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Kruka [31]
3 years ago
7

A circular plate has circumferrence 30.8 inches. What is the area of this plate ? Use 3.14 for π.

Mathematics
1 answer:
erastova [34]3 years ago
8 0

Answer:

75.39 in^2

Step-by-step explanation:

circumference=2πr

30.8=2(3.14)r

30.8=6.28r

4.9=r

Now that we know the radius, we can plug it in for the area equation

area=πr^2

area=(3.14)(4.9)^2

area=(3.14)(24.01)

area=75.39 in^2

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he section of paper shown in the pattern below is 1/4 of a circle. It will be wrapped around a cone. The wrapper will then be pa
uranmaximum [27]

The volume of the cone based on the figure illustrated wil be 196cm³.

<h3>Host illustrate the information?</h3>

The information is incomplete and the complete question wast found online. An overview will be given.

Let's assume that the height is 4cm and the radius of the cone is 7cm. The volume of the cone will be:

= 1/3πr²h

= 1/3 × 3.14 × 7² × 4

= 196cm³

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6 0
2 years ago
Can someone help me find the diameter? <br><br> also what is a diameter even?
lukranit [14]

Answer:

24

Step-by-step explanation:

area of circle = πr²

πr² = 144π

Cancel out π

r² = 144

r = √144

r = 12

diameter = 2 x radius

diameter = 2 x 12

diameter = 24m

* diameter is the "width of a circle", the distance from the widest points of the circle

3 0
2 years ago
Number 13 least to greatest please put an order
IgorLugansk [536]
5.67, 5.68, 5.87, 6.57
6 0
3 years ago
Read 2 more answers
Mrs. Purdue needs to pick three random students in her class to be representatives on the student council. Explain how Mrs. Purd
Xelga [282]

Answer:

See below

Step-by-step explanation:

She could have each student in the class write his or her name on a slip of paper and put it in a bowl. Then, she could shake the bowl and pull out any slip of paper with a student's name on it. After 3 times, the random sample would be over, and Mrs. Purdue would have her 3 representatives.

5 0
2 years ago
Try to fill in the missing numbers
GrogVix [38]

9514 1404 393

Answer:

  [[274][895][136]]

Step-by-step explanation:

Starting with the middle row, we need a product of two single-digit numbers that is between 53-1 = 52 and 53-9 = 44. Possible products are 5×9=45 and 6×8=48. This means the number in the middle position in the left column must be 8 or 5.

The middle number in the left column cannot be 5, because we must be able to get -5 by subtracting that number from a sum that is at least 3 = 1+2. So, the middle number in the left column is 8, the other two numbers in that column are 1 and 2, and the other two numbers in the middle row are 5 and 9.

There is no product of single-digit numbers that is 30-1 = 29, so the upper left number must be 2, and the bottom left number must be 1. The other two numbers on the top row must be 4 and 7, so that row's equation is 2+4×7=30.

The only remaining digits are 3 and 6. In order to have -3 on the bottom row, the equation there must be 1×3-6 = -3. Then the middle digit must be divisible by 3, so must be 9.

Our solution is ...

row 1: 2 + 7 × 4 = 30

row 2: 8 + 9 × 5 = 53

row 3: 1 × 3 - 6 = -3

And that makes the column equations be ...

col 1: 2 - 8 + 1 = -5

col 2: 7 + 9 / 3 = 10

col 3: 4 × 5 - 6 = 14

6 0
3 years ago
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