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kompoz [17]
3 years ago
13

The ac is a leg of the right triangle True False

Mathematics
1 answer:
Vinil7 [7]3 years ago
3 0

Answer:

False

Step-by-step:

The hypotenuse of the triangle is the side opposite to the right angle. The legs are the other two sides of the triangle. Since AC is on the opposite side of the right angle, it is the hypotenuse and therefore not a leg. Hope this helps!

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Rewrite as a sum<img src="https://tex.z-dn.net/?f=%282x%5E2-11%29%28x%5E2%2B2%29" id="TexFormula1" title="(2x^2-11)(x^2+2)" alt=
VashaNatasha [74]

Answer:

Step-by-step explanation:

(2x^2-11)(x^2+2)

2x^4-7x^2-22

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3 years ago
Is this complementary, linear pair, vertical, or adjacent?
Lorico [155]
I believe it is Complimentary
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Olivia wrapped a string around a circle and found it was 30 mm long. Approximately, what is the radius of this circle
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Answer:7.5

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3 years ago
Evaluate the numericial expression (32 - 20) ÷ 4
SashulF [63]

Answer:

3

Step-by-step explanation:

(32-20)÷4

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3 0
3 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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