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rjkz [21]
3 years ago
12

2c(b+15c)+(b–6c)(5c+2b)

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
7 0

Answer: 2b^2 - 5bc

Work is shown below

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If the point (-1,-5) is reflected across the x - axis, what is the new location of the point
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Unknown g (x) = ax + b and g (g (x)) = 16x - 15. Calculate the value of a and b.
goblinko [34]
G(g(x))=16x-15
means that when you subsituted g(x) for x in g(x), you got 16x-15

a(ax+b)+b=16x-15
distribute
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4 years ago
You have ​$11.40 to buy drink boxes for the football team. Each drink box costs ​$0.40. How many drink boxes can you​ buy? Shoul
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Step-by-step explanation:

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Let's prove

\boxed{\sf 1+tan^2\theta(1-sin^2\theta)=1}

LHS

\\ \sf\longmapsto 1+tan^2\theta(1-sin^2\theta)

\\ \sf\longmapsto 1+\dfrac{sin^2\theta}{cos^2\theta}(1-sin^2\theta)

\\ \sf\longmapsto \dfrac{cos^2\theta+sin^2\theta}{cos^2\theta}(1-sin^2\theta)

\\ \sf\longmapsto \dfrac{1}{cos^2\theta}(1-sin^2\theta)

\\ \sf\longmapsto \dfrac{1-sin^2\theta}{cos^2\theta}

\\ \sf\longmapsto \dfrac{1-sin^2\theta}{1-sin^2\theta}

\\ \sf\longmapsto 1

Hence provee

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