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CaHeK987 [17]
3 years ago
12

PLEASE HELP, THANK YOU! :)

Mathematics
1 answer:
elena-s [515]3 years ago
8 0

Steps to solve:

-4z + 4 + 12z = 3z + 4 + 5z

~Combine like terms

8z + 4 = 8z + 4

~Subtract 4 to both sides

8z = 8z

~Divide 8 to both sides

0 = 0

All real numbers are solutions.

Set builder notation: {x | R}

This is an identity since every value can be a solution to the equation.

Best of Luck!

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S_A_V [24]
Taking y=y(x) and differentiating both sides with respect to x yields

\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0

Solving for the first derivative, we have

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y

Differentiating again gives

\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0

Solving for the second derivative, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}

Now, when x=1 and y=2, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63
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Answer:

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Step-by-step explanation:

here's the solution : -

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Answer:

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Step-by-step explanation:

I think this is right, good luck!

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