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3 years ago

Please help and show work i need 17 19 and 21

Mathematics

1 answer:

Nat2105 [25]3 years ago

5 0

Answer:

Step-by-step explanation:

(17). g(x) = x³ + 4x

f(x) = 4x + 1

( f × g )( x ) = ( x³ + 4x )( 4x + 1 ) = 4  $x^{4}$  + x³ + 16x² + 4x

(19). f(t) = 4t - 4

g(t) = t - 2

( 4f + 3g )( t ) = 4(4t - 4) + 3(t - 2) = 16t - 16 + 3t - 6 = 19t - 22

(21). h(t) = t + 3

g(t) = 4t + 1

h(t - 2) + g(t - 2) = ( t - 2 ) + 3 + 4( t - 2 ) + 1 = t + 4t - 2 + 3 - 8 + 1 = 5t - 6

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Suppose a certain population satisfies the logistic equation given by dP

Ksenya-84 [330]

Answer:

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Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

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$\frac{dP}{10P-P^2}=dt$

Integrate both sides.

$\int \frac{dP}{10P-P^2}=\int dt$ .... (1)

Using partial fraction

$\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}$

$A=\frac{1}{10},B=\frac{1}{10}$

Using these values the equation (1) can be written as

$\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt$

$\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt$

On simplification we get

$\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C$

$\frac{1}{10}(\ln \frac{P}{10-P})=t+C$

We have P(0)=1

Substitute t=0 and P=1 in above equation.

$\frac{1}{10}(\ln \frac{1}{10-1})=0+C$

$\frac{1}{10}(\ln \frac{1}{9})=C$

The required equation is

$\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})$

Multiply both sides by 10.

$\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}$

$e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}$

$\frac{P}{10-P}=\frac{1}{9}e^{10t}$

Reciprocal it

$\dfrac{10-P}{P}=9e^{-10t}$

$P(t)=\dfrac{10}{1+9e^{-10t}}$

The population when t = 3 is

$P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}$

Using calculator,

$P=9.999\approx 10$

Therefore, the population when t = 3 is 10.

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3 years ago