A population of 100 frogs increases at an annual rate of 22%. How many frogs will there be in 5 years? Exponential growth or decay?: Write a function that represents the situation: - Initial amount: -Growth/Decay rate: -

270 frogs

Step-by-step explanation:

Initial amount = 100

Growth / Decay rate = 122% = 1.22

y = 100 (1.22)^5

y = 100 x 2.70271

y = 270.271

Exponential growth

a = 100 initial value

b = 1.22

y = 100 ( 1.22) ^x

in 5 years= 270 frogs

Step-by-step explanation:

Exponential growth is in the form

y = ab^x  where a is the initial value and b is the growth rate

a = 100

b = 1+ 22% = 1.22

y = 100 ( 1.22) ^x

Let x = 5 years

y = 100 ( 1.22)^5

 =270.2708 frogs

Rounded to the nearest frog

270 frogs

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation:

d. you cannot tell because there are no values shown.