Average leaf width, w (in mm), in tropical Australia¹ is a function of the average annual rainfall, r (in mm), so w = f(r). We have f'(1500) = 0.0218. (a) What are the units of 1500? The units of 1500 are Choose one (b) What are the units of the 0.0218? The units of 0.0218 are Choose one (c) About how much difference in average leaf width would you find in two forests whose average annual rainfalls are near 1500 mm but differ by 210 mm? NOTE: Round your answer to 2 decimal places. The difference in leaf width is mm.
Average leaf width, w (in mm), in tropical Australia¹ is a function of the average annual rainfall, r (in mm), so w = f(r). We have f'(1500) = 0.0218. (a) What are the units of 1500? The units of 1500 are Choose one (b) What are the units of the 0.0218? The units of 0.0218 are Choose one (c) About how much difference in average leaf width would you find in two forests whose average annual rainfalls are near 1500 mm but differ by 210 mm? NOTE: Round your answer to 2 decimal places. The difference in leaf width is mm.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Average leaf width, w (in mm), in tropical Australia¹ is a function of
= f(r). We have
=
the average annual rainfall, r (in mm), so w
ƒ'(1500) = 0.0218.
(a) What are the units of 1500?
The units of 1500 are Choose one
(b) What are the units of the 0.0218?
The units of 0.0218 are Choose one
(c) About how much difference in average leaf width would you find
in two forests whose average annual rainfalls are near 1500 mm
but differ by 210 mm?
NOTE: Round your answer to 2 decimal places.
The difference in leaf width is
mm.
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