SOLUTION: In a company that designs pollution control equipment, the equation {{{d=k^3 -8k^2 +26k +1}}} where d is the number of thousands of dollars per month and k is the number of million

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: In a company that designs pollution control equipment, the equation {{{d=k^3 -8k^2 +26k +1}}} where d is the number of thousands of dollars per month and k is the number of million      Log On


   

Question 864732: In a company that designs pollution control equipment, the equation d=k%5E3+-8k%5E2+%2B26k+%2B1 where d is the number of thousands of dollars per month and k is the number of millions of kilowatt hours per month used in the business:
a) find the cost of 1.5 million kwh - I got 25.38 thousand
b) According to the model, how much would you pay if no electricity were used in a given month and why? -I got 1 (thousand) +1 is a constant in the equation
c) how much electricity can be used without the bill exceeding $35 000 per month? I got 2.6 (million kwh) I'm not sure if this is right.

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

d=k%5E3+-8k%5E2+%2B26k+%2B1 

a)Yes, 25.38 thousand 0r = $25,380

b)Yes, $1000

c)Using graphing calculator, 3.22 million kwh

 k = 3.22 ⇒  d = 35