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____ [38]
4 years ago
8

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

Mathematics
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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5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l
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

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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

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

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kogti [31]

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Step-by-step explanation:

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d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

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