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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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    Getting the table, one moment

  2. anonymous
    • one year ago
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    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

  3. anonymous
    • one year ago
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    how are you supposed to do this (just asking because you have to "show work")

  4. anonymous
    • one year ago
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    I have no clue lol I typed everything the question asked and the table

  5. DecentNabeel
    • one year ago
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    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

  6. DecentNabeel
    • one year ago
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    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

  7. anonymous
    • one year ago
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    @DecentNabeel i am not sure that would work, since it is not exact

  8. anonymous
    • one year ago
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    im confused

  9. anonymous
    • one year ago
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    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

  10. DecentNabeel
    • one year ago
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    @satellite73 many way for solve i solve another but this is true

  11. anonymous
    • one year ago
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    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

  12. DecentNabeel
    • one year ago
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    yes is correct

  13. anonymous
    • one year ago
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    also you do not have to solve a system of three equations, only two

  14. anonymous
    • one year ago
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    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 \(b\)

  15. anonymous
    • one year ago
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    So lets write step one so i can start writing it so i can show our work lol

  16. anonymous
    • one year ago
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    the part you wrote above is correct

  17. DecentNabeel
    • one year ago
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    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

  18. anonymous
    • one year ago
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    what @DecentNabeel said solve that system but really it is not as hard as it looks since the first one tells you \(c=5.1\)

  19. anonymous
    • one year ago
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    I dont know how to plug all that in

  20. anonymous
    • one year ago
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    oops second one is \[3.03=a+b=5.1\] so \[a+b=-2.07\]

  21. anonymous
    • one year ago
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    do you see how @DecentNabeel got this one \[5.1 = a*0^2 + b*0 + c\]

  22. anonymous
    • one year ago
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    if the answer is "no" that is fine, just say so

  23. DecentNabeel
    • one year ago
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    yes c=5.1 @satellite73

  24. anonymous
    • one year ago
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    no i dont get it

  25. anonymous
    • one year ago
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    ok do you get that you are trying to find \(a,b\) and \(c\) to make up \[y=ax^2+bx+c\]?

  26. anonymous
    • one year ago
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    yes

  27. anonymous
    • one year ago
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    ok and from the table you see that \((0,5.1)\) is on the graph right?

  28. anonymous
    • one year ago
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    correct

  29. anonymous
    • one year ago
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    that means if \(x=0\) then \(y=5.1\) right ?

  30. anonymous
    • one year ago
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    correct

  31. DecentNabeel
    • one year ago
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    What @LynFran ????

  32. anonymous
    • one year ago
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    @satellite73 are you back? The serve kicked me off

  33. anonymous
    • one year ago
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    yea me too

  34. DecentNabeel
    • one year ago
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    @EllenJaz17 what are you said @LynFran

  35. anonymous
    • one year ago
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    they interrupted me learning lol

  36. anonymous
    • one year ago
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    lol

  37. DecentNabeel
    • one year ago
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    lol

  38. anonymous
    • one year ago
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    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 \[5.1=a\times 0^2+b\times 0+c\]

  39. anonymous
    • one year ago
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    is that ok?

  40. anonymous
    • one year ago
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    yes i get that one because its easy because x is 0 lol

  41. anonymous
    • one year ago
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    right so that tells us right away that \(c=5.1\) yes?

  42. anonymous
    • one year ago
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    correct

  43. anonymous
    • one year ago
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    ok now \(a\) and \(b\) to go

  44. anonymous
    • one year ago
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    @LynFran send him a message, im trying to get hw help, not get interrupted

  45. anonymous
    • one year ago
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    ok im ready

  46. anonymous
    • one year ago
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    the next point we have is \((1,3.03)\) right ?

  47. anonymous
    • one year ago
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    yes

  48. anonymous
    • one year ago
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    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 \[3.03=a\times 1^2+b\times 1+5.1\]

  49. anonymous
    • one year ago
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    this also cleans up nicely, since \(1^2=1\) we have \[3.03=a+b+5.1\]

  50. anonymous
    • one year ago
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    good so far?

  51. anonymous
    • one year ago
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    yes

  52. anonymous
    • one year ago
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    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

  53. anonymous
    • one year ago
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    how we doing so far?

  54. anonymous
    • one year ago
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    good

  55. anonymous
    • one year ago
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    |dw:1435898261868:dw|

  56. anonymous
    • one year ago
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    ok now one more equation to go you can pick any point you like but i will pick \((2,1.72)\)

  57. anonymous
    • one year ago
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    Sounds good

  58. DecentNabeel
    • one year ago
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    good @satellite73

  59. anonymous
    • one year ago
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    again we have \[y=ax^2+bx+5.1\] put \(x=2,y=1.72\) and get \[1.72=a\times 2^2+b\times 2+5.1\]

  60. anonymous
    • one year ago
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    ok with that?

  61. anonymous
    • one year ago
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    yes

  62. anonymous
    • one year ago
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    now we clean it up actually you can what do you get in terms of \(a,b\) and a number ?

  63. anonymous
    • one year ago
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    \[-2.07^{2}+1.72+5.1\] ? Im sure that's wrong

  64. anonymous
    • one year ago
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    quite wrong, it is much easier than that

  65. anonymous
    • one year ago
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    haha oh jeez. I needed this class apparently very much lol and your help

  66. anonymous
    • one year ago
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    lets go back to \[1.72=a\times 2^2+b\times 2+5.1\]

  67. anonymous
    • one year ago
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    your equation will have an \(a\) and a \(b\) in it all you have to do is compute \(2^2\)

  68. anonymous
    • one year ago
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    So a=4?

  69. anonymous
    • one year ago
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    no just \(a\times 4\) or \(4a\)

  70. anonymous
    • one year ago
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    lets back up a second

  71. anonymous
    • one year ago
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    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 \[a+b=-2.07\]

  72. anonymous
    • one year ago
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    we are trying to accomplish the same thing here with \(x=2\) and \(y=1.72\) first step is to write \[1.72=a\times 2^2+b\times 2+5.1\]

  73. anonymous
    • one year ago
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    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

  74. anonymous
    • one year ago
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    if you are still confused, say so and i will spell it out

  75. anonymous
    • one year ago
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    Yeah im confused

  76. anonymous
    • one year ago
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    thought so i am just going to rewrite \[1.72=a\times 2^2+b\times 2+5.1\]

  77. anonymous
    • one year ago
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    since \(2^2-4\) this is the same as \[1.72=a\times 4+b\times 2+5.1\] clear?

  78. anonymous
    • one year ago
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    yes

  79. anonymous
    • one year ago
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    i meant "since \(2^2=4\)

  80. anonymous
    • one year ago
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    ok now we just make it look better \[1.72=4a+2b+5.1\] go that ?

  81. anonymous
    • one year ago
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    yes

  82. anonymous
    • one year ago
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    again we subtract \(5.1\) from both sides and arrive finally at \[4a+2b=-3.38\]

  83. anonymous
    • one year ago
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    ok

  84. anonymous
    • one year ago
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    now lets get those two equations together \[a+b=-2.07\\ 4a+2b=-3.38\]

  85. anonymous
    • one year ago
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    damn \[a+b=-2.07\\ 4a+2b=-3.38\] thats better

  86. anonymous
    • one year ago
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    we have to solve this "system of equations" to find \(a\) and \(b\)

  87. anonymous
    • one year ago
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    sounds good

  88. anonymous
    • one year ago
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    do you know how to solve this system?

  89. anonymous
    • one year ago
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    no

  90. anonymous
    • one year ago
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    i hate decimals, at this point i would cheat, but you can do whatever you like

  91. anonymous
    • one year ago
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    do you have to show ALL your steps?

  92. anonymous
    • one year ago
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    no it just says show work

  93. anonymous
    • one year ago
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    well we have come this far, lets grind it out

  94. anonymous
    • one year ago
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    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 \[4a+2(-2.07-a)=-3.38\]

  95. anonymous
    • one year ago
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    ok im just writing as you go

  96. anonymous
    • one year ago
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    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 \(a=.33\)

  97. anonymous
    • one year ago
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    so all thats left to find is b, than we do the equation, correct?

  98. anonymous
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    now we know that \(a=.33\) and \[a+b=-2.07\] you have \[.33+b=-2.07\] so \[b=-2.4\]

  99. anonymous
    • one year ago
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    yeah that is it , find \(a\) and \(b\) we get \(a=.33\) and \(b=-2.4\)

  100. anonymous
    • one year ago
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    and we recall that \(c=5.1\)

  101. anonymous
    • one year ago
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    dang i made a mistake somewheres \(b=-2.45\) actually but whatever

  102. anonymous
    • one year ago
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    So .33^2-2.45+5.1

  103. anonymous
    • one year ago
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    oh damn damn damn i made a stupid typo

  104. anonymous
    • one year ago
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    everywhere you see \(.33\) replace it by \(.38\)

  105. anonymous
    • one year ago
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    i told you i hate decimals when you solve \[4a+2(-2.07-a)=-3.38\] you get \(a=.38\)

  106. anonymous
    • one year ago
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    and since \(a=.38\) we get \(b=-2.45\) final answer \[y=.33x^2-2.45x+5.1\]

  107. anonymous
    • one year ago
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    your math teacher must hate you to give you such a mess

  108. anonymous
    • one year ago
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    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

  109. anonymous
    • one year ago
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    should it say y=.38 or y=.33 lol

  110. anonymous
    • one year ago
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    Final Answer \[\huge \color{red}{y=.38x^2-2.45x+5.1}\]

  111. anonymous
    • one year ago
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    lol great thank you so much!

  112. anonymous
    • one year ago
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    |dw:1435900099694:dw|

  113. anonymous
    • one year ago
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    Do you wanna help me with another question that relates to this table lmao @satellite73

  114. anonymous
    • one year ago
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    hell no not without a shot of tequila

  115. anonymous
    • one year ago
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    ok post i will take a look, maybe it will be easier

  116. anonymous
    • one year ago
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    ooh to

  117. anonymous
    • one year ago
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    THIS TABLE ok

  118. anonymous
    • one year ago
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    |dw:1435900241301:dw|

  119. anonymous
    • one year ago
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    There you go lmao

  120. anonymous
    • one year ago
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    good, no lime or salt (for sissies)

  121. anonymous
    • one year ago
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    bahaha

  122. anonymous
    • one year ago
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    what is the question i am about to turn in to a pumpkin

  123. anonymous
    • one year ago
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    @satellite73 sorry are you still there

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