Suppose the fish population in a local lake increases at a rate proportional to the population each moment. there were 1000 fish 5 years ago and 2000 fish 4 years ago. How many fish are there right now?

6000

Step-by-step explanation:

If the fish population in a local lake increases at a rate proportional to the population each moment, this can be expressed as;

P = k/t (the scenario is an inverse proportionality because as the time is reducing, the population is increasing by the same factor)

If there are 1000 fish 5 years ago and there are 2000 fish 4 years ago

Yearly percentage increase will be;

2000-1000/1000 × 100%

1000/1000×100%

= 100%

This means that the population increases by 100% on a yearly basis (by a factor of 1000)

The population 3 years ago will be population

P = 2000+1000

P = 3000

The population 2 years ago will be population

P = 3000+1000

P = 4000

The population 1 year ago will be population

P = 4000+1000

P = 5000

The population now will be population

P = 5000+1000

P = 6000 populations

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms

3/5

step-by-step explanation:

double number line is not a standard concept, so it must have been defined in your textbook. assume top line is numerator.