An animal population is increasing at a rate of 11+62t per year (where t is measured in years). By how much does the animal population increase between the fourth and tenth years?
2007-04-27 07:57:26 · 5 answers · asked by Liz 1 in Science & Mathematics ➔ Mathematics
let P(t) be the population of animal at time t
P'(t) = 11+62t
The net increase between the fourth and tenth year
..........10
= INT (11+62t) dt ............... (INT - the integration sign)
..........4
............................10
= 11t + 62t^2/2 |
.............................4
= 11(10) + 62(10)^2/2 - 11(4) - 62(4)^2/2
= 110 +3100 - 44 -496
= 2670
2007-04-27 08:15:33 · answer #1 · answered by Phoenix S 2 · 0⤊ 1⤋
If the equation is understood by me correctly, we have
P = 11 + 62t where t is time in years.
So we have P4 = 11 + 62 X 4 = 11 + 248 = 259
P10 = 11 + 620 = 631
So the population increase from the 4th to the 10th year is 631 - 259 = 372
2007-04-27 08:04:39 · answer #2 · answered by Swamy 7 · 0⤊ 1⤋
If you INCLUDE the 4th and 10th years,
Increase = 11(7) + 62(4 + 5 + ... + 10)
Increase = 77 + 62(7/2)(14)
Increase = 3115
If you DON'T include those years,
Increase = 11(5) + 62(5 + 6 + 7 + 8 + 9)
Increase = 55 + 62(5/2)(14)
Increase = 2225
2007-04-27 08:08:02 · answer #3 · answered by Philo 7 · 1⤊ 0⤋
Since between 4th and 10th year is a total of 6 years, you put 6 in as t. 62(6)= 372 then you add 11 to get 383.
2007-04-27 08:13:41 · answer #4 · answered by Emily 5 · 0⤊ 1⤋
it is increased by 6*62
372 is the total number of increase in population
2007-04-27 08:02:05 · answer #5 · answered by Siddharth A 2 · 0⤊ 1⤋