An initial population of 460 quail increases at an annual rate of 21%. Write an exponential function to model the quail population. What will the approximate population be after 3 years?

the expected population after 3 years is 863.7009

Step-by-step explanation:

The computation of the expected population after 3 years is shown below

P = P_0 × e^rt

where,

r = 21% = 0.21

t = 3 years

P_0 = 460

Now place these values to the above formula

P = 460 × exp(0.21 × 3)

= 863.7009

hence, the expected population after 3 years is 863.7009

We simply applied the above formula so that the correct value could come

And, the same is to be considered

y=x+9

?

step-by-step explanation:

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places