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Nikolay [14]
3 years ago
5

Due to changing wind conditions, a pilot is able to fly 150 mph for the first third of the flying time of a trip, 175 mph for th

e middle third of the trip, and 190 mph for the final third of the trip. The average ground speed for the trip was ___ mph. Round your answer to the nearest whole number.
Mathematics
2 answers:
yan [13]3 years ago
8 0

Answer: Hello there! The average MPH that the pilot was able to fly the plane at was 171.66666666667 mph. If you round it to the nearest whole number, it equals out to 172 mph. This is 100% Verified! :) Please mark me as Brainliest! Thank you! :)

Step-by-step explanation:

Tema [17]3 years ago
3 0

Answer:

172 mph

Step-by-step explanation:

You're welcome :)

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garik1379 [7]

(\frac{1}{1-cos(x)})-(\frac{cos(x)}{1+cos(x)})

cscx is  1/sinx

maybe they want us to use pythagorean identity

cos^2(x)+sin^2(x)=1

I notice we have 1-cos(x) and 1+cos(x), if we multiply them, we get 1-cos^2(x)

and if we look at the pythagorean identity and minus cos^2(x) from both sides, we get

sin^2(x)=1-cos^2(x)

since csc(x)=\frac{1}{sin(x)}, csc^2(x)=\frac{1}{sin^2(x)}=\frac{1}{1-cos^2(x)}

(recall that (a-b)(a+b)=a²-b²)

match the denomenators of the original fraction

multiply first fraction by \frac{1+cos(x)}{1+cos(x)} and the 2nd by \frac{1-cos(x)}{1-cos(x)}


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

((csc^2(x))(1+cos(x)))-((csc^2(x))(cos(x)-cos^2(x)))=

csc^2(x)+csc^2(x)cos(x)-(csc^2(x)cos(x)-csc^2(x)cos^2(x)=

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csc^2(x)+csc^2(x)cos^2(x)=

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look to the pythagorean identity again


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1+cos^2(x)=2-\frac{1}{csc^2(x)}

subsituting


csc^2(x)(1+cos^2(x))=

(csc^2(x))(2-\frac{1}{csc^2(x)})= distributing

2csc^2(x)-\frac{csc^2(x)}{csc^2(x)}=

2csc^2(x)-1 is the simplified expression

8 0
3 years ago
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