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kvv77 [185]
3 years ago
13

Triangle ABC and DEF are similar what is the length of segment DF?

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
5 0

Answer:

Proportionate to whatever length the corresponding segment for the similar triangle.

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Factor: 8ab - 16ac + 12a
aksik [14]
The GCF we factor out is 4a

8ab = 4a*2b
16ac =  4a*4c
12a = 4a*3

Each term has been factored so that 4a is a factor

Using those factorizations, we can use the distributive property in reverse
8ab - 16ac + 12a = 4a*2b - 4a*4c + 4a*3
8ab - 16ac + 12a  = 4a*(2b - 4c + 3)

Answer is choice A<span />
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3 years ago
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Prove that<br>{(tanθ+sinθ)^2-(tanθ-sinθ)^2}^2 =16(tanθ+sinθ)(tanθ-sinθ)
USPshnik [31]

First, expand the terms inside the bracket you will get

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( 4 \tan(x)  \sin(x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x)  \sin {}^{2} (x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x) (1 -  \cos {}^{2} (x) ) = 16 (\tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan {}^{2} (x)  -   \frac{  \sin {}^{2} (x) \cos {}^{2} ( {x}^{} )  }{ \cos {}^{2} (x) }

16( \tan {}^{2} (x)  -  \sin {}^{2} (x) ) = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

5 0
2 years ago
Lilyworks for a florist. She worked 20hours last week and earned $200.00. At that rate, how much will she earn if she works for
aniked [119]

Answer:

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Step-by-step explanation:

8 0
4 years ago
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It will take 12*648/864=9 hours
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3 years ago
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Answer:

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