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
solving for 7t^2-14t+10>1000, we get:
7t^2-14t-990>0
use quadratic equation to get:
t> (7+ sqrt(6979))/7; thus:
t> ~12.93 years
explain in more detail plz
To determine when Mr. Jones will have a profit of more than $1000, we need to solve the equation "p > 1000" using the profit formula given:
p = 7t^2 - 14t + 10
So, we will substitute "1000" for "p" in the equation:
1000 = 7t^2 - 14t + 10
Now we have a quadratic equation. To solve it, we can rearrange it into the standard form:
7t^2 - 14t + 10 - 1000 = 0
Simplifying:
7t^2 - 14t - 990 = 0
To solve the quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring may not be straightforward, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
Plugging in the values of "a," "b," and "c" from our equation:
t = (-(-14) ± √((-14)^2 - 4(7)(-990))) / (2(7))
Simplifying:
t = (14 ± √(196 + 27720))/14
t = (14 ± √27916)/14
Now, we can calculate the values of t:
t = (14 + √27916)/14 ≈ 15.47
t = (14 - √27916)/14 ≈ -0.47
Since "t" represents the number of years, it cannot be negative. Therefore, we consider the positive solution, which is approximately 15.47.
Hence, Mr. Jones will have a profit of more than $1000 after approximately 15.47 years since 2006.