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kaybaby23
12.08.2020 •
Mathematics
An animal population is increasing at a rate of 13 51t13 51t per year (where t is measured in years). By how much does the animal population increase between the fourth and tenth years.
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Ответ:
ΔP = 567
Step-by-step explanation:
The increasing rate of the population is 13,51*t.
That rate by definition is:
dP/dt where P is the population therefore
dP/dt = 13,51*t
dt = 13,51*t*dt
Integrating on both sides of the equation we get:
∫dp = ∫ 13,51*t*dt
P = 13,51*t²/2 + K ( K is population for t = 0 )
Now the population in 10 years P(₁₀)
P(₁₀) = 13,51* (10)² /2 + K
P(₁₀) = 675,5 + K (1)
And P(₄) is
P(₄) = 13,51*(4)²/2 * K
P(₄) = 108,08 + K (2)
Then substracting
P(₁₀) - P(₄) = ( 675,5 + K ) - ( 108,08 + K )
ΔP = 567,42
But we don´t have fraction of animal, then
ΔP = 567
Ответ:
what is a dummy.
Step-by-step explanation: