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seraphim [82]
3 years ago
8

Solve the Absolute Value Equation. |9x+1| = 46

Mathematics
1 answer:
Sedaia [141]3 years ago
5 0
X = 5 (you can set it up as a regular equation)
9x + 1= 46
9x = 45 (subtracted 1 from both sides)
x = 5 (divided 45 by 9)
hope this helps :)
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Cos^4x csc^2x = csc^2 x sin^2 -2
lawyer [7]

Answer:

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Step-by-step explanation:

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2 years ago
Solve the triangle.
saw5 [17]

Answer:

Step-by-step explanation:

because <u>c</u> is shorter than<u> a</u> we know C is a smaller angle than A

so the first choice is out.

use the law of cosines to find b

b^{2} = a^{2} +c^{2} -2*a*c*cos(B)

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2 years ago
A man earned x pesos in 10 days and spent y pesos during each of those days. Write an expression to determine how many pesos he
Vsevolod [243]

Answer:  \dfrac{x}{10}-y

Step-by-step explanation:

Given : A man earned x pesos in 10 days and spent y pesos during each of those days.

i.e. Total earning in 10 days = x

Earning per day =\dfrac{x}{10}    [By unitary method]    (1)

Money spent per day =y      (2)

We know that

\text{Savings = Money earned - Money spent}

i.e. Subtract (2) from (1), we get

\text{Savings = }\dfrac{x}{10}-y

Hence, the expression to determine how many pesos he saved per day will be:-

\dfrac{x}{10}-y

3 0
3 years ago
What is the average rate of change of the function over the interval x = 0 to x = 6? f(x)=2x−1 3x+5 Enter your answer, as a frac
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-11

Step-by-step explanation:

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8 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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