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3 years ago

Find the circumference of a circle whose area is /. A. B. C. D.

Mathematics

2 answers:

notsponge [240]3 years ago

8 0

I just took the test and the correct answer that i got was B

Archy [21]3 years ago

3 0

You missed to write the value of the area of circle so just for explanation purpose, I will assume that given area is $25 \pi$

Now question says to find the circumference. To find circumference we first need to find the radius (r)

Hence plug the value of area into formula of the area of circle and find the radius.

$Area=\pi r^2$

$25 \pi=\pi r^2$

divide both side by pi

$25 =r^2$

take square root of both sides

5=r

now we can plug value or radius r=5 into formula of circumference of the circle.

$circumference = 2 \pi r$

$circumference = 2 \pi *5$

$circumference = 10 \pi$

Hence final answer is $circumference = 10 \pi$ .

You can repeat same procedure with given value of area to find correct answer.

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Find the circumference of a circle if the radius is 15 ft. Use 3.14 for pi.

Korvikt [17]

Circumference = 2 x radius x Pi

Circumference = 2 x 15 x 3.14

Circumference = 30 x 3.14

Circumference = 94.2 feet

8 0

3 years ago

Read 2 more answers

Audrey and her family are discussing how to pay for her college

Elan Coil [88]

Answer:

$137.50

Step-by-step explanation:

11,000 x .15 (100-.85)=$1.650 amount she will have to pay annually.

$1,650/12months = $137.50 is minimum to save monthly.

4 0

2 years ago

A square based prism and a cylinder both have

Anna11 [10]

Answer:

Parameter of base = 70.88 inch (Approx.)

Step-by-step explanation:

Given:

Volume of cylinder = 3768 cubic inch

Height of Prism = 12 inch

Find;

Parameter of base

Computation:

Volume of cylinder = Volume of Prism

3768 = (L)(B)(H)

3768 = (L)(B)(12)

(L)(B) = 314

L = B

So.

L = 17.72 inch

B = 17.72 inch

Parameter of base = 2[l+b]

Parameter of base = 2[17.72 + 17.72]

Parameter of base = 70.88 inch (Approx.)

8 0

2 years ago

PLEASE HELP ITS A TEST<br> Solve for x : - 1/2 (x + 3) - 10 = -6.5

Ksju [112]

Answer hey:

Step-by-step explanation:

4 0

2 years ago

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

DENIUS [597]

Answer:

The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.

The probability that a randomly chosen plate contains the sub-string HI is 0.002548 approximately.

Step-by-step explanation:

Consider the provided information.

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

Part (A)

Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.

Here repetition is allow. The number of alphabets are 26 and the number of distinct digits are 10.

The numbers of ways in which three letters followed by three digits can be chosen is: $26\times 26\times 26 \times 10 \times 10 \times10$

$26^3\times 10^3=17576000$

The numbers of ways in which two letters followed by four digits can be chosen is: $26\times 26\times 10 \times 10 \times 10 \times10$

$26^2\times 10^4=6760000$

Hence, the total number of ways are 17576000 + 6760000 = 24336000

Randomly chosen plate contains the number 2222 that means the first two letter can be any alphabets but the rest of the digit should be 2222.

Thus, the total number of ways that a randomly chosen plate contains the number 2222 number are: $26^2=676$

The probability that a randomly chosen plate contains the number 2222 is:

$\frac{676}{24336000} \approx 0.000028$

Part (B)

The number of ways in which chosen plate contains the sub-string HI:

If three letters followed by three digits plate contains the sub-string HI, then the number of possible ways are:

$26\times 1\times10^3+1\times 26\times10^3$

If two letters followed by four digits plate contains the sub-string HI, then the number of possible ways are:

$1\times 10^4$

Thus, the total number of ways that a randomly chosen plate contains the sub-string HI are:

$1\times 10^4+26\times 1\times10^3+1\times 26\times10^3$

$62000$

From part (A) we know that the total number of ways to chose a number plate is 24336000.

The probability that a randomly chosen plate contains the sub-string HI is:

$\frac{62000}{24336000} \approx 0.002548$

7 0

3 years ago