The annual consumption of beef per person was about 64.8 lb in 2000 and about 60.1 lb in 2006. Assume B(t), the annual beef consumption t years after 2000, is decreasing according to the exponential decay model. a) Find the value of k, and write the equation b) Estimate the consumption of beef in 2011 c) In what year (theoretically) will the consumption of beef be 10 lb?

a) B(t)= 64.8e^(-0.01255t), k=-0.0125 b) 56.5 c)2150

Step-by-step explanation:

a) B(t)= Ae^(-kt)

at t=0, B(t)=64.8

A=64.8

at t=6, B(t)=60.1

60.1=64.8e^(-6k)

k=0.0125

b) B(t)=64.8e^(-0.0125×11)

B(t)= 56.5

c) 10=64.8e^(-0.0125t)

0.15432=e^(-0.0125t)

-0.0125t=ln(0.15432)

-0.0125t=-1.869

t=149.5 or 150

Year= 2150

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places

your answer   is c

step-by-step explanation: