Find an equation of variation in which y varies directly as x and y = 12 when x = 6

1 answer:

8 0

Direct variation:-

y = kx where k is a constant

y=12 when x=6 so we have:-

12 = k*6

k = 12/6 = 2

so our equation of variation is y = 2x Answer

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I don't need you to work it out, just theorem(s) i need to reach the answer :)

VikaD [51]

Not sure why such an old question is showing up on my feed...

Anyway, let $x=\tan^{-1}\dfrac43$ and $y=\sin^{-1}\dfrac35$ . Then we want to find the exact value of $\cos(x-y)$ .

Use the angle difference identity:

$\cos(x-y)=\cos x\cos y+\sin x\sin y$

and right away we find $\sin y=\dfrac35$ . By the Pythagorean theorem, we also find $\cos y=\dfrac45$ . (Actually, this could potentially be negative, but let's assume all angles are in the first quadrant for convenience.)

Meanwhile, if $\tan x=\dfrac43$ , then (by Pythagorean theorem) $\sec x=\dfrac53$ , so $\cos x=\dfrac35$ . And from this, $\sin x=\dfrac45$ .

So,

$\cos\left(\tan^{-1}\dfrac43-\sin^{-1}\dfrac35\right)=\dfrac35\cdot\dfrac45+\dfrac45\cdot\dfrac35=\dfrac{24}{25}$

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If 2a+ 2/a =4, what is the value of 2a^2− 2/a^2 ?

aliina [53]

Answer:

Step-by-step explanation:

2a+ 2/a =4

Multiply each side by a to clear the fraction

a( 2a+ 2/a) =4*a

Distribute

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Subtract 4a from each side