Drag the tiles to the boxes to form correct pairs. not all tiles will be used. match each situation to its corresponding expression. there are 7 trout fish in a pond, and the population doubles every year. find the population after t years. arrowboth a company buys a machine for $3,000. the value of the machine depreciates by 7% every year. find the value of the machine after t years. arrowboth the initial population of a colony of ants is 300. the number of ants increases at a rate of 1.5% every month. find the population of ants after t months. arrowboth a research laboratory is testing a new vaccine on 300 infected cells. the decay rate is 1.5% per minute. find the number of infected cells after t minutes. arrowboth

Step-by-step explanation:

We will use the pattern f(x)= a(b)^t where a is the initial value, b is the base of the exponent. All these questions are about exponent function

A) Number of trout fish in the pound = 7 , it means a =7

population increases double every year. It means b=2

f(x)= a(b)^t

f(x)=7(2)^t

B) Cost of machine = $3000

The value depreciated every year = 7%

It means 100%-7%= 93% which is equal to 0.93

Therefore,

a = 3000

b = 0.93

f(x)= a(b)^t

f(x)=3000(0.93)^t

C) Initial population of a colony of ants = 300

The number of ants increase at a rate of 1.5%

It means 100%+1.5%=101.5%

101.5% = 1.015

Therefore,

a= 300

b = 1.015

f(x)= a(b)^t

f(x)=300(1.015)^t

D) A research laboratory is testing a new vaccine on 300 infected cells

The decay rate is 1.5% per minute

It means 100%-1.5% =98.5%

98.5% = 0.985

Therefore,

a = 300

b = 0.985

f(x)= a(b)^t  

f(x)= 300(0.985)^t

answer: haha (: i’ll go see what it is

step-by-step explanation:

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?