A population of bacteria can be modeled by the function f(t)= 400(0.98)t, where t is time in hours. What is the rate of change in the function?

the rate of change is f'(t) = (-8.08)*( 0.98^t )

Step-by-step explanation:

If the function is

f(t)= 400* (0.98)^t

then the rate of change of f(t) , that is f'(t)=df(t)/dt is

f'(t) = 400* d ( 0.98^t ) / dt

since the derivative of a function of the type g(x) = a^x is g'(x)= ln(a) * a^x , then

f'(t) = 400* d ( 0.98^t ) / dt = 400*ln (0.98) * ( 0.98^t )

f'(t) = (-8.08)*( 0.98^t )

Step-by-step explanation:

decrease by 2% per hour

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )