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IRISSAK [1]
2 years ago
15

If y = 12 x – 12, determine the value of y when x = –16.

Mathematics
1 answer:
GrogVix [38]2 years ago
4 0

Answer:

Step-by-step explanation: y=180

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What is the value of the function when x = −1?<br><br> Is my answer right?
kap26 [50]
The answer would be negative 2
7 0
3 years ago
A rumor spreads through a small town. Let y ( t ) be the fraction of the population that has heard the rumor at time t and assum
Ivan

Answer:

Differential equation

\frac{dy}{dt} =ky(1-y)

Solution

y=\frac{1}{1+4e^{-0.327t}}

Value of constant k=0.327 days^(-1)

The rumor reaches 80% at 8.48 days.

Step-by-step explanation:

We know

y(t): proportion of people that heard the rumor

y'(t)=ky(1-y), rate of spread of the rumor

Differential equation

\frac{dy}{dt} =ky(1-y)

Solving the differential equation

\frac{dy}{y(1-y)}=k\cdot dt \\\\\int \frac{dx}{y(1-y)} =k \int dt \\\\-ln(1-\frac{1}{y} )+C_0=kt\\\\1-\frac{1}{y} =Ce^{-kt}\\\\\frac{1}{y} =1-Ce^{-kt}\\\\y=\frac{1}{1-Ce^{-kt}}

Initial conditions:

y(0)=0.2\\y(3)=0.4\\\\y(0)=0.2=\frac{1}{1-Ce^0}\\\\1-C=1/0.2\\\\C=1-1/0.2= -4\\\\\\y(3)=0.4=\frac{1}{1+4e^{-3k}} \\\\1+4e^{-3k}=1/0.4\\\\e^{-3k}=(2.5-1)/4=0.375\\\\k=ln(0.375)/(-3)=0.327\\\\\\y=\frac{1}{1+4e^{-0.327t}}

Value of constant k=0.327 days^(-1)

At what time the rumor reaches 80%?

y(t)=0.8=\frac{1}{1+4e^{-0.327t}} \\\\1+4e^{-0.327t}=1/0.8=1.25\\\\e^{-0.327t}=(1.25-1)/4=0.0625\\\\t=ln(0.0625)/(-0.327)=8.48

The rumor reaches 80% at 8.48 days.

8 0
3 years ago
Solve the following absolute value equation. ​
cestrela7 [59]

Answer:

x= 8

x= -8

Step-by-step explanation:

|x+1| /3 = 3

|x+1| = 9

|x| + |1| = 9

|x| = 8

x= 8

x= -8

4 0
2 years ago
Identify the fraction as a proper improper and mixed however if the fraction is equivalent to one indicate so 7/3
tekilochka [14]
Improper because the top is supposed to be lower than the bottom 
6 0
3 years ago
Idk wby it sets it up so weird in word form so imma just put this here
Anit [1.1K]

Answer:

141/2, 279/4

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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