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3 years ago

Find the surface area. Do not round your answer

Mathematics

1 answer:

Tom [10]3 years ago

8 0

Answer:

1,672.6

Step-by-step explanation:

In order to find the surface area of a figure, you must first find all the bases you must find the area for. You must find the areas of two pentagons and five rectangles.

We know that the area of the base is 386.3. That means that the area of one pentagon is 386.3. We can multiply that value by 2 in order to find the area of both pentagons. That gets you 772.6.

Now, we have to find the areas of the five rectangles. Luckily, they're all the same size, so we can use the expression a=bh or 12x15 to find the area of one rectangle. That gets you 180. Now, multiply that by 5 in order to find the area of all the rectangles. That gets you 900.

Now, add the two areas together (bolded for clarity) to get the surface area.

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2 years ago

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SOMEONE HELP?!! ASAP !! I’ll give brainliest!! Please and ty !!

dsp73

Answer:

$(\pi 4^2 (9))-(\pi 2^2 (9))$

and

$339.292 cm^3$

Step-by-step explanation:

This question is simple once you break it down.

Let's concentrate on the outer cylinder.

First, find the volume of the outer cylinder.

The volume of a cylinder is expressed as: $V = \pi r^2 h$

Gathering information from the diagram..

d = 8 cm

h = 9 cm

You can get the radius from the diameter by just dividing the diameter by 2.

$r = \frac{8}{2}=4$

r = 4

Plug everything into the volume equation.

$V = \pi 4^2 (9)$

Don't solve just yet!

Now, find the volume of the inner cylinder.

Do the exact same thing as you did for the outer cylinder.

d = 4 cm

h = 9 cm

$r = \frac{4}{2}=2$

r = 2

Plug everything into the volume equation..

$V = \pi 2^2 (9)$

The question asks for an expression to represent the volume of the large cylinder after the cylindrical hole is removed. This means you need to subtract the volume of the cylindrical hole from the volume for the large cylinder.

Again, the volume for the large cylinder is $V = \pi 4^2 (9)$ and the volume for the cylindrical hole is $V = \pi 2^2 (9)$

This means..

$(\pi 4^2 (9))-(\pi 2^2 (9))$

That is your answer to the first part of the question.

For the second part, just solve.

$(\pi 4^2 (9))-(\pi 2^2 (9))$

$(\pi*16*9)-(\pi*4*9)$