Suppose that the number of cars, C, on 1st Avenue in a city over a period of time t, in months, is graphed on a rectangular coordinate system where time is on the horizontal axis. Suppose that the number of cars driven on 1st Avenue can be modeled by an exponential function, C= p * a^t where p is the number of cars on the road on the first day recorded. If you commuted to work each day along 1st Avenue, would you prefer that the value of be between 0 and 1 or larger than 1? Explain your reasoning.
2007-09-22 06:25:36 · 5 answers · asked by darrington2001 1 in Science & Mathematics ➔ Mathematics
Hi,
You would prefer that of "a" would be between 0 and 1 because that means there would be "p" cars or FEWER. If "a" was greater than 1, there would be more traffic and it would be HARDER to get to work.
I hope that helps!! :-)
2007-09-22 06:34:35 · answer #1 · answered by Pi R Squared 7 · 0⤊ 0⤋
Think about how the exponential function works. Graph it if you have never done so. If a > 1 then the function gets bigger VERY fast, if 0 < a < 1 then it gets smaller very fast.
Example: Suppose a = 2 and p = 10. Then at t = 1 there will be 10(2^1) = 10(2) = 20 cars, at t = 1 there will be 10(2^2) = 40, at t = 3 there will be 10(2^3) = 80 etc. and at each month there will be twice as many as the month before. That's how the exponential function works!
Now suppose a = 1/2 and p = 10. Then at t = 1 there will be 10(1/2^1) = 5, at t = 2 there will be 10(1/2^2) = 2.5, and each month there will be twice as FEW as the month before.
The problem is designed to give you a feel for how the exponential function works in a "practical" situation. Hope this helps.
2007-09-22 06:37:02 · answer #2 · answered by TurtleFromQuebec 5 · 0⤊ 0⤋
You would like a to be 0 and 1 because then the traffic would be less than the 1st day. If a > 1 the traffic will be greater.
If a = .5 and t =2, then, a^2 = .5^2 = .25
If a = 1.5 and t =2 then a^2 = 1.25^2 = 2.25
In the 1st case traffic on the 2nd day would be 1/4 of traffic on first day. In the second case ,traffick
on the 2nd day would be 2.25 times traffic on the 1st day.
2007-09-22 06:43:36 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
If a is less than 1 the value of C will be less than p because t would be raised to a fractional power That is the traffic is less than on the first day. You would probably prefer this because most people like less traffic.
2007-09-22 06:36:58 · answer #4 · answered by bignose68 4 · 0⤊ 0⤋
2 + 2 / [ 3x^(a million/3) ] = 0 multiplying and dividing the left hand side by using [ 6x^(a million/3) ] supplies us : [ 12x^(a million/3) + 4 ] / [6x^(a million/3)] = 0 Now divide the two components by using 2 to get [ 6x^(a million/3) + 2 ] / 6x^(a million/3) = 0 this gives 6x^(a million/3) + 2 = 0 => 6x^(a million/3) = - 2 => x^(a million/3) = -2/6 = -a million/3 => x = (-a million/6)^3 = -a million/27
2016-12-17 07:40:46 · answer #5 · answered by Anonymous · 0⤊ 0⤋