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Marina CMI [18]
2 years ago
8

Find the area of each circle to the nearest tenth. Use 3.14

Mathematics
2 answers:
Oksi-84 [34.3K]2 years ago
8 0

Answer:

50.24

Step-by-step explanation:

Area formula: Pi x R^2 or Pi x R x R

3.14x4x4=50.24

<u><em>Can I get Brainlest pls</em></u>

svetlana [45]2 years ago
7 0

Area of Circle: \pi r^{2}

  • radius 'r' = 4ft        (radius is half of diameter)
  • \pi =3.14

Area equals = 3.14 (4)^2 = 50.24 square feet = 50.2 square feet

Hope that helps!

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Answer: The answer is A.


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3 years ago
Read 2 more answers
Are -4 ( x - 6) and -4 • x - 4 • 6 equivalent
tekilochka [14]

Answer:

 no

Step-by-step explanation:

Lets solve the first half.

-4 ( x - 6) = -4x + 24

Second one:

-4 times x - 4 times  6 = -4x -24

Now compare.....Not the same right??

So your answer is:  no

Hope this helps :)

7 0
3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

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