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sleet_krkn [62]
3 years ago
9

Find the surface area, to the nearest square centimeter, of a cylinder with diameter 6 cm and height 8 cm. A.48cm^2 B.452cm^2 C.

207cm^2 D. 72cm^2
Mathematics
1 answer:
Black_prince [1.1K]3 years ago
7 0

The right answer is Option C.

Step-by-step explanation:

Given,

Diameter = 6 cm

Radius = \frac{Diameter}{2}=\frac{6}{2}=3\ cm

Height = 8 cm

π = \frac{22}{7}

Surface area = 2πrh+2πr²

Surface area = 2*\frac{22}{7}*3*8+2*\frac{22}{7}*(3)^2\\

Surface area = \frac{1056}{7}+\frac{396}{7}=\frac{1056+396}{7}

Surface area = \frac{1452}{7}

Surface area = 207.4 cm²

Rounding off to nearest whole number

Surface area = 207 cm²

The surface area of cylinder is 207cm².

The right answer is Option C.

Keywords: surface area, cylinder

Learn more about cylinder at:

  • brainly.com/question/12849439
  • brainly.com/question/12860445

#LearnwithBrainly

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They are quadrilaterals that have opposite sides equal in length.

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jonny [76]
4 - 2(x + 3) = 5
4 - 2x - 6 = 5
-2x - 2 = 5
-2x = 5 + 2
-2x = 7
x = -7/2 or = -3 1/2
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3 years ago
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number between 1 and 10, inclusive, is randomly chosen. Events A and B are defined as follows. A: {The number is even} B: {The n
Rufina [12.5K]

Answer:

A U B = {1,2,3,4,5,6,8}

Step-by-step explanation:

Let's define the sets.

Let ¢ = universal set ( Contains all the elements in the set)

¢ = {1, 2, 3, 4, 5, 6, 7, 8, 9 , 10}

A= {2, 4, 6, 8}

B= {1,2,3,4,5,6}

Therefore: A U B (A union B):

A U B = {1, 2, 3, 4, 5, 6, 8}

6 0
3 years ago
The sum of 5 times a larger number and twice a smaller is 5. The difference of 4 times the larger and the smaller number is 4. F
Gemiola [76]

Answer:

1 and 0

Step-by-step explanation:

Let's call the smaller number x and the larger number y.

This means 5y+2x=5.

Also, 4y-x=4.

This is a system of equations. There are two main ways two solve this.

Method 1: Substitution

5y+2x=5

4y-x=4

Solve for one of the variables, then substitute it back into one of the equations to solve for the other variable.

I will take 4y-x=4 and solve for x.

4y=4+x

x=4y-4

Then substitute for x in the other equation and solve for y.

5y+2x=5

5y+2(4y-4)=5

5y+8y-8=5

13y=13

y=1

Now we can go back to our equation for x and substitute for y.

x=4y-4

x=4(1)-4=0

y, the larger number, is 1 and

x, the smaller number, is 0

Method 2: Elimination

5y+2x=5

4y-x=4

Change the equations so that the coefficient of one of the variables in an equation is opposite the one in the other equation. Then add the equations and solve for each variable.

If I multiply both sides of the second equation, 4y-x=4, by 2, the coefficient of x will become -2. In the first equation, the coefficient is 2, so they will become opposites.

4y-x=4

becomes

8y-2x=8

Now we have 5y+2x=5 and 8y-2x=8.

Add the sides of both equations to cancel the x values.

5y+2x+8y-2x=5+8

13y=13

y=1

like last time, we can substitute for y in any of the original equations to get x.

5y+2x=5

5(1)+2x=5

5+2x=5

2x=0

x=0

y, the larger number, is 1 and

x, the smaller number, is 0

6 0
3 years ago
Can someone help me please?
marishachu [46]
(6x + 30)+(2x+6)= 180
8x +36= 180
8x= 144
X= 18
So 2x+6 = 42
And 6x+30 = 138
8 0
3 years ago
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