In a study by the texas a & m university, found there were 23,000 prairie dogs in texas in 2000. by 2010, the number has increased to 40,000. the exponential growth function d=2.3e^kt describes the prairie dog population, d in ten thousands, t years after 2000.
a) solve for k
b)by this model how many prairie dogs existed last year?
c)in which year will the population double?
!

let x=k

D=dog

T=Taxes

I=increase P=population

solve k

d=23,000

t=2000

d=2.3e^

divide both sides by 2

after year is,2000+k2-2.3

=1.999.7

k2- 23,000=

k-x

or

k2-2

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions