Answer:
Import in 2012 was the greatest
Step-by-step explanation:
Given
l(t) = -33t² + 800t - 3000
9 ≤ t ≤ 13
Required
Determine the year with highest import.
To do this we simply substitute the values of t from 9 to 13 in the given function.
When t = 9
l(t) = -33t² + 800t - 3000
l(9) = -33 * 9² + 800 * 9 - 3000
l(9) = -33 * 81 + 800 * 9 - 3000
l(9) = -2673 + 7200 -3000
l(9) = 1527
When t = 10
l(10) = -33 * 10² + 800 * 10 - 3000
l(10) = -33 * 100 + 800 * 10 - 3000
l(10) = -3300 + 8000 - 3000
l(10) = 1700
When t = 11
l(11) = -33 * 11² + 800 * 11 - 3000
l(11) = -33 * 121 + 800 * 11 - 3000
l(11) = -3993 + 8800 - 3000
l(11) = 1807
When t = 12
l(12) = -33 * 12² + 800 * 12 - 3000
l(12) = -33 * 144 + 800 * 12 - 3000
l(12) = -4752 + 9600 - 3000
l(12) = 1848
When t = 13
l(13) = -33 * 13² + 800 * 13 - 3000
l(13) = -33 * 169 + 800 * 13 - 3000
l(13) = -5577 + 10400 - 3000
l(13) = 1823
Comparing the values of l(t) for the range of t = 9 to 13,
The highest value of l(t) is:
l(12) = 1848
Hence, the year with the highest import is 2012
is the distance between points
and
are the coordinates of the second point
