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Shkiper50 [21]
3 years ago
11

If the expression 2x/a plus b = c is solved for x in terms of a, b, and c, then x = what?

Mathematics
1 answer:
Doss [256]3 years ago
3 0

Answer:

The value of x in terms of a, b ans c is x=\frac{ac-ab}{2}

Step-by-step explanation:

We have given the expression \frac{2x}{a}+b=c

We have to find the value of x in terms of a, b and c

So \frac{2x}{a}+b=c

\frac{2x}{a}=c-b

2x=a(c-b)

2x=ac-ab

x=\frac{ac-ab}{2}

So the value of x in terms a ,b and c is x=\frac{ac-ab}{2}

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Novay_Z [31]

Answer:

$ \frac{\sqrt{3} - 1}{2\sqrt{2}} $

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Step-by-step explanation:

Given $ \frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6} $

(A) $ sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4}  + \frac{\pi}{6}) $

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$ sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2} $

$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}

(B) Cos(A + B) = CosAcosB - SinASinB

$ cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}}) $

$ \implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2} $

$ \implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}) $

$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}

(C) Tan(A + B) = $ \frac{Sin(A +B)}{Cos(A + B)} $

From the above obtained values this can be calculated and the value is $ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $.

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