In an experiment a petri dish with a colony of bacteria is exposed to cold temperatures and then warmed again. Find a quadratic model for the data in the table. Show work

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In an experiment a petri dish with a colony of bacteria is exposed to cold temperatures and then warmed again. Find a quadratic model for the data in the table. Show work

Mathematics
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Getting the table, one moment
Time (hours) 0 1 2 3 4 5 6 Population (1000s) 5.1 3.03 1.72 1.17 1.38 2.35 4.08
how are you supposed to do this (just asking because you have to "show work")

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Other answers:

I have no clue lol I typed everything the question asked and the table
The general formula for quadratic eqn is y = ax^2 + bx + c take any 3 points from the table (0, 5.1) (1, 3.03) (5, 2.35) plug these points into the above equation and you will get 3 different equations use substitution or any method that you know to solve these 3 eqns with 3 unknowns
then use a calculator http://www.wolframalpha.com/input/?i=quadratic+%280%2C5.1%29%2C%281%2C3.03%29%2C%282%2C1.72%29%2C%283%2C1.17%29%2C%284%2C1.38%29%2C%285%2C2.35%29%2C%286%2C4.08%29
a point is written in the form (x,y) so plug 0 for x and 5.1 for y to get 1st eqn. follow same method to get other 2 eqns
@DecentNabeel i am not sure that would work, since it is not exact
im confused
So this part is correct? The general formula for quadratic eqn is y = ax^2 + bx + c take any 3 points from the table (0, 5.1) (1, 3.03) (5, 2.35) plug these points into the above equation and you will get 3 different equations use substitution or any method that you know to solve these 3 eqns with 3 unknowns
@satellite73 many way for solve i solve another but this is true
if it is exact you can do it using only three points \[(0,5.1),(1,3.03),(2,2.35)\] or any other 3
yes is correct
also you do not have to solve a system of three equations, only two
you want \[ax^2+bx+c\] but sine \((0,5.1\) is on the graph, you have \(c=5.1\) right away, only need \(a\) and \(b\)
So lets write step one so i can start writing it so i can show our work lol
the part you wrote above is correct
5.1 = a*0^2 + b*0 + c 3.03 = a*1^2 + b*1 + c 2.35 = a*5^2 + b*5 + c solve of a, b and c
what @DecentNabeel said solve that system but really it is not as hard as it looks since the first one tells you \(c=5.1\)
I dont know how to plug all that in
oops second one is \[3.03=a+b=5.1\] so \[a+b=-2.07\]
do you see how @DecentNabeel got this one \[5.1 = a*0^2 + b*0 + c\]
if the answer is "no" that is fine, just say so
yes c=5.1 @satellite73
no i dont get it
ok do you get that you are trying to find \(a,b\) and \(c\) to make up \[y=ax^2+bx+c\]?
yes
ok and from the table you see that \((0,5.1)\) is on the graph right?
correct
that means if \(x=0\) then \(y=5.1\) right ?
correct
What @LynFran ????
@satellite73 are you back? The serve kicked me off
yea me too
@EllenJaz17 what are you said @LynFran
they interrupted me learning lol
lol
lol
ok so to resume our story, were were here: we want \(a,b,c\) so make \(y=ax^2+bx+c\) and we know if \(x=0\) then \(y=5.1\) plug them in and get \[5.1=a\times 0^2+b\times 0+c\]
is that ok?
yes i get that one because its easy because x is 0 lol
right so that tells us right away that \(c=5.1\) yes?
correct
ok now \(a\) and \(b\) to go
@LynFran send him a message, im trying to get hw help, not get interrupted
ok im ready
the next point we have is \((1,3.03)\) right ?
yes
now don't forget, we already know \(c=5.1\) so we have \[y=ax^2+bx+5.1\] put \(x=1\) and \(y=3.03\) to get \[3.03=a\times 1^2+b\times 1+5.1\]
this also cleans up nicely, since \(1^2=1\) we have \[3.03=a+b+5.1\]
good so far?
yes
now to make it neater, subtract \(5.1\) from both sides that gives \[-2.07=a+b\] or \[a+b=-2.07\] and we put that aside for the moment
how we doing so far?
good
|dw:1435898261868:dw|
ok now one more equation to go you can pick any point you like but i will pick \((2,1.72)\)
Sounds good
again we have \[y=ax^2+bx+5.1\] put \(x=2,y=1.72\) and get \[1.72=a\times 2^2+b\times 2+5.1\]
ok with that?
yes
now we clean it up actually you can what do you get in terms of \(a,b\) and a number ?
\[-2.07^{2}+1.72+5.1\] ? Im sure that's wrong
quite wrong, it is much easier than that
haha oh jeez. I needed this class apparently very much lol and your help
lets go back to \[1.72=a\times 2^2+b\times 2+5.1\]
your equation will have an \(a\) and a \(b\) in it all you have to do is compute \(2^2\)
So a=4?
no just \(a\times 4\) or \(4a\)
lets back up a second
remember that when we replace \(x\) by \(1\) and \(y\) by \(3.03\) we ended up with an equation with both an \(a\) and a \(b\) in it it was \[a+b=-2.07\]
we are trying to accomplish the same thing here with \(x=2\) and \(y=1.72\) first step is to write \[1.72=a\times 2^2+b\times 2+5.1\]
so don't do too much work just see what the numbers give you your equation should still have an \(a\) and a \(b\) in it
if you are still confused, say so and i will spell it out
Yeah im confused
thought so i am just going to rewrite \[1.72=a\times 2^2+b\times 2+5.1\]
since \(2^2-4\) this is the same as \[1.72=a\times 4+b\times 2+5.1\] clear?
yes
i meant "since \(2^2=4\)
ok now we just make it look better \[1.72=4a+2b+5.1\] go that ?
yes
again we subtract \(5.1\) from both sides and arrive finally at \[4a+2b=-3.38\]
ok
now lets get those two equations together \[a+b=-2.07\\ 4a+2b=-3.38\]
damn \[a+b=-2.07\\ 4a+2b=-3.38\] thats better
we have to solve this "system of equations" to find \(a\) and \(b\)
sounds good
do you know how to solve this system?
no
i hate decimals, at this point i would cheat, but you can do whatever you like
do you have to show ALL your steps?
no it just says show work
well we have come this far, lets grind it out
probably the easiest thing to do is rewrite \[a+b=-2.07\] as \[b=-2.07-a\] and substitute that in the second equation get \[4a+2(-2.07-a)=-3.38\]
ok im just writing as you go
multiply out, combine like terms, work all the magic, get \[a=.33\] i will let you grind that out for yourself, but the answer is \(a=.33\)
so all thats left to find is b, than we do the equation, correct?
now we know that \(a=.33\) and \[a+b=-2.07\] you have \[.33+b=-2.07\] so \[b=-2.4\]
yeah that is it , find \(a\) and \(b\) we get \(a=.33\) and \(b=-2.4\)
and we recall that \(c=5.1\)
dang i made a mistake somewheres \(b=-2.45\) actually but whatever
So .33^2-2.45+5.1
oh damn damn damn i made a stupid typo
everywhere you see \(.33\) replace it by \(.38\)
i told you i hate decimals when you solve \[4a+2(-2.07-a)=-3.38\] you get \(a=.38\)
and since \(a=.38\) we get \(b=-2.45\) final answer \[y=.33x^2-2.45x+5.1\]
your math teacher must hate you to give you such a mess
at any rate, all the steps are there you can copy them down and look at them at your leisure took a while, hope you learned something
should it say y=.38 or y=.33 lol
Final Answer \[\huge \color{red}{y=.38x^2-2.45x+5.1}\]
lol great thank you so much!
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Do you wanna help me with another question that relates to this table lmao @satellite73
hell no not without a shot of tequila
ok post i will take a look, maybe it will be easier
ooh to
THIS TABLE ok
|dw:1435900241301:dw|
There you go lmao
good, no lime or salt (for sissies)
bahaha
what is the question i am about to turn in to a pumpkin
@satellite73 sorry are you still there

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