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Makovka662 [10]
2 years ago
9

Marble Slab Creamery is creating a cone-shaped paper wrapper for its new green Baby Yoda waffle cone. The wrapper should have a

radius of 5 cm and a slant height of 10 cm. One location wants to offer the cones a weekend earlier, so they plan to make the cone-shaped wrappers for first 25 waffle cones sold that Saturday. About how many square cm of wrapper paper will they need? (Round up to the nearest cm.)
Mathematics
1 answer:
mote1985 [20]2 years ago
3 0

Answer:

3927 cm² wrapping paper will be required.

Step-by-step explanation:

Lateral surface area of a cone shaped wrapper = πrl

where, r = radius of the waffle cone

and l = slant height of the cone

It has been given in the question, radius 'r' = 5cm

and lateral height 'l' = 10 cm

Lateral surface area of the cone = π(5)(10)

                                                      = 50π

                                                      = 157.08 cm²

If they plan to make 25 waffles then wrapper paper required

= 25×157.08

= 3927 cm²

Therefore, 3927 cm² wrapping paper will be required.

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<h3>How to estimate the common difference of an arithmetic progression?</h3>

let the nth term be named x, and the value of the term y, then there exists a function y = ax + b this formula exists also utilized for straight lines.

We just require a and b. we already got two data points. we can just plug the known x/y pairs into the formula

The 9th and the 12th term of an arithmetic progression exist at 50 and 65 respectively.

9th term = 50

a + 8d = 50 ...............(1)

12th term = 65

a + 11d = 65 ...............(2)

subtract them, (2) - (1), we get

3d = 15

d = 5

If a + 8d = 50 then substitute the value of d = 5, we get

a + 8 * 5 = 50

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a = 50 - 40

a = 10.

Therefore, the first term is 10 and the common difference is 2.

To learn more about common differences refer to:

brainly.com/question/1486233

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4 0
2 years ago
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Answer:

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Step-by-step explanation:

The complete question in the attached figure

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In this problem

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x

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