I'm getting something a little different. Is this the correct equation?
The second root should be rather than 3800. For the second part, you can get an exact solution by solving the quadratic
My bounds come out differently than yours.
I think I have found the right answer but if someone could have a look at it for me, that would be awesome. Thanks.
The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p, in dollars, that it charges. The revenue R, in dollars is -.05p^2 + 1900p
At what prices p is revenue zero? Round to four decimal places. I have when revenue is zero, then p=0 or 3800. (0,0) and (3800,0)
For what range of prices will revenue exceed $1,300,000? Round to four decimal places.
886<p<2904 The price must be between or equal to the values of $886 and $2900 for revenue to exceed $1,300,000.
For this part, I just looked on the table of my calculator. Should I be more precise and use a formula?
I'm getting something a little different. Is this the correct equation?
The second root should be rather than 3800. For the second part, you can get an exact solution by solving the quadratic
My bounds come out differently than yours.
My mistake. it is .5 instead of .05
With that being the case, it looks like I am right then. 1900/.5 is 3800 and 1900/.05 is 38,000
I added the extra zero so that gave you an extra zero.
The second part is what is messing with me. The way it states, I need to find what p values (or x) will make the R values (or y) exceed $1,300,000. Looking at the table, that that is 886 to 2900. I am unsure how to get those values via the equation.
I know I need to do -.5p^2+1900p > 1,300,000 but I am kinda lost from there. Should I just move the 13m over with everything else and then factor it?