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Anni [7]
3 years ago
5

Please help thanks very much!

Mathematics
1 answer:
Drupady [299]3 years ago
4 0
1st one is AZ for the allitude
2nd one is BX for the median i think
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Given equations A and B as 2/5x+y=12 and 5/2y-x=6, respectively, which expression will eliminate variable x?
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The answer is C.) A+2/5B.
Adding equation A to 2/5B will result in canceling the x terms, i.e. 2/5x + 2/5(-x) = 0.
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4 years ago
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A triangle with the coordinates X(3, 2), Y(5, -1), and Z(-2, 3) is reflected across both the x- and y-axes. The coordinates afte
nata0808 [166]

Answer:

The coordinates of the reflected triangle are X'(x,y) = (-3,-2), Y' (x,y) = (-5,1) and Z'(x,y) = (2,-3).

Step-by-step explanation:

Let be a point (a,b)\in\mathbb{R}^{2}, reflections across x- and y-axes are represented by the following operations:

x-Axis reflection:

d = d'

b - 0 = 0 - b'

b' = -b

y-Axis reflection:

d = d'

a - 0 = 0 - a'

a' = -a

If we know that X (x,y) = (3,2), Y(x,y) = (5,-1) and Z (x,y) = (-2, 3), the coordinates after both reflections are, respectively:

X'(x,y) = (-3,-2)

Y' (x,y) = (-5,1)

Z'(x,y) = (2,-3)

The coordinates of the reflected triangle are X'(x,y) = (-3,-2), Y' (x,y) = (-5,1) and Z'(x,y) = (2,-3).

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4 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

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Let n be one side of the square wood block. Then:
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Factor -4x^2+16x-24 completely
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