A population of bacteria P is changing at a rate of dP/dt = 3000/1+0.25t where t is the time in days. The initial population (when t=0) is 1000. Write an equation that gives the population at any time t. Then find the population when t = 3 days.

- At any time t, the population is:

P = 375t² + 3000t + 1000

- At time t = 3 days, the population is:

P = 13,375

Step-by-step explanation:

Given the rate of change of the population of bacteria as:

dP/dt = 3000/(1 + 0.25t)

we need to rewrite the given differential equation, and solve.

Rewriting, we have:

dP/3000 = (1 + 0.25t)dt

Integrating both sides, we have

P/3000 = t + (0.25/2)t² + C

P/3000 = t + 0.125t² + C

When t = 0, P = 1000

So,

1000/3000 = C

C = 1/3

Therefore, at any time t, the population is:

P/3000 = 0.125t² + t + 1/3

P = 375t² + 3000t + 1000

At time t = 3 days, the population is :

P = 375(3²) + 3000(3) + 1000

= 3375 + 9000 + 1000

P = 13,375

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i ran through it by the other answer is wrong

answer is 8x+24

area of vanessa's patio = length × width

given, length = (3x² + 5x + 10)

width = (x² - 3x - 1)

area of vanessa's patio = (3x² + 5x + 10)(x² - 3x - 1)

= 3x²(x² - 3x - 1) + 5x(x² - 3x - 1) + 10(x² - 3x - 1)

= 3x⁴ - 9x³ - 3x² + 5x³ - 15x² - 5x + 10x² - 30x - 10

= 3x⁴ - 4x³ - 8x² - 35x - 10

hence, expression of vanessa's patio = (3x⁴ - 4x³ - 8x² - 35x - 10)