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MrRissso [65]
3 years ago
11

I want to cover the curved surface of a cylinder with orange paper. My cylinder is 14” tall. The diameter is 3.5”. I have listed

the sizes of paper I have available. Which sheet of paper will cover the cylinder without overlapping?
A. 5”x14”

B. 1K4”x20”

C. 14”x30”

D. 11”x14”

Now I want to cover the lid and bottom of the cylinder with blank paper. I need to know the total area of the bases. What is the area of the bases rounded to the nearest inch ?

A 19”

B. 4”

C. 38”

D. 10”
Mathematics
1 answer:
Dominik [7]3 years ago
7 0
Circumference of the cylinder = 3.14 x 3.5 = 10.99 = 11 inches
 so D. 11 x 14 is correct for the first part

area of the top  =  PI x r^2 = 3.14 x 1.75^2 = 9.62 inches

top and bottom = 9.62 x 2 = 19.23 = A. 19 inches




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Hoochie [10]

9514 1404 393

Answer:

  the listed sequences have no common difference(s)

Step-by-step explanation:

The differences in the first sequence are ...

  8, 10, -1, 19

These values are not equal, so there is no common difference.

__

The differences in the second sequence are ...

  -3, 3, -3, 3

These values are not equal, so there is no common difference.

_____

The difference is the difference between a term value and the one before.

  1 -(-7) = 8

  3 -6 = -3

7 0
3 years ago
Evaluate d – f if d = 7 and f = –15
KonstantinChe [14]

Answer:

The answer is 22!!!!!!!

3 0
3 years ago
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Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1
belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

            F(x) = \int\limits^a_b {f(x)} \, dx

By using integration formula

\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

 \int\limits^6_^-6} (x^{5}  + 8 x^{4}  + 2 x^{2}  + 5 x + 15) )dx

<em>On integration , we get</em>

=   (\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}

F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)

= (\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))

After simplification and cancellation we get

 =  \frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6

on calculation , we get

= \frac{124,416}{5} + \frac{864}{3} + 180

On L.C.M  15

= \frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}

= 25 351.2 units²

<u><em>Conclusion</em></u>:-

<em>The area of the region between the graph of the given function and the x-axis = 25,351 units²</em>

6 0
3 years ago
75 + 13 + ?? = 140 i dont know what it is.
mixer [17]

Answer:

52

Step-by-step explanation:

140 - 13 = 127

127 - 75 = 52

Check your work:

75 + 13 + 52

= 140

8 0
3 years ago
Read 2 more answers
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,
Andreyy89

Answer:

1) the planning value for the population standard deviation is 10,000

2)

a) Margin of error E = 500, n = 1536.64 ≈ 1537

b) Margin of error E = 200, n = 9604

c) Margin of error E = 100, n = 38416

3)

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

Step-by-step explanation:

Given the data in the question;

1) Planning Value for the population standard deviation will be;

⇒ ( 50,000 - 10,000 ) / 4

= 40,000 / 4

σ = 10,000

Hence, the planning value for the population standard deviation is 10,000

2) how large a sample should be taken if the desired margin of error is;

we know that, n = [ (z_{\alpha /2 × σ ) / E ]²

given that confidence level = 95%, so z_{\alpha /2  = 1.96

Now,

a) Margin of error E = 500

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 500 ]²

n = [ 19600 / 500 ]²

n = 1536.64 ≈ 1537

b) Margin of error E = 200

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 200 ]²

n = [ 19600 / 200 ]²

n = 9604

c)  Margin of error E = 100

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 100 ]²

n = [ 19600 / 100 ]²

n = 38416

3) Would you recommend trying to obtain the $100 margin of error?

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

7 0
3 years ago
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