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3 years ago

15

in mr.kim's history class, there are sixteen male students and twelve female students . wich of the following expresses the rati

o of female students to male students to simplest form?

2 answers:

FromTheMoon [43]3 years ago

8 0

The ratio of females to males would be 12:16, but reduced it would be 3:4

babymother [125]3 years ago

7 0

it would be: 12/16 = 6/8 = 3/4

So, 3/4 is your answer.

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Nana76 [90]

Answer:

The only point (0,0) lies inside the shaded region and hence it gives a solution for the set of inequalities.

Step-by-step explanation:

See the graph attached to this question.

The solution of the set of inequalities is given by the shaded region on the graph.

Now, the point (0,5) is outside this shaded region, hence it can not be the solution.

The point (3,0) also is outside this shaded region, hence it can not be the solution.

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aliya0001 [1]

Answer:

y = -5/2x - 13

Step-by-step explanation:

First, we have to find the slope of the original equation

Subtract 5x from both sides

5x + 2y = 1

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2y = -5x + 1

Divide both sides by 2

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y = -5/2x + 1/2

The slope of this equation is -5/2, so -5/2 has to be in our new parallel equation

y = -5/2x + b

Plug in the given x and y

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Subtract 10 from both sides

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- 10 - 10

b = -13

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3 years ago

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:

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

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

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3 years ago