Answer:
12
Step-by-step explanation:
For the top triangle, he has 3 options
3C1 = 3
For the middle one, he can't use the top triangle's colour, so 2 options
2C1 = 2
For the ones on sides of the middle one, 2 options each
2C1 × 2C1 = 4
3 × 2 × 2 × 2 = 24
These include all three outer ones same, so subtract 3
Also, these include 2 of the three outer ones same, so subtract:
3C2 × 3 = 9
24 - 3 - 9 = 12
According to the identity if a+b+c=0
then a3+b3+c3=3abc
a3+b3+c3/abc=3
a2*a/bc*a+b2*b/ca*b+c2*c/ab*c=3
cancel a,b,c in all the fraction then you get
<span>a²/bc+b²/ca+c²/ab=3.
</span>hence proved
Answer:
pretty good tired of the virus stuff but good
Step-by-step explanation:
The answer is a 12% increase.

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Write the divide fraction horizontally:
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Factorise the numerators and denominators when possible:
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Convert the divide fraction to multiplication fraction
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Cancel the factors
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Combine to single fraction
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