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11Alexandr11 [23.1K]
3 years ago
11

You start at (3, 0). You move down 5 units and right 1 unit. Where do you end?

Mathematics
1 answer:
nevsk [136]3 years ago
8 0

Answer:

Pretty sure it's (4, -5)

Step-by-step explanation:

If not, shoot me I guess.

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Goran is riding his bicycle. He rides for 22.5 kilometers at a speed of 9 kilometers per hour. For how many hours does he ride?
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Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1
aalyn [17]

Answer:

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First, differentiate the first time, term by term:

{3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\

3y^{2}.\frac{d^{2} y}{dx^{2}} =  - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x}{y^{5}}

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