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3 years ago

If A + B = 2X -2 and A -B equals 4X -8, find A and B

1 answer:

8 0

A=3X-3
B=-X+3
If A+B=2X-2, then using equalivant exchange A=2X-2-B and
B=2X-2-A
If A-B=4X-8, then A=4X-8+B and -B=4X-8-A(or B=-4X+8+A)
Since A equals both 2x-2-B and 4X-8+B we can add both equations to equal 2A=6X-6 now we devide that by 2 to get
A=3X-3
Do the exact same for B, so b equals both 2X-2-A and -4X+8+A
We can add them both to get 2B=-2X+6 and decide that by 2 to get B=-X+3

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Step-by-step explanation:

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Therefore,

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Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

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$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

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