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Inessa [10]
3 years ago
11

don walked 3 3/5 miles on Friday, 3.7 miles on saturday, and 3 5/8 miles on sunday. list the distances from least to greatest

Mathematics
1 answer:
astraxan [27]3 years ago
8 0

Solution:

we are given that

Don walked 3 3/5 miles on Friday

It can be re-written as

3\frac{3}{5}miles=\frac{18}{5}= 3.6miles

3 5/8 miles on sunday.

It can be re-written as

3\frac{5}{8}miles=\frac{29}{8}= 3.625miles

an d He walk 3.7 miles on satarday.

The distances from least to greatest is  3.6, 3.625, 3.7

or

The distances from least to greatest is  3\frac{3}{5}miles, 3\frac{5}{8}miles, 3.7 miles.

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This is one pathway to prove the identity.

Part 1

\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 2

\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 3

\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\

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We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity \sin^2(\theta)+\cos^2(\theta) = 1 in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

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