Which relation is a function?

1 answer:

4 0

{(2, 3), (1, 5), (2, 7)} - NOT because 2->3 and 2->7
{(11, 9), (11, 5), (9, 3)} - NOT because 11->9 and 11->5
{(3, 8), (0, 8), (3, -2)} - NOT because 3->8 and 3->-2

{(-1, 5), (-2, 6), (-3, 7)} - YES

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Given the graph of a line y=−x. Write an equation of a line which is perpendicular and goes through the point (8,2).

coldgirl [10]

Answer:

y = x - 6

Step-by-step explanation:

The slope of y = -x is -1, so the slope of any line perpendicular to y = -x is +1. Thus,

y = mx + b becomes 2 = 1(8) + b, so that b = -6.

The desired equation is y = x - 6.

Check: Does (8,2) lie on this line? Is 2 = 8 - 6 true? YES.

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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$