Thread: Polynomial and rational functions

POWER FUNCTIONS!


A 30-second commercial during Super Bowl XL in 2006 cost advertisesrs $2.5 million. For the first Super Bowl in 1967, an advertiser could have purchased approximately 28.699 minutes of advertising time for the same amount of money

a) Assuming that cost is proportional to time, find the cost of advertising, in dollars/second, during the 1967 and 2006 super bowls

b) How many times more expensive was super bowl advertising in 2006 than in 1967

setting this up i tried y= kx^n and dividing eachother, solving for N, but i realized that i couldn't becuase i chose x as time and i put it as .3 forgetting that it really isn't 30 seconds, this one has stumped me for awhile, and i can't figure it out.

Originally Posted by getevlcted

POWER FUNCTIONS!


A 30-second commercial during Super Bowl XL in 2006 cost advertisesrs $2.5 million. For the first Super Bowl in 1967, an advertiser could have purchased approximately 28.699 minutes of advertising time for the same amount of money

a) Assuming that cost is proportional to time, find the cost of advertising, in dollars/second, during the 1967 and 2006 super bowls

b) How many times more expensive was super bowl advertising in 2006 than in 1967

setting this up i tried y= kx^n and dividing eachother, solving for N, but i realized that i couldn't becuase i chose x as time and i put it as .3 forgetting that it really isn't 30 seconds, this one has stumped me for awhile, and i can't figure it out.

You don't need the "n" for a. "Assuming that cost is proportional to time" means just y= kx. In 1967, "an advertiser could have purchased approximately 28.699 minutes of advertising time for $2.5 million". y= 2500000, x= 29.699: 250000= k(28.699) so k= 250000/28.699. The fact that you were asked give the answer "in dollars/second" should have told you that you needed to divide "2.5 million dollars" by "30 seconds".

Do the same, with x= 30, for 2006.