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ratelena [41]
4 years ago
8

Factor completely 81m^2-169

Mathematics
1 answer:
wel4 years ago
7 0
<span> 81m^2-169 = (8m + 13)(8m - 13)
</span>
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Dara has to solve 35 math problems for homework. She has completed 14 of them.what fraction of the problems does she have left t
RSB [31]

Answer: 3/5

Step-by-step explanation: thats your answer.

6 0
3 years ago
Carol has 88 cm of string and cuts it in the ratio 5 : 6. What is the total parts?​
Mrrafil [7]

Answer:

5+6= 11

88/11= 8

5x8= <u>40</u>

6x8= <u>48</u>

6 0
4 years ago
Read 2 more answers
Which scenario best matches the linear relationship expressed in the equation y=-14+1700
Elden [556K]

Question:

Which scenario best matches the linear relationship expressed in the equation y = –14x + 1,700?

1) Kent has $1,700 in his bank account and spends $14 each week.

2) Kent has $1,700 in his bank account and deposits $14 each week.

3) Kent had $1,700 in his bank account and deposited another $14.

4) Kent has $14 in his bank account and spent $1,700.

Answer:

Option  1) Kent has $1,700 in his bank account and spends $14 each week.

Step-by-step explanation:

Let x denote the number of weeks.

Option 1: Kent has $1,700 in his bank account and -14x as the negative sign denotes how much amount he spend each week.

Thus, Option 1 is the correct answer.

Option 2: Kent has $1,700 in his bank account and deposits $14 each week.

Writing it as equation, we have, y=14x+1700 which is not the expressed linear equation.

Hence,  Option 2 is not the correct answer.

Option 3: Kent had $1,700 in his bank account and deposited another $14.

Writing it as equation, we have, y=14+1700 which is not the expressed linear equation.

Hence,  Option 3 is not the correct answer.

Option 4: Kent has $14 in his bank account and spent $1,700.

Writing it as equation, we have, y=14-1700 which is not the expressed linear equation.

Hence,  Option 4 is not the correct answer.

Thus, the scenario that matches the linear equation relationship expressed in the equation is Option 1.

The answer is Kent has $1,700 in his bank account and spends $14 each week.

4 0
3 years ago
Find the domain of the function y = 3 tan(23x)
solmaris [256]

Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

4 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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