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kolezko [41]
4 years ago
8

The circumference of a circle is 6.28. What is the area of the circle.

Mathematics
2 answers:
Rom4ik [11]4 years ago
6 0
Hello there! Thank you for asking your question here at Brainly. I will be assisting you today with how to handle this problem, and will teach you how to handle it on your own in the future.

First, let's evaluate the question.
"The circumference of a circle is 6.28. What is the area of a circle?"

Now, let's remember the different formulas for area and circumference.
The circumference is "2•3.14•r", while the area is "3.14•r•r".

We have our circumference, 6.28.
However, we are looking for the area. Since we have the circumference, we need to narrow down to the radius (so we can solve for the area).
Let's set this up as an equation;
C = 2 • 3.14 • r
Plug in the value for our circumference.
6.28 = 2 • 3.14 • r
Multiply 2 by 3.14 and r to simplify the right side of the equation.
2 • 3.14 • r = 6.28 • r = 6.28r

We're now left with:
6.28 = 6.28r
Divide both sides by 6.28 to solve for r.
6.28 / 6.28 = 1
6.28r / 6.28 = r

We are now left with the radius:
R = 1.

Now, we can solve for the area.
Remember our formula for the area.
A = r • r • 3.14.
Plug in 1 for r.
A = 1 • 1 • 3.14
A = 3.14.

Your area is 3.14 units^2.

I hope this helps, and has prepared you for your future problems in relation to this topic!
Ann [662]4 years ago
3 0
Circumfrence= πd
6.28=3.14*d
6.28/3.14=d
D=2
Area= πr*r
=3.14*1*1
=3.14
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Add the 3 angles together to equal 180:


X + (5x-20) + 4x = 180

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3 years ago
The table shows the average annual cost of tuition at 4-year institutions from 2003 to 2010.
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Answer: 1) The best estimate for the average cost of tuition at a 4-year institution starting in 2020 =$ 31524.31

2) The slope of regression line b=937.97 represents the rate of change of  average annual cost of tuition at 4-year institutions (y) from 2003 to 2010(x).  Here,average annual cost of tuition at 4-year institutions is dependent on school years .

Step-by-step explanation:

1) For the given situation we need to find linear regression equation Y=a+bX for the given situation.

Let x be the number of years starting with 2003 to 2010.

i.e. n=8

and y be the average annual cost of tuition at 4-year institutions from 2003 to 2010.  

With reference to table we get

\sum x=36\\\sum y=150894\\\sum x^2=204\\\sum xy=718418

By using above values find a and b for Y=a+bX, where b is the slope of regression line.

a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}=\frac{150894(204)-(36)718418}{8(204)-(36)^2}=\frac{30782376-25863048}{1632-1296}=\frac{4919328}{336}\\\\=14640.85

and

b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}=\frac{8(718418)-(36)150894}{8(204)-(36)^2}=\frac{5747344-5432184}{1632-1296}=\frac{315160}{336}\\\\=937.97


∴ To find average cost of tuition at a 4-year institution starting in 2020.(as n becomes 18 for year 2020 if starts from 2003 ⇒X=18)

So, Y= 14640.85 + 937.97×18 = 31524.31

∴The best estimate for the average cost of tuition at a 4-year institution starting in 2020 = $31524.31


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