A pond contains 100 insect larvae. Suppose the population triples every 2 days. How many insect larvae are there after 10 days?
I know you have to make an equation, but I don't know what it would be. Can you please help me with the equation?
2007-12-19 03:34:09 · 8 answers · asked by Tofuie 3 in Science & Mathematics ➔ Mathematics
the equation is :
number of larvae after x days = 100 * 3^(x/2)
substituting x=10 in the previous equation :
number of larvae after 10 days= 100 *3^(10/2)
=100* 3^5
=100* 243
= 24300
2007-12-19 05:31:09 · answer #1 · answered by incognito 3 · 2⤊ 0⤋
As population P increases exponentially,
dP/dt = kP
=> dP/P = k dt
Integrating,
ln P = kt + c
P = 100 when t = 0 => c = ln 100
=> ln (P/100) = kt
P = 300 when t = 2 dys
=> ln (300/100) = k*2
=> k = ln (√3)
=> ln (P/100) = t ln (√3)
=> P = 100 (√3)^t
When t = 10 days,
P = 100(√3)^(10) = 24300 insect larvae
2007-12-19 03:46:29 · answer #2 · answered by Madhukar 7 · 0⤊ 0⤋
no. of larvae after 2 days = 100 x 3
no. of larvae after 4 days = (100 x 3) x 3
no. of larvae after 6 days = [(100 x 3) x 3] x 3
no. of larvae after 8 days = [[(100 x 3) x 3] x 3] x 3
no. of larvae after 10 days = [[[(100 x 3) x 3] x 3] x 3] x 3 = 100 x 3^5
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i hope this helps (:
2007-12-19 03:44:54 · answer #3 · answered by balikbayanista 2 · 0⤊ 0⤋
100(3)^(t/2), so for t = 10, the value of the expression is 24,300 larvae
2007-12-19 03:45:00 · answer #4 · answered by Anonymous · 0⤊ 0⤋
properly, attempt to no longer worry with reference to the negatives in the initiating too lots. basically bear in mind that after a power is raised to a power, you multiply the numbers jointly. shall we initiate with the numerator: (2a^[2] b^[-3])^3 simplifies into (8a^6)(b^-9) And the denominator is: (9a^[-3] b^[2]) simplifies into (9a^-3)(b^2) So (8a^6)(b^-9) ------------------ (9a^-3)(b^2) Now, this is an significant rule. you additionally could make exponents advantageous by moving them to the different fringe of the branch sign. So, you may flow a^-3 to the magnificent to make it a^3. and you will flow b^-9 to the backside to make it b^9. So after doing this, you get a simplified expression: [(8a^9)/(9b^11)]^a million/2 which ought to be the basically precise answer.
2016-11-23 15:02:05 · answer #5 · answered by anuj 3 · 0⤊ 0⤋
100+ 300 +1200 + 4800 + 19200+76800 = 102,400
2007-12-19 03:41:15 · answer #6 · answered by ironduke8159 7 · 0⤊ 0⤋
I agree with the answerer above.
2007-12-19 09:00:20 · answer #7 · answered by ღ Scent of a Woman ღ 4 · 1⤊ 0⤋
never mind wrong equation sorry
2007-12-19 03:42:09 · answer #8 · answered by orkillies 2 · 0⤊ 0⤋