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Ugo [173]
3 years ago
6

Does anybody know how to solve this equation: 5+4x-7=4x-2-x?

Mathematics
2 answers:
GrogVix [38]3 years ago
5 0

5 + 4x - 7 = 4x - 2 - x

5 - 7 = -2 -x

-2 = -2 - x

0 = -x

-x = 0

x = 0

kakasveta [241]3 years ago
4 0

5+4x-7=4x-2-x -2+4x=4x-2-x 4x-4x-x=-2+2 -x=0 X=0   To check 5+4(0)-7=4(0)-2-0 5-7=-2 -2=-2

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adoni [48]

Answer: The required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

Step-by-step explanation:

Since we have given that

y=\ln[x(2x+3)^2]

Differentiating log function w.r.t. x, we get that

\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}

Hence, the required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

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Step-by-step explanation:

6 0
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