The fish population of lake collins is decreasing at a rate of 3% per year. in 2004 there were about 1,300 fish. write an exponential decay function to model this situation. then find the population in 2010.

1083 fish

Step-by-step explanation:

The fish population of Lake Collins is decreasing at a rate of 3% per year.

If starting population were about 1,300 in 2004 so the sequence will be

1,300, (1300 - 3% of 1300)..............

or 1300, 1261,.........

We know exponential function is modeled by

Where         = population at time t.

                  = initial population

                   r = common ratio

                   n = time in years.

Common ratio (r) =

                            =

Then the function will be

Then population in year 2010 (after 6 years) will be

   = 1300(0.8329)

  = 1082.86 ≈ 1083 fishes

correct answer is infinitely many solutions

step-by-step explanation:

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

solving the above equation:

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

combining like terms

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

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1