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4 years ago

8

How to find the solution of |a-4|=3a-6

1 answer:

Anika [276]4 years ago

6 0

For
|a|=b, solve for
a=b and a=-b

so

|a-4|=3a-6
solve
a-4=3a-6 and a-4=-(3a-6)

first one
a-4=3a-6
minus a from both sides
-4=2a-6
add 6 to both sides
2=2a
divide by 2
1=a

other

a-4=-(3a-6)
a-4=-3a+6
add 3a both sides
4a-4=6
add 4 to both sides
4a=10
divide by 4
a=10/4
a=5/2

a=5/2 and 1

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4 years ago

Read 2 more answers

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sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

$A(t)=100-80e^{-0.02t}$

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Step-by-step explanation:

The rate of change of the amount of salt can be written as

$\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})$

Then we can rearrange and integrate

$\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50} \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}$

Then we have the model of A(t) like

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt reaches 50 lb is

$A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5$

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kogti [31]

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