A population of cattle is increasing at a rate of 800 + 70t per year, where t is measured in years. By how much does the population increase between the 7 th and the 12 th years.
Total Increase =
2007-11-26 16:08:27 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
P 12 = 800 + 8400 =9200
P 7 = 800 + 490 = 1290
Increase = 7910
2007-11-26 20:36:09 · answer #1 · answered by Como 7 · 0⤊ 1⤋
This is a Calculus problem, noobs. Pay attention. Cattle is INCREASING at a RATE of _______. That means c'(t)=800+70t. Integrate the rate to find the actual number.
int(f(x),x,a,b) denotes the definite integral of f(x) over [a,b] with respect to x. (that means it's dx at the end)
int(800+70t,t,7,12)
=[800t + 35t^2](from t=7 to t=12)
=800(12) + 35(144) - (800(7) + 35(49))
=9600 + 5040 - 5600 - 1715 (distribute the negative)
=7325 cattle gained between year 7 and year 12
2007-11-26 16:17:00 · answer #2 · answered by Anonymous · 2⤊ 0⤋
When t=7 population=800+70(7)=1290
When t=12 population=800+70(12)=1640.
Therefore, when t- between 7 and 12, population=1640-1290=350.
2007-11-26 16:15:08 · answer #3 · answered by nivik 3 · 0⤊ 0⤋
Input 12 then 7 into the equation then subtract the values you yet
800+70(12)=1640
800+70(7)=1290
1640 - 1290 = 350
Total Increase= 350
2007-11-26 16:13:46 · answer #4 · answered by Matty B 3 · 1⤊ 1⤋
Plug 7 and 12 into the formula. Calculate and then subtract the year 7 total from the year 12 total. Not hard at all. Stop being lazy and figure it out for yourself.
2007-11-26 16:11:31 · answer #5 · answered by martymar1222 5 · 1⤊ 2⤋
p(t) = 800 + 70t , integrate
P(t) = 800t + 35t^2
P(12) = 800(12) + 35(12)^2
P(12) = 14640
P(7) = 7315
14640 - 7315 = 7325 increase
2007-11-26 16:17:01 · answer #6 · answered by ? 3 · 0⤊ 0⤋
(800+(70x12)-800+(70x7)=350
2007-11-26 16:15:42 · answer #7 · answered by john 4 · 0⤊ 0⤋
dugaru32 is right, with the correct increase.
2007-11-26 16:18:06 · answer #8 · answered by z32486 3 · 1⤊ 0⤋