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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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?

Mathematics
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You have to find x such that\[10.8 \times 10^{12}=0.0012x^3-0.09x^2+1.5x+13\]
Have you learnt techniques for estimation?
Actually, it should be\[10.8=0.0012x^3-0.09x^2+1.5x+13\]

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sortof im confused
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
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.
Hang on, chai. I'll finish this first.
You'll see that this polynomial spits out 10.8 if x=-1.35, x=27.98, x=48.38.
Now, we're not going backward in time, so x=-1.35 can be discarded.
We have two other solutions, though. \[x \approx 28\]and\[x \approx 48\]These would correspond to the years\[1960+28 = 1988\]and\[1960+48 = 2008\]
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.
AH I see where I was messing up I didnt take out the 1.35
Yeah, and if this question was one that was created before 2008, the 2008 solution wouldn't even be looked at either.
ok, well thank you .. wow its so easy to forget things that need to be eliminated
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.
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 :)
no probs...good luck :)
okay, im stay here
lol...ok
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.
ok
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\[m_1=\frac{-1}{-1}=1\]
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:\[M_{AB}=\left( \frac{-2+6}{2},\frac{3+(-5)}{2} \right)=\left( 2,-1 \right)\]
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 \[y=x-3\]
and ,m here too! =) sorry for being a nag but i still have some questions for the solution you stated .
This is the equation of your line, chai :)
OK

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