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anonymous

  • 5 years ago

The polynominal .0012x^3-.09x^2+1.5x+13 models natural gas consumption in trillions cubic feet, where x=0 corresponds to 1960, x=1 to 1961 and so on. Between 1970 and 1990 gas was about 10.8 trillion cubic feet. Natural gas consumption was 10.8 trillion cubic feet in year?

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  1. anonymous
    • 5 years ago
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    You have to find x such that\[10.8 \times 10^{12}=0.0012x^3-0.09x^2+1.5x+13\]

  2. anonymous
    • 5 years ago
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    Have you learnt techniques for estimation?

  3. anonymous
    • 5 years ago
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    Actually, it should be\[10.8=0.0012x^3-0.09x^2+1.5x+13\]

  4. anonymous
    • 5 years ago
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    sortof im confused

  5. anonymous
    • 5 years ago
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    Okay, if you go to this site, I've plugged in the equation and you should see a plot, along with estimated solutions: http://www.wolframalpha.com/input/?i=10.8%3D0.0012x^3%E2%88%920.09x^2%2B1.5x%2B13

  6. anonymous
    • 5 years ago
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    lokisan Hero, its me chai. can i ask you some help to solve this problem Find the equation of the perpendicular bisector of the line joining the points A and B in the following exercises 2. The points A(-2,3), B(6,-5), and C(8,5) are vertices of a triangle. Find the equation of the median.

  7. anonymous
    • 5 years ago
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    Hang on, chai. I'll finish this first.

  8. anonymous
    • 5 years ago
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    You'll see that this polynomial spits out 10.8 if x=-1.35, x=27.98, x=48.38.

  9. anonymous
    • 5 years ago
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    Now, we're not going backward in time, so x=-1.35 can be discarded.

  10. anonymous
    • 5 years ago
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    We have two other solutions, though. \[x \approx 28\]and\[x \approx 48\]These would correspond to the years\[1960+28 = 1988\]and\[1960+48 = 2008\]

  11. anonymous
    • 5 years ago
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    It seems as if the question wants you to find when consumption was 10.8 trillion between the years 1970 and 1990...well, if this is the case, the answer is 1988.

  12. anonymous
    • 5 years ago
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    AH I see where I was messing up I didnt take out the 1.35

  13. anonymous
    • 5 years ago
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    Yeah, and if this question was one that was created before 2008, the 2008 solution wouldn't even be looked at either.

  14. anonymous
    • 5 years ago
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    ok, well thank you .. wow its so easy to forget things that need to be eliminated

  15. anonymous
    • 5 years ago
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    Yeah. If you need to estimate solutions manually, you should look into the Newton-Raphson method, or bisection method...you can probably find clips on these on YouTube.

  16. anonymous
    • 5 years ago
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    I really appreciate that you explain the problems to me, it helps me to understand so the next one I have with different figures I will be able to do correctly - Ok I will look into those :)

  17. anonymous
    • 5 years ago
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    no probs...good luck :)

  18. anonymous
    • 5 years ago
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    okay, im stay here

  19. anonymous
    • 5 years ago
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    lol...ok

  20. anonymous
    • 5 years ago
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    You have to find the equation of the line that is 1) perpendicular to the line that joins A and B, and 2) cuts the middle of A and B.

  21. anonymous
    • 5 years ago
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    ok

  22. anonymous
    • 5 years ago
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    Now, a line that is perpendicular to AB will have a slope, m, such that\[m_1m_{AB}=-1 \rightarrow m_{AB}=-\frac{1}{m_1}\] where m_{AB} is the slope of AB. The slope of AB is found by\[m_{AB}=\frac{-5-3}{6-(-2)}=\frac{-8}{8}=-1\]so the slope of the line perpendicular to AB must be\[m_1=\frac{-1}{-1}=1\]

  23. anonymous
    • 5 years ago
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    Now, in order to find the line, we need one more thing - a point. We're told we want the line to cut through the middle of AB, so we have to find the mid-point of AB. This is found by taking the average of the x and y coordinates of A and B as such:\[M_{AB}=\left( \frac{-2+6}{2},\frac{3+(-5)}{2} \right)=\left( 2,-1 \right)\]

  24. anonymous
    • 5 years ago
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    The equation of the line can then be found using the point-gradient formula:\[y-y_1=m_{AB}(x-x_1)\]that is,\[y-(-1)=(1)(x-2) \rightarrow y+1=x-2\]so \[y=x-3\]

  25. anonymous
    • 5 years ago
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    and ,m here too! =) sorry for being a nag but i still have some questions for the solution you stated .

  26. anonymous
    • 5 years ago
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    @lokisan

  27. anonymous
    • 5 years ago
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    This is the equation of your line, chai :)

  28. anonymous
    • 5 years ago
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    OK

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