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victus00 [196]
3 years ago
7

Which inequality correctly compares One-third, Five-sixths, and Three-fifths? One-third < Three-fifths < Five-sixths One-t

hird < Five-sixths < Three-fifths Five-sixths < One-third < Three-fifths Five-sixths < Three-fifths < One-third
Mathematics
1 answer:
lilavasa [31]3 years ago
5 0

Answer:

\frac{1}{3}   \: <  \:  \frac{3}{5}   \: <  \:  \frac{5}{6}

Step-by-step explanation:

We want to compare the fractions

\frac{1}{3}, \frac{5}{6} , \frac{3}{5}

First, we collect LCM

The LCM of 3,6, and 5 is 30

\frac{1}{3} =  \frac{10}{30} ,  \\  \frac{5}{6} =  \frac{25}{30} , \\   \frac{3}{5}  = \frac{18}{30}

\frac{10}{30} , \frac{18}{30} ,  \frac{25}{30}

This means that

\frac{1}{3} , \frac{3}{5} ,  \frac{5}{6}

Hence,

\frac{1}{3}   \: <  \:  \frac{3}{5}   \: <  \:  \frac{5}{6}

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Step-by-step explanation:

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Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A
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Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
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D = cos7A + cosA - (cos5A + cos3A)
   = 2cos4A*cos3A - 2cos4A*cosA
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Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
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Step-by-step explanation:

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2 years ago
Help me asap please, very much appreciated :)​
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Answer:

155°

Step-by-step explanation:

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Hope it will help :)

8 0
2 years ago
Read 2 more answers
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