The number of a particular fish in thousands x years after 1974 can be modeled by f(x)=233(0.885)^x. estimate the year when the number of these fish reached 90 thousand. the number of these fish reached 90 thousand in the year?

-51.4

Step-by-step explanation:

The solution for this problem is obtained by equating this equation to 90,000 and getting the value of x:

233(0.885)^x = 90,000

This x is the value of years that get you the value 90,000.

The first you need to get X is divide both sides between 233 so you can eliminate the 233 on the left side:

[233(0.885)^x] / 233 = 90,000 / 233

0.885^x = 386.27

Then, if you apply logarithm to both sides, the value of x "comes down" as Ln(a^b)=b*Ln(a) for every a and b

So:

x Ln (0.885) = Ln (386.27)

x (-0.05) = 2.59

Dividind by -0.05 in both sides:

x (-0.05) / (-0.05) = 2.59 / (-0.05)

x = -51.8 And this is your solution.

The problem here is your x measures thousands x years after 1974, and a negative value would imply years previous to 1974. Here the answer is:

51.4 thousand years BEFORE 1974

your answer is  

a.

5

step-by-step explanation:

6-3=3

3*3=9

9+12=21

21/3=7

7*8=56

56 is your answer