12. In an experiment, a petri dish with a colony of bacteria is exposed to cold temperatures and then warmed again. 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 a. Find a quadratic model for the data in the table. Type your answer below. o b. Use the model to estimate the population of bacteria at 9 hours. Type your answer below.
\( \begin {cases} 0a+0b+c=5.1\\ 1a+1b+c=3.03\\ 2^2a+2b+c=1.72 \end{cases}\) Solve this system of equations for a, b and c. Then your answer will be in the form of \(y=a^2x+bx+c\)
i mean \(y=ax^2+bx+c\)
I have to think about this cause I'm not very good in equations like this...
ok im super confused and I can't figure this out can you help me ...
sure, what's the problem?
aren't the equations to help find the y=ax^2+bx+c right? Cause if that's so I am not really sure how to do that. Also wouldn't it be easier to take the population and find the difference between the first two and continuing on through the list to find the population for hour 9?
That will not work because it's not a linear equation.
Oh ok that would make sense ... so in order to find the population then finding y=ax^2+bx+c is going to be difficult for me... I'm very confused about how to get there..
do you know how to solve equations like 2y+4x=7 and -3y+x=2?
system of equations*
No I had notes on it but I can't find them..
Hmm, alright, because that's what we need to do here.
Ok
Anyway, this one isn't particularly hard, luckily
The first equation is simply 0a+0b+c=5.1 Since the a and b are both multiplied by 0, they can be ignored. So it basically just sais c=5.1
you get that?
Yes
ok, now then let's substitude everywhere where there is a c for 5.1, since we know c=5.1 \(\begin {cases} \\ 1a+1b+\color{red}{5.1}=3.03\\ 2^2a+2b+\color{red}{5.1}=1.72 \end{cases}\)
So from that we can conclude a+b+5.1=3.03 while 4a+2b+5.1=1.72 or, if we move the numbers to one side \( \begin {cases} \\ 1a+1b=3.03- \color{red}{5.1} \\ 2^2a+2b=1.72-\color{red}{5.1} \end{cases} \) makes \( \begin {cases} \\ 1a+1b=-2.07 \\ 2^2a+2b=3.83 \end{cases} \)
\( \begin {cases} \\ 1a+1b=-2.07 \\ 2^2a+2b=-3.83 \end{cases}\)
Ok so then we figure out what b is?
\( \begin{array} 22a+2b=-4.14\\ 4a+2b=-3.83~-\\ \hline\\ -2a+0= -0.31\end{array} \)
where did the -4.14 come from?
multiplying the first equation by 2
oh ok continue XD
it seems that i have made a mistake somewhere though as i don't get the answer t should give
And that mistake would be where?
thats what i'm trying to figure out
Do you think that -3.83 should've been doubled aswell??
oh no wait nvm >.<
this is strange.
maybe
@zimmah what was the answer to my question
I made a mistake with the c, let's do it again
\( \begin {cases} \\ 1a+1b+\color{red}{5.1}=3.03\\ 2^2a+2b+\color{red}{5.1}=1.72 \end{cases} \) let's work from here.
I think it was when 1.72 -5.1
Because I did it and got -4.62
HEY !!!!!!!!@zimmah WHAT WAS IT
Edge, not now
what was the answer
haven't gotten one yet
\(\begin{array} 22a+2b+10.2=6.06\\ 4a+2b+5.1=1.72~-\\ \hline\\ -2a+5.1= 4.34\end{array}\)
so -2a=4.34-5.1 \[a=-\frac{ 4.34+c }{ 2 }=0.38\]
and since a+b+c=3.03 we can fill it in to know 0.38+b+5.1=3.03 so b is -2.45
sorry for the confusion earlier, i must have messed up with subtracting with the 5.1 somewhere.
you did x3 and its ok we all make mistakes
Ok and could hint on how to find population for hour 9 ??
just plug in 9 for x in the formula you just created with your a, b and c
oh ok :3 thank you !
I really appreciate the help Zim :D
so my population number for hour 9 would be 13.83?
@DecentNabeel Maiṁ tumasē pyāra karatā hūm̐. Gus'sā na karēṁ.
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