2. 2.
Which of the following functions would give the population Plt) as a function
of t, in years, given that the initial population is 2,500 and that the population
is growing at an annual rate of 4%?
(1) P(t)=2500(0.96)
(3) P(t)=2,500(1.04)
(2) P(t)=(1.04) +2,500
(4) P(t)= 1.04t +2,500

Step-by-step explanation:

None of your proposed answers has an exponent t.

Here's the logic behind the

Year 0: pop. = 2500

Year 1:  pop. = 2500(1.04)  That is 104% of the previous year's pop.

Year 2: pop. = 2500(1.04)(1.04) = 2500(1.04)^2

Year 3: pop. = 2500(1.04)^2 x (1.04) = 2500(1.04)^3

See the pattern?  The exponent matches the year number.

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.