The table shows data for sulfur dioxide emissions from electricity generation (in millions of short tons) for selected years from 2000 and projected to 2035. These data can be modeled bythe functionE(x) = (0.001x - 0.062)(-0.18x2 + 8.2x - 200)where x is the number of years past 2000.Short TonsYear(in millions)200011.4200510.220087.620154.720204.220253.820303.720353.8(a) Find the function that models the rate of change of these emissions. (Use exact numerical values. Do not round.)E'(X) = (0.001x – 0.062)[2(-0.18)x+ 8.2] + (–0.18x² + 8.2x –(b) Find E'(20). (Give an exact answer. Do not round.)E'(20) = 0Interpret E'(20).O In 2020, emissions from electricity generation will be increasing.O In 2020, emissions from electricity generation will be constant.O In 2020, emissions from electricity generation will be decreasing.

Answered: The table shows data for sulfur dioxide…

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter6: Vector Spaces

Section6.7: Applications

Problem 16EQ

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The table shows data for sulfur dioxide emissions from electricity generation (in millions of short tons) for selected years from 2000 and projected to 2035. These data can be modeled by
the function
E(x) = (0.001x - 0.062)(-0.18x2 + 8.2x - 200)
where x is the number of years past 2000.
Short Tons
Year
(in millions)
2000
11.4
2005
10.2
2008
7.6
2015
4.7
2020
4.2
2025
3.8
2030
3.7
2035
3.8
(a) Find the function that models the rate of change of these emissions. (Use exact numerical values. Do not round.)
E'(X) = (0.001x – 0.062)[2(-0.18)x+ 8.2] + (–0.18x² + 8.2x –
(b) Find E'(20). (Give an exact answer. Do not round.)
E'(20) = 0
Interpret E'(20).
O In 2020, emissions from electricity generation will be increasing.
O In 2020, emissions from electricity generation will be constant.
O In 2020, emissions from electricity generation will be decreasing.

Expert Solution

Step 1

(b) To find $E' (20)$ , substitute $x = 20$ in $E' (x) = (0.001 x - 0.062) [2 (- 0.18) x + 8.2] + (- 0.18 x^{2} + 8.2 x - 200) (0.001)$ .

$\begin{array}{rcl} E' (20) & = & (0.001 (20) - 0.062) [2 (- 0.18) (20) + 8.2] + (- 0.18 {(20)}^{2} + 8.2 (20) - 200) (0.001) \\ = & (0.02 - 0.062) (- 7.2 + 8.2) + (- 72 + 164 - 200) (0.001) \\ = & (- 0.042) (1) + (- 108) (0.001) \end{array}$

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