mr jones estimates his profit starting in 2006 will be given by the formula p=7t^2-14t+10 where t is number of years and p is the profit.if this formula is correct,when will mr.jones have a profit of more than $1000. please help
Just plug in the numbers and solve for t:
7t^2 - 14t + 10 > 1000
7t^2 - 14t - 990 > 0
The roots of 7t^2 - 14t - 900 = 0 are
t = (7±√6979)/7 = -10.934,12.934
So, t>12.934 will produce > 1000 profit
That would be in 2019
To find out when Mr. Jones will have a profit of more than $1000, we need to solve the equation given by the formula p = 7t^2 - 14t + 10 > 1000.
First, let's rewrite the equation as 7t^2 - 14t + 10 - 1000 > 0.
Simplifying further, we have 7t^2 - 14t - 990 > 0.
To solve this quadratic inequality, we can use the quadratic formula: t = (-b ± sqrt(b^2 - 4ac))/(2a), where a = 7, b = -14, and c = -990.
Using the quadratic formula, we get t = (-(-14) ± sqrt((-14)^2 - 4 * 7 * -990))/(2*7) = (14 ± sqrt(196 + 27720))/14.
Simplifying further, we have t = (14 ± sqrt(27916))/14.
Now, let's calculate the values of t:
1. For t = (14 + sqrt(27916))/14, we get t ≈ 13.246.
2. For t = (14 - sqrt(27916))/14, we get t ≈ -0.75.
Since the number of years cannot be negative, we will discard the solution t ≈ -0.75.
Therefore, Mr. Jones will have a profit of more than $1000 after approximately 13.246 years starting from 2006.