## 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?

1. anonymous

You have to find x such that$10.8 \times 10^{12}=0.0012x^3-0.09x^2+1.5x+13$

2. anonymous

Have you learnt techniques for estimation?

3. anonymous

Actually, it should be$10.8=0.0012x^3-0.09x^2+1.5x+13$

4. anonymous

sortof im confused

5. anonymous

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

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

Hang on, chai. I'll finish this first.

8. anonymous

You'll see that this polynomial spits out 10.8 if x=-1.35, x=27.98, x=48.38.

9. anonymous

Now, we're not going backward in time, so x=-1.35 can be discarded.

10. anonymous

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

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

AH I see where I was messing up I didnt take out the 1.35

13. anonymous

Yeah, and if this question was one that was created before 2008, the 2008 solution wouldn't even be looked at either.

14. anonymous

ok, well thank you .. wow its so easy to forget things that need to be eliminated

15. anonymous

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

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

no probs...good luck :)

18. anonymous

okay, im stay here

19. anonymous

lol...ok

20. anonymous

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

ok

22. anonymous

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

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

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

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

26. anonymous

@lokisan

27. anonymous

This is the equation of your line, chai :)

28. anonymous

OK