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Westkost [7]
3 years ago
8

Which relation is a function? {(3,6), (-2,4), (6,4), (8.2); (3.5), (-48), (3, 2), (7-2)}

Mathematics
1 answer:
frutty [35]3 years ago
5 0

Answer:

The second one

Step-by-step explanation:

For every input, there is one output.

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The vertical line test can be used to determine whether a graph represents a function. ... If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that x value has more than one output. A function has only one output value for each input value.

Step-by-step explanation:

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\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right)  (xy+1)^2 = 0
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\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
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\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
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\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
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\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y


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