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kramer
3 years ago
13

.164 round to the nearest tenth

Mathematics
1 answer:
timurjin [86]3 years ago
8 0

Answer:

.2

Step-by-step explanation:

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Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1
aalyn [17]

Answer:

y³ + x³ = 1

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}}

3 0
3 years ago
Approximate the unknown side x in the diagram. <br> A. 3.356<br> B. 5.222<br> C. 3.064<br> D. 4.737
Marianna [84]

Answer:

The answer to your question is x = 3.356 cm

Step-by-step explanation:

This is a right triangle so we can use the Trigonometric functions to solve this problem.

Data

Ф = 40°

Adjacent side = 4 cm

Opposite side = x

Process

1.- We must know the lenght of the opposite side so we must use the tangent to solve it.

    tangent Ф = \frac{Opposite side}{Adjacent side}

2.- Solve it for Opposite side

    Opposite side = tanФ x Adjacent side

     Opposite side = tan 40° x 4

     Opposite side = 0.839 x 4

     Opposite side = 3.356

     Opposite side = x = 3.356 cm

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