City A had a population of 10000 in the year 1990. City A’s population grows at a constant rate of 3% per year. City B has a population that is growing exponentially. In the year 2000, there were half as many people in B as in A. In the year 2010, the population of A was 20% more than the population of B. When will the populations be equal? Give your answer in years after 1990.

City A and city B will have equal population 25years after 1990

Step-by-step explanation:

Given

Let

years after 1990

population function of city A

population function of city B

City A

---- initial population (1990)

--- rate

City B

----- t = 10 in 2000

---- t = 20 in 2010

Required

When they will have the same population

Both functions follow exponential function.

So, we have:

Calculate the population of city A in 2000 (t = 10)

Calculate the population of city A in 2010 (t = 20)

From the question, we have:

 and  

Solve for B20

and can be used to determine the function of city B

For:

We have:

For:

We have:

Divide by

Apply law of indices

--- (1)

Take 10th root of both sides

Subtract 1 from both sides

To calculate , we have:

Recall that:

So:

Hence:

The question requires that we solve for t when:

Where:

and

becomes

Divide both sides by 10000

Divide both sides by

Take natural logarithm of both sides

Rewrite as:

Solve for t

Approximate

the answer is (-4,-4)

hope this ! : d

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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