Answer :
To determine the difference in the number of bacteria between species [tex]\( A \)[/tex] and species [tex]\( B \)[/tex] after [tex]\( t \)[/tex] hours, we need to look at the given functions and find [tex]\( N(t) = A(t) - B(t) \)[/tex].
The given functions are:
[tex]\[ A(t) = 5 + (0.25t)^3 \][/tex]
[tex]\[ B(t) = 2 + 8(1.06)^t \][/tex]
To find [tex]\( N(t) \)[/tex], we subtract [tex]\( B(t) \)[/tex] from [tex]\( A(t) \)[/tex]:
[tex]\[ N(t) = A(t) - B(t) \][/tex]
[tex]\[ N(t) = \left( 5 + (0.25t)^3 \right) - \left( 2 + 8(1.06)^t \right) \][/tex]
Next, distribute the negative sign and combine like terms:
[tex]\[ N(t) = 5 + (0.25t)^3 - 2 - 8(1.06)^t \][/tex]
[tex]\[ N(t) = (0.25t)^3 + 5 - 2 - 8(1.06)^t \][/tex]
[tex]\[ N(t) = 3 + (0.25t)^3 - 8(1.06)^t \][/tex]
Thus, the function that describes the difference in the number of bacteria [tex]\( N(t) \)[/tex] between species [tex]\( A \)[/tex] and species [tex]\( B \)[/tex] after [tex]\( t \)[/tex] hours is:
[tex]\[ N(t) = 3 + (0.25t)^3 - 8(1.06)^t \][/tex]
Therefore, the correct answer is:
A. [tex]\( N(t) = 3 + (0.25t)^3 - 8(1.06)^t \)[/tex]
The given functions are:
[tex]\[ A(t) = 5 + (0.25t)^3 \][/tex]
[tex]\[ B(t) = 2 + 8(1.06)^t \][/tex]
To find [tex]\( N(t) \)[/tex], we subtract [tex]\( B(t) \)[/tex] from [tex]\( A(t) \)[/tex]:
[tex]\[ N(t) = A(t) - B(t) \][/tex]
[tex]\[ N(t) = \left( 5 + (0.25t)^3 \right) - \left( 2 + 8(1.06)^t \right) \][/tex]
Next, distribute the negative sign and combine like terms:
[tex]\[ N(t) = 5 + (0.25t)^3 - 2 - 8(1.06)^t \][/tex]
[tex]\[ N(t) = (0.25t)^3 + 5 - 2 - 8(1.06)^t \][/tex]
[tex]\[ N(t) = 3 + (0.25t)^3 - 8(1.06)^t \][/tex]
Thus, the function that describes the difference in the number of bacteria [tex]\( N(t) \)[/tex] between species [tex]\( A \)[/tex] and species [tex]\( B \)[/tex] after [tex]\( t \)[/tex] hours is:
[tex]\[ N(t) = 3 + (0.25t)^3 - 8(1.06)^t \][/tex]
Therefore, the correct answer is:
A. [tex]\( N(t) = 3 + (0.25t)^3 - 8(1.06)^t \)[/tex]