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Semmy [17]
3 years ago
11

I-Ready

Mathematics
1 answer:
Sindrei [870]3 years ago
3 0

Answer:

2

Step-by-step explanation:

(f+g)(x)

f(x)= x-3, g(x)= 2x+8

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Your answer is 0.605
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3 years ago
Q.3 Write the factors of 28 in circle A and the factors of 32 in circle B. Write
marissa [1.9K]

Answer: The largest common factor of 28 and 32 is 4.

Step-by-step explanation:

The factors of 28 are 1, 2, 4, 7, 14, 28

The factors of 32 are 1, 2, 4, 8, 16, 32

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2 years ago
How do you write 0 .25 in a fraction?
tatuchka [14]
0.25 in a fraction would be 1/4
4 0
3 years ago
Read 2 more answers
Cedric and doug each had an equal amount of money. After cedric spent $35 and doug spent $28, the ratio of cedrics money to doug
uysha [10]

Answer:

Each of them had $49 at first.

Step-by-step explanation:

Given:

Cedric and Doug each had an equal amount of money.

Let the number of money they had at first be x

Cedric spent = $35

Doug spent = $28

Cedric Money left = x-35

Doug Money left = x-28

the ratio of Cedric money to Doug money was 2:3

Hence the expression can be made as;

\frac{x-35}{x-28}=\frac{2}{3}

Now by cross multiplication method we get;

3(x-35)=2(x-28)

Now we will use distributive property to elaborate the same;

3x-105=2x-56

Combining common terms we get;

3x-2x=105-56

Using Subtraction property we get;

x= 49

Hence $49 they each had at first.

8 0
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

8 0
3 years ago
Read 2 more answers
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