Aspecies of extremely rare, deep water fish rarely have children. if there are 821 of this type of fish and their growth rate is 2% each month, how many will there be in half of a years, in 2 years, and in 10 years?

There are 925 in half of a year

There are 1321 in 2 years

There are 8838 in 10 years

Step-by-step explanation:

* Lets revise the exponential function

- The original exponential formula was y = ab^x, where a is the initial

 amount and b is the growth factor

- The new growth and decay functions is y = a(1 ± r)^x. , the b value

 (growth factor) has been replaced either by (1 + r) or by (1 - r).

- The growth rate r is determined as b = 1 + r

* Lets solve the problem

∵ The number of fish is growth every month

∴ We will use the growth equation y = a(1 + r)^x, where a is the initial

  amount of the fish, r is the rate of growth every month and x is the

  number of months

- There are 821 of a type of fish

∴ The initial amount is 821 fish

∴ a = 821

- Their growth rate is 2% each month

∴ The rate of growth is 2% per month

∴ r = 2/100 = 0.02

- We want to find how many of them be in half year

∵ There are 6 months in half year

∴ y = 821(1 + 0.02)^6

∴ y = 821(1.02)^6 = 924.58 ≅ 925

* There are 925 in half of a year

∵ There are 24 months in 2 years

∴ y = 821(1 + 0.02)^24

∴ y = 821(1.02)^24 = 1320.53 ≅ 1321

* There are 1321 in 2 years

∵ There are 120 months in 10 years

∴ y = 821(1 + 0.02)^120

∴ y = 821(1.02)^120 = 8838.20 ≅ 8838

* There are 8838 in 10 years

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: