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Delvig [45]
3 years ago
9

What topic is this and how do you go about it?

Mathematics
2 answers:
shepuryov [24]3 years ago
6 0
Finding an angle of a triangle
Luden [163]3 years ago
6 0
Im gessing it is angles
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What is the area of the rectangle? (area = length x width)
dimaraw [331]

Answer:

C. 10 Units

Step-by-step explanation:

Multiply two and five together to get ten. You can get these numbers by counting the amount of space between each dot.

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The radius of a circle is 3 miles. What is the circle's diameter?
Gemiola [76]

Answer: 6 miles

Step-by-step explanation:

The diameter is 2x the radius

2 x 3 = 6

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Select one of the factors of x2y3 − 11x2y + 6y2 − 66
EleoNora [17]
We try to factor by maybe grouping
experiment

(x²y³-11x²y)+(6y²-66)
factor
x²y(y²-11)+6(y²-11)
undistribute (y²-11) from each
(x²y+6)(y²-11)
we can force a factor out of the 2nd group in the form of a difference of 2 perfect squares
(x²y+6)(y-√11)(y+√11)
either of those 3 are factors
5 0
4 years ago
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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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