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Andreas93 [3]
3 years ago
5

Please help me with this question please

Mathematics
1 answer:
dmitriy555 [2]3 years ago
3 0

Answer:

The solution is (2, 1/2).

Step-by-step explanation:

The question is asking you to solve a system of equations using elimination.  The elimination method means that when you add the two equations together, one of the variables will be eliminated so that you are able to solve for the other variable.  In this case, if we multiply the first equation (6/x - 2/y = -1) by a factor of 2 we get: (12/x - 4/y = -2).  We can then add the two equations together to eliminate the 'y' variable and solve for x using inverse operations.  When we add (12/x - 4/y = -2) + (2/x + 4/y = 9) = (14/x = 7) or x = 2.  In order to solve for 'y', we must now plug in the value of x = 2 into one of the equations and solve for 'y'.  So, 2/2 + 4/y = 9 becomes 1 + 4/y = 9.  Using inverse operations, be subtract 1 from both sides and then multiply to get y = 1/2.

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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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