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3 years ago

What is the area of the circle below if it is 18 m

Mathematics

1 answer:

DiKsa [7]3 years ago

6 0

Answer:

A≈1017.88 if r=18m or A≈254.47 if r=9

Step-by-step explanation:

A=Pi*r^2, plugging in 18 m or 9 m since lack of info but i assume you are giving 18m total length of circle, gives us answer.

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What is the area of the circle of radius 9

erma4kov [3.2K]

The equation for the area of a circle is: $\pi r^2$ . Insert 9 as your radius (r) into the equation and solve:
\pi 9^2 = 254.469

4 0

2 years ago

Read 2 more answers

Consider the following sets of sample data: A: $29,400, $30,900, $21,000, $33,200, $21,300, $24,600, $29,500, $22,500, $35,200,

Lana71 [14]

Answer:

CV for A = 21.8%

CV for B = 15.5%

Step-by-step explanation:

The formula for coefficient of variation is:

CV = Standard Deviation / Mean

So,

For A:

Mean = Sum/No. of items

= 391300/14

=$27950

and

SD = $6085.31

CV for A = 6085.31/27950 * 100

=21.77%

Rounding off to one decimal

CV for A = 21.8%

For B:

Mean = Sum/No. of items

= 43.58/11

=3.96

and

SD = 0.615

CV for B = 0.615/3.96 * 100

=15.53%

=15.5% ..

6 0

2 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$