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solong [7]
3 years ago
6

Which equation represents the line that passes through the points (6,9) and (3.5)?

Mathematics
1 answer:
sukhopar [10]3 years ago
5 0

the answer is y=2x-3

You can look at my work if you really need to

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two bicycle riders start a race at the same time. the function f(x) and g(x) give the distance in miles that each rider has trav
iragen [17]

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https://jbarrueta.weebly.com/uploads/5/3/2/9/53297595/lesson2.2.2.pdf

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here's a link

4 0
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Find out what make 30 with only the number 1,3,5,7,9,11,13,15
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Step-by-step explanation:

7 + 3 + 5 + 15 = 30

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3 years ago
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Type the correct answer in each box.
svlad2 [7]

Answer:

heads 4 H-4

tails 2 T-2 and 6 is T-6

Step-by-step explanation:

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3 years 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

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