Respuesta :
The number of bacteria in species A at time t is given by: [tex]A(t)=5+(.25t) ^{3} [/tex].
The number of bacteria in species B at time t is given by [tex]B(t)=2+8(1.06)t[/tex]
You are asked to find N(t) which is the difference in the species at time t. Difference refers to the answer is a subtraction problem. Thus, we are asked to find N(t) = A(t)-B(t)
We subtract and obtain: [tex]N(t)=5+(.25t) ^{3}-(2+8(1.06)t )=5-2+(.25t) ^{3}-(8(1.06)t )[/tex]
That is, [tex]N(t)=3+(.25t) ^{3}-(8(1.06)t )[/tex] which is the last answer choice.
With respect to notation, it is common to denote an exponent using ^. So if we want to write [tex](.25t) ^{3} [/tex] we can write "(.25t)^3"
The number of bacteria in species B at time t is given by [tex]B(t)=2+8(1.06)t[/tex]
You are asked to find N(t) which is the difference in the species at time t. Difference refers to the answer is a subtraction problem. Thus, we are asked to find N(t) = A(t)-B(t)
We subtract and obtain: [tex]N(t)=5+(.25t) ^{3}-(2+8(1.06)t )=5-2+(.25t) ^{3}-(8(1.06)t )[/tex]
That is, [tex]N(t)=3+(.25t) ^{3}-(8(1.06)t )[/tex] which is the last answer choice.
With respect to notation, it is common to denote an exponent using ^. So if we want to write [tex](.25t) ^{3} [/tex] we can write "(.25t)^3"