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vovikov84 [41]
3 years ago
11

If x2 = 81, what are the TWO answers?

Mathematics
2 answers:
Sphinxa [80]3 years ago
8 0
Your answer is 9, -9
nika2105 [10]3 years ago
4 0

Answer:

9 and -9

Step-by-step explanation:

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D. The equation has a variable multiplied by itself. A squared variable.

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A quadratic equation will always start with a term being squared. For example, 8x^{2} + 5x + 7 = 44. Another example would be 4x^{2} + 2x - 10 = 12. These are just random numbers, but the equation will always look the same! Hope this helps!

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The granola summer buys used to cost $6.00 per pound but it has been marked up %15 A. How much did it cost summer to buy 2.6 pou
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3 years ago
On a recent trip to the movies , you see a advertisement for 2 adult tickets and 6 child tickets for $60. Write and equation usi
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2A + 6C = $60

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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
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