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zaharov [31]
3 years ago
11

A sphere has a radius of 4x + 1. Which polynomial in standard form best describes the total surface area of the sphere? Use the

formula S = 4pir2 for the surface area of a sphere.
Mathematics
1 answer:
vampirchik [111]3 years ago
7 0
Answer:
area = pi * (64x² + 32x + 4)
Using pi = 3.14:
area = 200.96 x² + 100.48 x + 12.56

Explanation:
We are given that:
area of sphere = 4 * pi * r²
radius = 4x + 1

Substitute with the radius in the given formula to get the area as follows:
area = 4 * pi * (4x+1)²
area = 4 * pi * (16x² + 8x + 1)
area = pi * (64x² + 32x + 4)

Using pi = 3.14:
area = 200.96 x² + 100.48 x + 12.56

Hope this helps :)

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Mrrafil [7]

Answer:

r=\sqrt{\frac{3V}{(\pi h)}}

Step-by-step explanation:

we know that

The volume of a right circular cone is equal to

V=\frac{1}{3}\pi r^{2}h

where

r is the radius of the base of the cone

h is the height

Solve for r-----> That means, isolate the variable r

so

step 1

Multiply by 3 both sides

3V=\pi r^{2}h

step 2

Divide by (\pi h) both sides

\frac{3V}{(\pi h)}=r^{2}

step 3

take square root boot sides

r=\sqrt{\frac{3V}{(\pi h)}}

7 0
3 years ago
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Answer:

1/11

Step-by-step explanation:

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What is the surface area ?
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I think the answer is 816cm
8 0
3 years ago
2X + 5y = -13<br> 3x - 4y = -8
almond37 [142]
Answer:

You will need to use simultaneous equations.

Method:

First we need to make one set of variable the same, in this case we will:

-Multiply the top equation by 3 to get 6x
-Multiply the bottom equation by 2 to get 6x

The result will be as follows:

1) 2x+5y=-13 x3
1) 6x+15y=-39

2) 3x-4y=-8 x2
2) 6x-8y=-16

Now we will make both equations equal to 6x:

1) 6x=-15y-39
2) 6x=8y-16

Now because they both equal 6x, they therefore equal each other:

-15y-39=8y-16
-23=23y
y=-1

Now that we have y, we can substitute it back into one of the equations to get x:

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3 0
4 years ago
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Digiron [165]

Answer:

B

Step-by-step explanation:

The y-intercept is 0,2 so that rules out C and D.

We then see that A cannot be true so it must be B.

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