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geniusboy [140]
2 years ago
7

Which set of fractions is NOT an example of reciprocal fractions? Group of answer

Mathematics
1 answer:
Alex2 years ago
7 0

It can be deduced that the set of fractions that is an example of reciprocal fractions include, 1/2 , 3/1, and 1/6.

<h3>What is a reciprocal fraction?</h3>

It should be noted that a reciprocal fraction simply means the number whose product with the given fraction is 1.

Here, the set of fractions that is an example of reciprocal fractions include, 1/2 , 3/1, and 1/6. This is because their product isn't equal to 1.

Learn more about fractions on:

brainly.com/question/78672

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Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0
andreev551 [17]

Recall the angle sum identity for cosine:

cos(<em>x</em> + <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)

cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) + sin(<em>x</em>) sin(<em>y</em>)

==>   sin(<em>x</em>) sin(<em>y</em>) = 1/2 (cos(<em>x</em> - <em>y</em>) - cos(<em>x</em> + <em>y</em>))

Then rewrite the equation as

sin(4<em>x</em>) sin(5<em>x</em>) + sin(4<em>x</em>) sin(3<em>x</em>) - sin(2<em>x</em>) sin(<em>x</em>) = 0

1/2 (cos(-<em>x</em>) - cos(9<em>x</em>)) + 1/2 (cos(<em>x</em>) - cos(7<em>x</em>)) - 1/2 (cos(<em>x</em>) - cos(3<em>x</em>)) = 0

1/2 (cos(9<em>x</em>) - cos(<em>x</em>)) + 1/2 (cos(7<em>x</em>) - cos(3<em>x</em>)) = 0

sin(5<em>x</em>) sin(-4<em>x</em>) + sin(5<em>x</em>) sin(-2<em>x</em>) = 0

-sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

Recall the double angle identity for sine:

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Rewrite the equation again as

sin(5<em>x</em>) (2 sin(2<em>x</em>) cos(2<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) sin(2<em>x</em>) (2 cos(2<em>x</em>) + 1) = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   2 cos(2<em>x</em>) + 1 = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   cos(2<em>x</em>) = -1/2

sin(5<em>x</em>) = 0   ==>   5<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   5<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   5<em>x</em> = 2<em>nπ</em>   <u>or</u>   5<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = 2<em>nπ</em>/5   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/5

sin(2<em>x</em>) = 0   ==>   2<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   2<em>x</em> = 2<em>nπ</em>   <u>or</u>   2<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = <em>nπ</em>   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/2

cos(2<em>x</em>) = -1/2   ==>   2<em>x</em> = arccos(-1/2) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -arccos(-1/2) + 2<em>nπ</em>

… … … … … …    ==>   2<em>x</em> = 2<em>π</em>/3 + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -2<em>π</em>/3 + 2<em>nπ</em>

… … … … … …    ==>   <em>x</em> = <em>π</em>/3 + <em>nπ</em>   <u>or</u>   <em>x</em> = -<em>π</em>/3 + <em>nπ</em>

<em />

(where <em>n</em> is any integer)

5 0
3 years ago
Find the AOS for y = x^2 – 2x – 15
aivan3 [116]

Answer:

Step-by-step explanation:

x=1

6 0
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