College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.1 - Composite Functions - 6.1 Assess Your Understanding - Page 410: 68

Answer

a) $K(C(F))=\frac{5}{9}(F-32)+273$ b) 80 degrees Fahrenheit is almost 300 degrees kelvin.

Work Step by Step

a) We find K(C(F)): $K(C(F))=\frac{5}{9}(F-32)+273$ b) To convert 80 degrees Fahrenheit to Kelvin, we solve K(C(80)): $K(C(80))=\frac{5}{9}(80-32)+273=$ $\frac{5}{9}(48)+273=$ $26.\overline{66}+273\approx$ $300$ K
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