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andrezito [222]
3 years ago
7

Please help me 5x4+37-25(5x26)=??????????/

Mathematics
2 answers:
Sav [38]3 years ago
7 0

Answer:

The answer is -3193

Step-by-step explanation:

5x4+37-25(5x26)=20+37-25(130)= -25(130)+57= -3193

can you make me brainliest please

Nimfa-mama [501]3 years ago
3 0

Answer:

-3193

Step-by-step explanation:

20+37-25(130)

-25(130)+57

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Read 2 more answers
Let P(t) be a population at time t. A simple population model supposes the rate of growth of the population is proportional to t
jeka94

Answer:

(a) \dfrac{dP}{dt} =k P(t)\\(b)P(t)=Ce^{kt}

(c)P(10)\approx 272

(ii)P(1000)\approx 26.88 \times 10^{44}\\

Step-by-step explanation:

(a)The rate of growth of the population is proportional to the population, this is written as:

\dfrac{dP}{dt} \propto P(t)\\$Introducing our proportionality constant, k\\ \dfrac{dP}{dt} =k P(t)

(b)

\dfrac{dP(t)}{P(t)} =k dt\\$Take the integral of both sides\\\int \dfrac{dP(t)}{P(t)} =\int k dt\\\ln P(t)=kt+C, $C a constant of integration\\Take the exponential of both sides\\e^{\ln P(t)}=e^{kt+C}\\P(t)=e^{kt}\cdot e^C  $, (Since e^C$ is a constant, we then have:)\\P(t)=Ce^{kt}

(c)

Suppose the net birthrate of the population is .1, and the initial population is 100.

k=0.1

P(0)=100

Substitution into P(t) gives:

100=Ce^{kX0}

C=100

Therefore:

P(t)=100e^{0.1t}

(i)When t=10

P(10)=100e^{0.1 \times 10}\\=271.8\\\approx 272

(ii)When t=1000

P(1000)=100e^{0.1 \times 1000}\\=26.88 \times 10^{44}\\

8 0
4 years ago
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