For 1985 through 1996, the number, C (in thousands), of videos rented each year in Moose Jaw can be modeled by C= 0.069(t^3+4t^2+37t+600) where t=0 represents 1990. During which year are 60.4 thousand movies projected to be rented?
I'm not sure how to solve this problem. Could someone help me?
To determine the year in which 60.4 thousand movies are projected to be rented, we need to find the value of t that satisfies the equation C = 60.4.
Given that C = 0.069(t^3 + 4t^2 + 37t + 600), we can solve for t as follows:
1. Substitute C with 60.4 in the equation:
60.4 = 0.069(t^3 + 4t^2 + 37t + 600)
2. Simplify the equation:
0.069(t^3 + 4t^2 + 37t + 600) = 60.4
3. Divide both sides of the equation by 0.069 to isolate the polynomial:
t^3 + 4t^2 + 37t + 600 = 60.4 / 0.069
4. Evaluate the right side of the equation:
t^3 + 4t^2 + 37t + 600 ≈ 876.81
5. Rearrange the equation:
t^3 + 4t^2 + 37t + 600 - 876.81 = 0
6. Solve the resulting cubic equation.
Unfortunately, solving cubic equations algebraically can be quite complex. However, there are numerical methods or graphing calculators that can help find approximate solutions. One common method is to use Newton's method or other iterative approximation techniques to find the root(s) of the equation.
Alternatively, if you have a graphing calculator, you can graph the equation and find the x-coordinate of the point where the graph intersects the line y = 60.4. This x-coordinate corresponds to the value of t when 60.4 thousand movies are projected to be rented.
By using either of these methods, you can find the year in which 60.4 thousand movies are projected to be rented in Moose Jaw.