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Alex
3 years ago
9

Solve the equation. 8n+15=25

Mathematics
2 answers:
Maslowich3 years ago
8 0

One way to find out is: 25 minus 15 divided by 8... this will give the answer 1.25

Aneli [31]3 years ago
3 0
N=1.25 or you can use n= 1 1/4
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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
Read 2 more answers
Please answer this asap!
777dan777 [17]
This is an exponential equation that can be represented by the following:

f(x) = a(b)^x

In this case...

25143 = a(0.66)^3

25143 is the population after 3 hours.

3 is the amount of time in hours.

0.66 represents the percent of the population remaining after each hour (66% as there is a 34% decline each hour).

We must solve for a, which is the initial population.

First, simplify (0.66)^3 to 0.2874.

25143 = 0.2874a

Now divide both sides by 0.2874 to isolate a.

a = 87455

There were initially 87,455 people within the city. I wouldn't want to be in that place!
7 0
2 years ago
Students were packed shoulder to shoulder in a gym for a pep rally. The students were in a rectangular shaped area of 7,500 ft2.
hram777 [196]

Answer: 1527

Step-by-step explanation:

Total Area = 7500 ft^2

Area covered by one student = Area of one circle

                                                   = π*r^2

r = radius of circle = 2.5/2 = 1.25 ft

Area covered by one student = π*1.25^2 = 4.91 ft^2

Number of students who can fit into total area = Total Area/Area covered by one student

Number of students who can fit into total area = 7500/4.91 = 1527.49

Hence the answer is 1527 students

7 0
2 years ago
Which algebraic expression is equivalent to 7(3x + 12) + 12x - 16
Sphinxa [80]

Answer:

fhcchjgcdhj cell

uyyjfgwfyrrfascfavczffbczxvvsfcf

5 0
2 years ago
Amistad deposited $2,163.27 in a savings account that earns 3.9% simple interest. What will Amistad's account balance be in nine
avanturin [10]

Given that Amistad deposited $2,163.27 in a savings account that earns 3.9% simple interest.

That means we need to use Simple interest formula to find the Amistad's account balance in nine months.


Simple interest formula is

A=P(1+RT)

Where P= principal amount = $2163.27

R= rate of interest = 3.9%= 0.039

T= time in years = 9 months = 9/12 years = 0.75


Now plug these values into above formula:

A=2163.27(1+0.039* 0.75)

A=2163.27(1+0.02925)

A=2163.27(1.02925)

A=2226.5456475


Hence Amistad's account balance in nine months will be approx $2226.55

8 0
3 years ago
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