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3 years ago

Find the surface area of each figure.

Mathematics

1 answer:

timurjin [86]3 years ago

6 0

Answer:

For a cylinder of radius R and height H, the surface area is given by:

A = 4*pi*R^2 + H*(2*pi*R)

Where pi = 3.14

We can just use that formula for each one of the given cylinders.

1) We can see that the diameter is 10 yd, and the radius is half of the diameter, then the radius is:

R = 10yd/2 = 5yd

And the height is 3yd, then H = 3yd

Replacing these in the area equation, we get:

A = 4*3.14*(5 yd)^2 + 3yd*(2*3.14*5yd) = 408.2 yd^2

2) Here we can see that the diameter is 24 cm, then the radius is:

R = 24cm/2 = 12cm

And the height is H = 10cm

Then the area is:

A = 4*3.14*(12 cm)^2 + 10cm*(2*3.14*12cm) = 2,562.24 cm^2

3) In this case we have a radius equal to 12 cm, and a height equal to 7 cm, then the area is:

A = 4*3.14*(12 cm)^2 + 7cm*(2*3.14*12cm) = 2,336.16cm^2

4) Here is hard to see the measures, I think that here we have:

diameter = 8m

Then R = 8m/2 = 4m

And the height is also 8m, H = 8m

Then the area is:

A = 4*3.14*(4 m)^2 + 8 m*(2*3.14*4m) =401.92 m^2

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7/9+6/8+6/3= I really need the answer

Neporo4naja [7]

Answer:

$\frac{1}{30}$

Step-by-step explanation:

Take it in steps. First, find 7/9+6. Then we'll find 8+6/3, and, finally, we'll divide the two answers.

7/9+6 = 7/15

8+6/3 = 14/3

$\frac{\frac{7}{15}}{\frac{14}{3}}$ or $\frac{7/15}{14/3}$

Then take that in chunks: 7/14 and 15/3.

7/14 = 1/2

15/3 = 5/1

Use those to rewrite it as $\frac{1/5}{2/3}$ .

1/5 = .2

2/3 ≈ .6667 so we'll keep writing it as 2/3

$\frac{\frac{.2}{2}}{3}$

.2/2 = .1, so:

$\frac{.1}{3}$ = $\frac{1}{30}$

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3 years ago

Read 2 more answers

The dimensions of a square are altered so that one dimension is increased by 7 feet and the other is decreased by 2 feet. The ar

Rufina [12.5K]

I got 64 square feet.

My work is shown in the image.

For possible questions:

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2: solved via factoring. Not sure if you've learned it or not.

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Step-by-step explanation:

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With the aid of an illustrative example, discuss the relationship between the area of a region and the definite integral. *

melisa1 [442]

Answer:

The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the a and b above and below) to represent an antiderivative. Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same. A definite integral is a number. An indefinite integral is a family of functions. Later in this chapter we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we’re working with a definite integral or an indefinite integral.

Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz, who is often considered to be the codiscoverer of calculus, along with Isaac Newton. The integration symbol ∫ is an elongated S, suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of the interval, \left[a,b\right]. The numbers a and b are x-values and are called the limits of integration; specifically, a is the lower limit and b is the upper limit. To clarify, we are using the word limit in two different ways in the context of the definite integral. First, we talk about the limit of a sum as n\to \infty . Second, the boundaries of the region are called the limits of integration.

We call the function f(x) the integrand, and the dx indicates that f(x) is a function with respect to x, called the variable of integration. Note that, like the index in a sum, the variable of integration is a dummy variable, and has no impact on the computation of the integral.

his leads to the following theorem, which we state without proof.

Step-by-step explanation:

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3 years ago