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Misha Larkins [42]
2 years ago
9

10 ft

Mathematics
1 answer:
Lelu [443]2 years ago
3 0

Answer:

P = 40.56 ft

Step-by-step explanation:

The figure is composed of a rectangle and half circle

The perimeter of the figure = the length of the borders round the figure

= W + 2L + ½(perimeter of full circle)

W = 8 ft

L = 10 ft

½(perimeter of circle) = ½(πd) = ½(3.14*8)

= 12.56 ft

Plug in the values

Perimeter of the figure = 8 + 2(10) + 12.56

= 8 + 20 + 12.56

= 40.56 ft

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Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
Mnenie [13.5K]
One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2

the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
D=\sqrt{(8+2)^2+(1)^2}
D=\sqrt{10^2+1}
D=\sqrt{100+1}
D=\sqrt{101}




r=3\sqrt{2}≈4.2
D=\sqrt{101}≈10.04

do r<D

(8,4) is outside the circle

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3 years ago
What is the GCF of 90 and 190?
IrinaVladis [17]
Let's first find out their factors.

90's factors are : 

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190's factors are :

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The common factors of 90 and 190 are :

1,2,5 and 10. 

The greatest one among these is, 10. So the greatest factor of 90 and 190 is 10.




7 0
3 years ago
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Slav-nsk [51]
1 day use 4 bags
? day use 1/2 bag

Cross multiply
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In 1/8 of the day 1/2 of bag of milk will be used.
3 0
3 years ago
Factor completely.<br> 5x4 + 35x3– 150x2
Igoryamba

5x^4+35x^3-150x^2

=  

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Answer:

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