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DochEvi [55]
2 years ago
5

What is the diameter of a circle with the circumference of 18 feet

Mathematics
2 answers:
Advocard [28]2 years ago
5 0

Answer: d ≈ 5.73 ft

Step-by-step explanation: Using the formulas

                                             C=2πr

                                             d=2r

                                             Solving for d

                                             d= C/π = 18/π ≈ 5.72958ft

Olegator [25]2 years ago
4 0

Answer:

5.732484076 =d

Step-by-step explanation:

The circumference is given by

C = pi*d

18 = 3.14 *d

Divide each side by 3.14

18/3.14 = 3.14d/3.14

5.732484076 =d

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2 years ago
In the figure above, are &lt;3 and &lt;7 alternate interior angles,
icang [17]

Answer:

Corresponding angles

Step-by-step explanation:

<3 and <7 lie on the same side of the transversal line. Each angle also is located on the outside of each parallel lines cut across by the transversal.

Therefore, <3 and <7 are corresponding angles.

7 0
3 years ago
(-31/2)-(-41/4)=-7/2-(17/4)=
Andrew [12]
Just forget about the first half, and just do the second half. It’s way easier. Answer is -7.75
6 0
2 years ago
A survey of 600 randomly selected high school students determined that 290 play organized sports. what is the probability that a
blsea [12.9K]

Answer:

29/60

Step-by-step explanation:

probability of playing is 290/600 which when simplified is29/60

5 0
1 year ago
Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma
pickupchik [31]

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]

M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....

M[N]=[10^(N+3)].............................................3

..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....

3.AFTER N=20 SECONDS , THE ANSWER IS ....

M[20]=10^(20+3)=10^23

8 0
3 years ago
Read 2 more answers
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