WEBVTT
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welcome to lecture number three in this n
p t e l online certification course on bioreactors
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at in our course in the first two lectures
we got introduced to what bioreactors where
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what kind of products they make what are the
common types of bioreactors that i used what
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are the modes of operation predominantly batch
continuous and semi batch or fed batch and
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then we started looking at some details the
first aspect that we looked at was how to
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have a clean slate because we said we kill
all organisms that are there we create a clean
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slate and then we introduce the organisms
of our interest into the bioreactor which
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produces the product of interest to us
to achieve the clean slate there are three
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methods possible three main methods possible
temperature based or chemicals based chemical
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weapons and so on or radiation based gamma
u b any things like that then we saw thermal
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sterilization the temperature based aspect
in some detail so that we had enough information
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to we able to design a priory this sterilization
process thats a whole point about analyzing
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and so on it also gives us a lot of insights
but the practical aspect is to be able to
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design appropriate processes let me open
this particular presentation this is what
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we saw the thermal sterilization the response
of viable cells to high temperature we have
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on the y axis the percent viable cell concentration
on a long scale versus time on the x axis
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and the relationship is linear when this on
a long scale also we said that the decrease
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the slope of the decease increases has
the temperature increases this is the kind
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of response that we were trying to analyze
we said that we could analyze this situation
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if we considered the rate of decrease in concentration
to be directly proportional to the concentration
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of viable cells a sort of a first order relationship
using that and the fact that this operation
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is happening in a batch by a reactor we wrote
a balance for it and arrived at get there
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this was the governing equation d d t of x
v is minus k d into x v and if you solve this
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we would get lon of x v not by x v equal k
d t that gives you the time there the time
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needed for the viable cell concentration to
go down to x v starting from x v naught is
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two point three naught three by k d log to
the base ten x v naught by x v as we can realize
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this is natural log log to the base e when
we convert that to log to the base ten you
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have a factor of two point three naught three
but otherwise this is essentially this equation
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transports the main thing here is that we
have the time that is necessary for the viable
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cell concentration to go down from x v naught
to x v under certain set of conditions which
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is characterized by this death constant k
d or this sterilization constant k d so this
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is what we arrived at and then we also looked
at something called a decimal reduction rate
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which is the time that is required for a ten
fold decrease in viable cell concentration
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at a given temperature
for ten fold reduction all we needed to do
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was replace x v naught by x v have to be ten
and then we got the decimal reduction time
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d as two point three naught three by k d at
this point a problem was assigned the practice
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problem one point one it red a bioreactor
needs to be sterilize before use the solution
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in the bioreactor consists of single cells
with similar thermal response characteristics
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at seventy degree c it takes five minutes
for the viable cell concentration to reduce
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to twenty percent of its original value a
determine the decimal reduction time and b
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how long would it take for the viable cell
concentration to reduce to point one percent
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of its original value under the same conditions
this was the question that was or the problem
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that was assigned when we pretty much finished
up the previous lecture lecture number two
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i hope you would have a chance to work this
out i will as promised earlier work
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out problems that are assigned in the beginning
of the next lecture we will start this lecture
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by solving this problem first
let me first talk a little bit about problem
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solving problem solving is a higher level
skill its a same way as somebody being able
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to either sing well or dance well or play
a game well or come what may you think of
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skills and you have a certain association
with it similarly problem solving is also
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a skill its acts is a higher level skill in
the cognitive domain some people would be
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naturally good at it as have some people who
a who have a natural ability in a music in
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dance and so on so forth if you have the natural
ability then this initial part or the details
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of the problem solving may not be very necessary
for you you are already had a good level
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i think you should continue with whatever
you are doing assign do the problems that
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are assigned and you should be fine
however in my experience i found that most
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people do not have problem solving as that
natural skill it is necessary for us to pick
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up that skill because it is necessary for
this profession and if you dont have a
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natural skill but you need to do something
then you need to pick it up to pick it up
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of course we practice we put an effort and
then we get to a certain level of be able
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to do problems i am sure with the consistent
good effort you would be able to solve problems
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we will leave it ourselves to what are called
closed ended problems in this course closed
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ended problems are problems in which everything
that is needed for the solution of the problem
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is either given in the problem statement or
is known through material in your notes and
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so on so forth this is distinct from open
ended problems that may not have all the
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information that is required state it in the
problem or known might need to extend a few
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things and then get information from other
sources literature and so on and then solve
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the problem
infact in my regular courses i do assign as
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long open ended problem solve a problem to
be solved over the entire length of the semester
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its a very good learning experience for the
students this format does not allow us do
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that effectively so let us at least look at
close ended problem solving for this course
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if you are interested in knowing or in
developing the problem solving skill in a
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systematic fashion you may want to look at
this book this is a the title of the book
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is strategies for creative problem solving
the authors are scott fogler and steven e
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leblanc this is a good book you might want
to take a look at that book if you want to
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know a little more about problem solving if
you really want to get into it and become
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an expert with sustained effort with that
let me go on to how we are going to go about
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closed ended problem solving the first question
to ask is what is needed what does the problem
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want us to find once that is clear write it
down then ask the question what is given or
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known through the problem through the material
that is relevant to the problem that was covered
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in the lectures and so on once both those
are clear then we will ask the question how
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do we connect the first one with the second
one how do we connect the needs of this
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particular problem with givens are knowns
in this problem and while doing that we will
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also ask are there any basic principles that
we can rely on to be able to solve this problem
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in fact while presenting the solution i
am going to use only the first three what
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is needed what is given or known how do we
connect the needs with givens and knowns and
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take this as a part of the third aspect thats
how we are going to do it now let us go about
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solving the problem that is we already read
this we need to find the decimal reduction
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time in part a and in part b how long would
it take for the viable cell concentration
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to reduce to point one percent of its original
value some information is given
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the solution in the bioreactor consist of
single cells with similar thermal response
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characteristics therefore you can assume
linear dependence of the log of the viable
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cell concentration with time as we saw in
class so in the lecture and also its says
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that at seventy degree c it takes five minutes
for the viable cell concentration to reduce
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to twenty twenty percent of its original value
ok having seen this let us go about the solution
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first question as we said was what is needed
let us take one part of it after an other
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part a the decimal reduction time so that
is clear so let us ask the second question
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what is known or given we know that it is
a single cell suspension we discuss this earlier
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too and also we know that at seventy degree
c it takes five minutes or three hundred seconds
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for the viable cell concentration to go from
hundred percent to twenty percent or in other
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words a five fold reduction takes a takes
the three hundred seconds we see that i have
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converted this five minutes into three hundred
seconds ok
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it is usually good to work in a consistent
system of units whatever that might be and
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for the purposes of this um course we will
mostly use the s i units and thats the reason
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why i have converted five minutes to three
hundred seconds it is good to do this a priory
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and then go about the actual solution or
before hand and go about actual solution then
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the third main question how to connect what
is needed to what is given if you think about
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it i am sure quite a lot of you would have
solved this you think about it we have derived
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the decimal reduction time which we also review
a few minutes ago as two point three naught
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three by k d therefore we need k d if we have
k d we can find out d which is the decimal
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reduction time which is what is you required
in part a
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how do you find k d to do that let us go back
and look at the basis for the derivation of
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decimal reduction time we i think we
went through it during the review process
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so let me just say this gives the time for
the reduction and viable cell concentration
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from x v naught to x v we derived the time
to be two point three naught three by k d
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log to the base ten x v naught by x v we have
the information on the time taken for x v
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naught by x v to be five a nother was the
five fold reduction form hundred percent to
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twenty percent so that if you substitute here
in in otherwords this is becomes five right
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so we we need five minutes or three hundred
seconds for a five fold reduction therefore
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three hundred equals two point three naught
three by k d log five to the base ten so k
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d is the only variable here and k d if we
tansforces or essential multiply both sides
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of the equation you can do the same things
to both side of both sides of the equation
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that you know if we multiply this side by
k d and this side also by k d we get
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and divide this side by three hundred and
this side also by three hundred we get k d
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equals two point three naught three by three
hundred log five to the base ten which turns
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out to be five point three seven into ten
power minus three seconded percent
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i would like you to verify this particular
answer if you have this answer fine but please
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verify i have deliberately not verified my
answers during the solution process alone
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i have done that for the exam questions and
so on but during the in a demonstration
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of the solutions of the problems the practice
problems that are assigned i have deliberately
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not verified my answers if you find an error
please point it out that will be a nice exercise
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for you now that we have k d we can find the
decimal reduction time as two point three
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naught three by k d which turns out to be
two point three naught three by five point
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three seven written power minus three or four
twenty eight point nine seconds because this
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was in second inverse we get this in seconds
if we divide this by sixty we get seven point
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two minutes ok therefore it takes seven point
two minutes for ten fold reduction in the
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viable cell concentration under these conditions
now let us consider part b again will ask
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the same questions what is needed the time
taken for the viable cell concentration to
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reduce to point one percent of its original
value under the same conditions what is known
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or given single cell suspension the same as
earlier at seventy degree c it takes three
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hundred seconds for the viable cell concentration
to go from hundred percent to twenty percent
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and also we found in part a the value of k
d and the value of the decimal reduction time
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this is what is known how do we connect what
is known to what is given the time for reduction
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and viable cell concentration from x v naught
x v we have already derived and seen earlier
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also that equals two point three naught three
by k d log to the base ten of x v naught by
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x v when the viable cell concentration reduces
to point one percent of its original value
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this is the condition that is of interest
to us in problem b we know that x v naught
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by x v sorry x v by x v naught is point one
if x v by x v naught is point one percent
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this is this is percent if we converted into
fraction this will turn out to be point one
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by hundred you know percentage is always out
of hundred so fraction is when we need
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the fraction we need to divided by hundred
this is point naught naught one or ten part
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minus three
so x v naught by x v if we flip it around
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it will be one by ten power minus three or
ten power three and thus the time required
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for the viable cell concentration to reduce
to point one percent of its original value
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would be this substitution of the various
know aspects now into this equation two point
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three naught three by five point three seven
to ten power minus three log to the base ten
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of ten power three that answer to be one two
eight six point six seconds or twenty one
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point four minutes this is a one way to solve
the problem i have shown you all these steps
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hopefully you understood this steps and
if you be acted if its not natural to you
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and if you be acted i am sure you will be
able to solve closed ended problems i think
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we will finish of lecture three with this
these lectures would be of variable times
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they will all add up to around ten hours the
each lecture would be a variable time we
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will meet in the next lecture lecture four
see you there