anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Getting the table, one moment
anonymous
  • anonymous
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
anonymous
  • anonymous
how are you supposed to do this (just asking because you have to "show work")

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More answers

anonymous
  • anonymous
I have no clue lol I typed everything the question asked and the table
DecentNabeel
  • DecentNabeel
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
anonymous
  • anonymous
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
DecentNabeel
  • DecentNabeel
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
anonymous
  • anonymous
@DecentNabeel i am not sure that would work, since it is not exact
anonymous
  • anonymous
im confused
anonymous
  • anonymous
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
DecentNabeel
  • DecentNabeel
@satellite73 many way for solve i solve another but this is true
anonymous
  • anonymous
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
DecentNabeel
  • DecentNabeel
yes is correct
anonymous
  • anonymous
also you do not have to solve a system of three equations, only two
anonymous
  • anonymous
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 \(b\)
anonymous
  • anonymous
So lets write step one so i can start writing it so i can show our work lol
anonymous
  • anonymous
the part you wrote above is correct
DecentNabeel
  • DecentNabeel
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
anonymous
  • anonymous
what @DecentNabeel said solve that system but really it is not as hard as it looks since the first one tells you \(c=5.1\)
anonymous
  • anonymous
I dont know how to plug all that in
anonymous
  • anonymous
oops second one is \[3.03=a+b=5.1\] so \[a+b=-2.07\]
anonymous
  • anonymous
do you see how @DecentNabeel got this one \[5.1 = a*0^2 + b*0 + c\]
anonymous
  • anonymous
if the answer is "no" that is fine, just say so
DecentNabeel
  • DecentNabeel
yes c=5.1 @satellite73
anonymous
  • anonymous
no i dont get it
anonymous
  • anonymous
ok do you get that you are trying to find \(a,b\) and \(c\) to make up \[y=ax^2+bx+c\]?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok and from the table you see that \((0,5.1)\) is on the graph right?
anonymous
  • anonymous
correct
anonymous
  • anonymous
that means if \(x=0\) then \(y=5.1\) right ?
anonymous
  • anonymous
correct
DecentNabeel
  • DecentNabeel
What @LynFran ????
anonymous
  • anonymous
@satellite73 are you back? The serve kicked me off
anonymous
  • anonymous
yea me too
DecentNabeel
  • DecentNabeel
@EllenJaz17 what are you said @LynFran
anonymous
  • anonymous
they interrupted me learning lol
anonymous
  • anonymous
lol
DecentNabeel
  • DecentNabeel
lol
anonymous
  • anonymous
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 \[5.1=a\times 0^2+b\times 0+c\]
anonymous
  • anonymous
is that ok?
anonymous
  • anonymous
yes i get that one because its easy because x is 0 lol
anonymous
  • anonymous
right so that tells us right away that \(c=5.1\) yes?
anonymous
  • anonymous
correct
anonymous
  • anonymous
ok now \(a\) and \(b\) to go
anonymous
  • anonymous
@LynFran send him a message, im trying to get hw help, not get interrupted
anonymous
  • anonymous
ok im ready
anonymous
  • anonymous
the next point we have is \((1,3.03)\) right ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
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 \[3.03=a\times 1^2+b\times 1+5.1\]
anonymous
  • anonymous
this also cleans up nicely, since \(1^2=1\) we have \[3.03=a+b+5.1\]
anonymous
  • anonymous
good so far?
anonymous
  • anonymous
yes
anonymous
  • anonymous
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
anonymous
  • anonymous
how we doing so far?
anonymous
  • anonymous
good
anonymous
  • anonymous
|dw:1435898261868:dw|
anonymous
  • anonymous
ok now one more equation to go you can pick any point you like but i will pick \((2,1.72)\)
anonymous
  • anonymous
Sounds good
DecentNabeel
  • DecentNabeel
good @satellite73
anonymous
  • anonymous
again we have \[y=ax^2+bx+5.1\] put \(x=2,y=1.72\) and get \[1.72=a\times 2^2+b\times 2+5.1\]
anonymous
  • anonymous
ok with that?
anonymous
  • anonymous
yes
anonymous
  • anonymous
now we clean it up actually you can what do you get in terms of \(a,b\) and a number ?
anonymous
  • anonymous
\[-2.07^{2}+1.72+5.1\] ? Im sure that's wrong
anonymous
  • anonymous
quite wrong, it is much easier than that
anonymous
  • anonymous
haha oh jeez. I needed this class apparently very much lol and your help
anonymous
  • anonymous
lets go back to \[1.72=a\times 2^2+b\times 2+5.1\]
anonymous
  • anonymous
your equation will have an \(a\) and a \(b\) in it all you have to do is compute \(2^2\)
anonymous
  • anonymous
So a=4?
anonymous
  • anonymous
no just \(a\times 4\) or \(4a\)
anonymous
  • anonymous
lets back up a second
anonymous
  • anonymous
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 \[a+b=-2.07\]
anonymous
  • anonymous
we are trying to accomplish the same thing here with \(x=2\) and \(y=1.72\) first step is to write \[1.72=a\times 2^2+b\times 2+5.1\]
anonymous
  • anonymous
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
anonymous
  • anonymous
if you are still confused, say so and i will spell it out
anonymous
  • anonymous
Yeah im confused
anonymous
  • anonymous
thought so i am just going to rewrite \[1.72=a\times 2^2+b\times 2+5.1\]
anonymous
  • anonymous
since \(2^2-4\) this is the same as \[1.72=a\times 4+b\times 2+5.1\] clear?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i meant "since \(2^2=4\)
anonymous
  • anonymous
ok now we just make it look better \[1.72=4a+2b+5.1\] go that ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
again we subtract \(5.1\) from both sides and arrive finally at \[4a+2b=-3.38\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
now lets get those two equations together \[a+b=-2.07\\ 4a+2b=-3.38\]
anonymous
  • anonymous
damn \[a+b=-2.07\\ 4a+2b=-3.38\] thats better
anonymous
  • anonymous
we have to solve this "system of equations" to find \(a\) and \(b\)
anonymous
  • anonymous
sounds good
anonymous
  • anonymous
do you know how to solve this system?
anonymous
  • anonymous
no
anonymous
  • anonymous
i hate decimals, at this point i would cheat, but you can do whatever you like
anonymous
  • anonymous
do you have to show ALL your steps?
anonymous
  • anonymous
no it just says show work
anonymous
  • anonymous
well we have come this far, lets grind it out
anonymous
  • anonymous
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 \[4a+2(-2.07-a)=-3.38\]
anonymous
  • anonymous
ok im just writing as you go
anonymous
  • anonymous
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 \(a=.33\)
anonymous
  • anonymous
so all thats left to find is b, than we do the equation, correct?
anonymous
  • anonymous
now we know that \(a=.33\) and \[a+b=-2.07\] you have \[.33+b=-2.07\] so \[b=-2.4\]
anonymous
  • anonymous
yeah that is it , find \(a\) and \(b\) we get \(a=.33\) and \(b=-2.4\)
anonymous
  • anonymous
and we recall that \(c=5.1\)
anonymous
  • anonymous
dang i made a mistake somewheres \(b=-2.45\) actually but whatever
anonymous
  • anonymous
So .33^2-2.45+5.1
anonymous
  • anonymous
oh damn damn damn i made a stupid typo
anonymous
  • anonymous
everywhere you see \(.33\) replace it by \(.38\)
anonymous
  • anonymous
i told you i hate decimals when you solve \[4a+2(-2.07-a)=-3.38\] you get \(a=.38\)
anonymous
  • anonymous
and since \(a=.38\) we get \(b=-2.45\) final answer \[y=.33x^2-2.45x+5.1\]
anonymous
  • anonymous
your math teacher must hate you to give you such a mess
anonymous
  • anonymous
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
anonymous
  • anonymous
should it say y=.38 or y=.33 lol
anonymous
  • anonymous
Final Answer \[\huge \color{red}{y=.38x^2-2.45x+5.1}\]
anonymous
  • anonymous
lol great thank you so much!
anonymous
  • anonymous
|dw:1435900099694:dw|
anonymous
  • anonymous
Do you wanna help me with another question that relates to this table lmao @satellite73
anonymous
  • anonymous
hell no not without a shot of tequila
anonymous
  • anonymous
ok post i will take a look, maybe it will be easier
anonymous
  • anonymous
ooh to
anonymous
  • anonymous
THIS TABLE ok
anonymous
  • anonymous
|dw:1435900241301:dw|
anonymous
  • anonymous
There you go lmao
anonymous
  • anonymous
good, no lime or salt (for sissies)
anonymous
  • anonymous
bahaha
anonymous
  • anonymous
what is the question i am about to turn in to a pumpkin
anonymous
  • anonymous
@satellite73 sorry are you still there

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