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amm1812
3 years ago
8

Someone could help me?​

Mathematics
1 answer:
Helen [10]3 years ago
4 0

Answer:

B= 3.14 * 4^4 = 50.24cm^2\\h = 16cm\\V=B*h=50.24*16=803.84cm^3

Step-by-step explanation:

The area of the base is the area of a circle with a radius equal to 4 cm. It means that the area can be calculated as:

B = 3.14 * r^2\\B= 3.14 * 4^4 = 50.24cm^2

The height of the cylinder is shown in the picture, it is equal to 16 cm.

Finally, the volume of the cylinder can be calculated as:

V = B*h=50.24*16 = 803.84cm^3

Where B is the base and h is the height of the cylinder.

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5 adult and 3 student tickets were purchased.

8 0
3 years ago
Please help me please please ASAP ASAP please ASAP
fgiga [73]

Answer:

<h2>47 degrees</h2>

Step-by-step explanation:

So m<ABD = 47 degrees

It appears to be congruent to m<BAC, so we can infer that it is 47 degrees

Hope that helps!

3 0
3 years ago
A machine that produces ball bearings has initially been set so that the mean diameter of the bearings it produces is 0.500 inch
natali 33 [55]

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10

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5 0
3 years ago
Wha is the area of this
Korvikt [17]

Octagon, stop sign.

Eight isoscles triangles.   It looks like we're told the side is 9.9 and the height to the side (also called the apothem) is 12.

So each isosceles triangle has area  (1/2)(9.9)(12) and we have eight of them,

area = 8(1/2)(9.9)(12) = 475.2

Answer: 475.2

Usually we wouldn't be told 9.9 -- this is the baby version.  We know each of those isoscles triangles has unique angle 360/8=45 degrees, so the apothem and half the side of the octagon are a right triangle with acute angle 22.5 degrees.

The area of the right triangle with long leg 12, short leg x,

tan 22.5 = x/12 or

x = 12 tan 22.5

Twice that is what we're told is 9.9; let's check:

2x = 24 tan 22.5 = 9.941125496954282

The area of the little right triangle is

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6 0
3 years ago
Point J is on line segment IK. Given JK = 2x, IJ = 5x, and IK = x + 6, determine the numerical length of JK.
Andre45 [30]

Answer:

JK = 2

Step-by-step explanation:

Given:

JK = 2x

IJ = 5x

IK = x + 6

Required

Solve for JK.

Since J is on IK, we have:

IK = IJ + JK

x + 6 = 5x + 2x

x + 6 = 7x

Collect Like Terms

7x - x = 6

6x = 6

Solve for x

x = 6/6

x = 1

Substitute 1 for x in JK = 2x

JK = 2 * 1

JK = 2

8 0
4 years ago
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