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Lunna [17]
3 years ago
14

The diameter of a circle is 4 kilometers. What is the circle's circumference?

Mathematics
2 answers:
iragen [17]3 years ago
7 0

Answer:

12.56km

Step-by-step explanation:

C=2pi r

2r=d

C=pi d=4pi=4x3.14=12.56km

scZoUnD [109]3 years ago
6 0

Answer:

12.57 km

Step-by-step explanation:

Radius = 4 / 2 = 2 km

Circumference = 2πr = 2π(2) = 4π = 12.57 km

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For A brainlist ** explain
IrinaVladis [17]

Answer:

So interval notation is with ( and [ where ( is exclusive and [ is inclusive.

Like (1,2) is between 1 and 2 exclusive. [1,2] is between 1 and 2 inclusive. (1,2] is between 1 and 2, 1 exclusive 2 inclusive.

at the point (6,0) you see that the graph goes from above 0 to below 0 (from positive to negative)

The values are positive when x is less than 6 and negative when x is greater than 6.

so the positive interval is

(-infinity, 6)

and with inifinity you always use exclusive

It's that because everything from all the way to the left (-infinity) to 6, is above the x-axis, which means it's positive

using this logic can you do the negative interval?

5 0
3 years ago
V/-5+8=9 what does v equal
Sergeeva-Olga [200]
Subtract 8 from both sides and then multiply both sides by -5
V = -5
6 0
3 years ago
What is the answer to this?
Svetach [21]

Answer: F: I only

Step-by-step explanation:

7 0
3 years ago
Write the missing number <br> 1/5 out of 15
Sedbober [7]

Answer:

3

Step-by-step explanation:

1/5 × 15

= 15/5

= 3 (would love if you could mark me the brainliest :))

5 0
2 years ago
AB=8 and CD =40 find EB ,BC and AC
SIZIF [17.4K]

Answer:

EB=20, BC=8, AC=16

Step-by-step explanation:

The symbols indicate that:

AB=BC  and AE=ED

EB and CD are parallels

AB=BC=8

AC= AB+BC

AC= 8+8

AC=16

To find EB we can use the Cosine Law

For the upper triangle x=∡EAB:

EB^2 = AB^2 + AE^2 -2*AB*AE*Cosx

AB*AE*Cosx= -(EB^2-AB^2 - AE^2)/2 (Part I)

For de big triangle:

DC^2= AC^2+AD^2 -2AC*AD*Cosx

Also:

AC=2*AB

AD=2*AE

DC^2= (2*AB)^2 + (2*AE)^2 -2(2*AB)(2*AE)*Cosx

DC^2= 4*AB^2 +4*AE^2- 8*AB*AE*Cosx

AB*AE*Cosx =-(DC^2-4*AB^2 -4*AE^2)/8 (Part II)

Part I= Part II

-(EB^2-AB^2 - AE^2)/2= -(DC^2-4*AB^2 -4*AE^2)/8

Extracting EB:

EB^2=DC^2/4

EB=DC/2

EB=40/2

EB=20

8 0
3 years ago
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