Mathematics: kay so i have this problem in math and it s find the length of a side of a square using a polynomial how do u solve that?

(a)(b)Step-by-step explanation:(a) In order to find the rate at which Glen Cove s population is changing with respect to time we need to calculate the derivative of P(t):Use the quotient rule: (1)Where:Replacing into the equation (1) and symplifying:(b) In order to find what s the population after 10 years, just evaluate P(t) at t=10Glen Cove s population is in thousands so:Finally, to find the rate at which the population is increasing, you need to evaluate the derivative of P(t) at t=10...

kay so i have this problem in math and it s find

Ответ:

deadmoney

30.05.2021

Mathematics

(a)

$\frac{dP(t)}{dt}=\frac{800(2t+5)}{(t^2+5t+40)^2}$

(b)

$P(10)\approx20789$

$\frac{dP(t)}{dt} \left \{ {{} \atop {t=10}} \right. \approx0.55$

Step-by-step explanation:

(a) In order to find the rate at which Glen Cove's population is changing with respect to time we need to calculate the derivative of P(t):

Use the quotient rule:

$\frac{d}{dt} (\frac{u}{v} )=\frac{v\frac{du}{dt} -u\frac{dv}{dt} }{v^2}$ (1)

Where:

$u=25t^2+125t+200\\v=t^2+5t+40$

$\frac{du}{dt} =50t+125\\\\\frac{dv}{dt} =2t+5$

Replacing into the equation (1) and symplifying:

$\frac{dP(t)}{dt}=\frac{800(2t+5)}{(t^2+5t+40)^2}$

(b)

In order to find what's the population after 10 years, just evaluate P(t) at t=10

$P(10)=\frac{25(10)^2+125(10)+200}{(10)^2+5(10)+40} =\frac{2500+1250+200}{100+50+40} =\frac{3950}{190} =20.78947368$

Glen Cove's population is in thousands so:

$P(10)=20.78947368\times10^3=20789.47368\approx 20789$

Finally, to find the rate at which the population is increasing, you need to evaluate the derivative of P(t) at t=10

$\frac{dP(t)}{dt} \left \{ {{} \atop {t=10}} \right. \approx0.55 =\frac{800(2(100)+5)}{((10)^2+5(10)+40)^2} =\frac{20000}{36100}=0.5540166205\approx0.55$

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kay so i have this problem in math and it's find the length of a side of a square using a polynomial how do u solve that?

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