area of a cricle is equal to pi*radius^2 , then you can just find that and divide by 2.

So, A = pi*2^2. = 4pi

We know that the radius is 2 because the length of the side of the rectangle is 4, meaning that the diameter of the semi-circle is 4, and so the radius is 2 as it is half of the diameter. 

We can easily calculate the area of the rectangle, which is
 Length * width = 6*4 = 24.

Next we divide 4pi by 2 in order to get the area of the semi-circle, giving us an area of 2pi

We can just subtract 2pi from 24 and get the area of the shaded region.

Area of the shaded region (answer): 17.7
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Need help real fast.
ira [324]

Answer:

option 4

Step-by-step explanation:

using the equation stated above

1st term = -6+(1-1=0)*(6)=-6

4th term= -6+(4-1=3)*6=12

10th term= -6+(10-1=9)*6=48

7 0
3 years ago
What is the image of (-9,3)(−9,3) after a reflection over the y-axis?
nirvana33 [79]

Answer:

(9, 3 )

Step-by-step explanation:

under a reflection in the y- axis

a point (x, y ) → (- x, y ) , then

(- 9, 3 ) → (9, 3 )

3 0
2 years ago
Who won the Superbowl in 1996
Sophie [7]

The Dallas Cowboys.

7 0
3 years ago
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The perimeter of a rectangular field is 60ft and its width is 20ft find the area of this field
Gennadij [26K]

Perimeter of rectangle = 2( length +width)

given: Perimeter=60 ft and width=20 ft

plugging these values in the formula,

60=2(20+ length)

dividing both sides by 2

30=20+length

length =30-20=10

length is 10 feet

width is 20 feet

Area of rectangle is given by:

Area = l*w

Area = 10 *20 = 200 square feet

Answer : Area of rectangle is 200 square feet.

4 0
3 years ago
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Algebra 1 CR-8
antiseptic1488 [7]

\mathrm{Domain\:of\:}\:x^2+2x-15\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}x^2+2x-15:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-16\:\\ \:\mathrm{Interval\:Notation:}&\:[-16,\:\infty \:)\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:x^2+2x-15:\quad \mathrm{X\:Intercepts}:\:\left(3,\:0\right),\:\left(-5,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:-15\right)

\mathrm{Vertex\:of}\:x^2+2x-15:\quad \mathrm{Minimum}\space\left(-1,\:-16\right)

4 0
3 years ago
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