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3 years ago

15

What’s the answer ?

1 answer:

Ivan3 years ago

8 0

75 is the answer to this equation

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alukav5142 [94]

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3 years ago

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yulyashka [42]

Hello,

$L=\sqrt{x^2+y^2} \\\\\dfrac{dx}{dt}=-4\\\\\dfrac{dy}{dt}=3\\\\x=4, y=3\\\\\dfrac{\partial L} {\partial x} =\dfrac{x}{\sqrt{x^2+y^2}} \\\\\dfrac{\partial L} {\partial y} =\dfrac{y}{\sqrt{x^2+y^2}} \\\\\dfrac{dL}{dt} =\dfrac{\partial L} {\partial x}*\dfrac{dx}{dt}+\dfrac{\partial L} {\partial y}*\dfrac{dy}{dt}\\\\=-4*\frac{-4}{\sqrt{4^2+3^2}} +3*\frac{3}{\sqrt{4^2+3^2}}\\\\=\dfrac{16}{5} +\dfrac{9}{5} \\\\=5\\$

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3 years ago

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Gre4nikov [31]

Answer:2

Step-by-step explanation:

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3 years ago