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Delicious77 [7]
3 years ago
13

What is the image of the point (-2, 4) under a 90° rotation about the origin?

Mathematics
1 answer:
Anon25 [30]3 years ago
5 0

Answer:

(- 4, - 2 )

Step-by-step explanation:

Under a rotation about the origin of 90°

a point (x, y ) → (- y, x ) , then

(- 2, 4 ) → (- 4, - 2 )

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Find the quotient using long division<br> 6x2 - 8x + 5/<br> X-2
joja [24]

Step-by-step explanation:

hope this helps have a great day!

3 0
3 years ago
Simone measures the width of one of cardboard strip as 1/2 yard. A second cardboard strip measures 5/6 yard in width. Estimate t
Zepler [3.9K]

Answer:

The correct option is C. About 1 on the side with the 1 on top and 4 on the bottom yard.

Step-by-step explanation:

Consider the provided information.

Simone measures the width of one of cardboard strip as 1/2 yard.

A second cardboard strip measures 5/6 yard in width.

We need to find the combined width.

The combined width of the cardboard strip:

\frac{1}{2} + \frac{5}{6}=\frac{8}{6}=1\frac{1}{3}

1\frac{1}{3} is closer to1\frac{1}{4} as compare to 1\frac{1}{2}

Hence, the correct option is C. About 1 on the side with the 1 on top and 4 on the bottom yard.

3 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
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Solve by substitution. Please show work<br> Y=-3x-14<br> -2x-y=9
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Answer:

I dont know math

Step-by-step explanation:

like bro i havent been paying attention to school that I havent learn stuff and I'm in 9th grade

8 0
2 years ago
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How do you find pi?
Crazy boy [7]
Pi is just 3.14--- math questions will most likely ask you to round to the nearest tenth or something of the sort -
4 0
3 years ago
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