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3 years ago

Which value of k will make this relationship NOT a function? {(−7, 2), (−1, 0), (4, 8), (k, 0)}

Mathematics

1 answer:

ruslelena [56]3 years ago

4 0

Answer:

In this example, if k = -7, this relationship is not a function, since x = -7 would have both y = 2 and y = 0.

Step-by-step explanation:

Requirement to be a function:

For a relationship to be a function, all values of x must have only one corresponding value of y.

In this question:

For x = -7, y = 2

For x = -1, y = 0

For x = 4, y = 0

For x = k, y = 0

A value of x must not have more than 1 value of y.

In this example, if k = -7, this relationship is not a function, since x = -7 would have both y = 2 and y = 0.

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Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

100 is the exterior angle.

( 2x + 3 ) and 51 are interior angles.

Formula : -

Exterior angles = Sum of two interior angles.

100 = 2x + 3 + 51

100 = 2x + 54

2x = 100 - 54

2x = 46

x = 46 / 2

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3 years ago

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Find the distance between the point (-2, -3) and the line with the equation y=4-2/3x

otez555 [7]

Answer:

$D = \frac{25\sqrt{13}}{13}$

Step-by-step explanation:

Given

$y = 4 - \frac{2}{3}x$

$(x_1,y_1) = (-2,-3)$

Required

Determine the distance

$y = 4 - \frac{2}{3}x$

Write the above equation in standard form:

$Ax + By + C = 0$

So, we have:

$\frac{2}{3}x+y - 4 = 0$

By comparison:

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

The distance is calculated using:

$D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Where:

$(x_1,y_1) = (-2,-3)$

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

This gives:

$D = \frac{|\frac{2}{3} * (-2) + 1 * (-3) - 4|}{\sqrt{(\frac{2}{3})^2 + 1^2}}$

$D = \frac{|-\frac{4}{3} -3 - 4|}{\sqrt{\frac{4}{9} + 1}}$

Take LCM

$D = \frac{|\frac{-4-9-12}{3}|}{\sqrt{\frac{4+9}{9}}}$

$D = \frac{|\frac{-25}{3}|}{\sqrt{\frac{13}{9}}}$

$D = |\frac{-25}{3}|/\sqrt{\frac{13}{9}}$

$D = \frac{25}{3}/\sqrt{\frac{13}{9}}$

Split the square root

$D = \frac{25}{3}/\frac{\sqrt{13}}{\sqrt{9}}$

Change / to *

$D = \frac{25}{3}*\frac{\sqrt{9}}{\sqrt{13}}$

$D = \frac{25}{3}*\frac{3}{\sqrt{13}}$

$D = \frac{25}{\sqrt{13}}$

Rationalize

$D = \frac{25}{\sqrt{13}} * \frac{\sqrt{13}}{\sqrt{13}}$

$D = \frac{25\sqrt{13}}{13}$

Hence, the distance is:

$D = \frac{25\sqrt{13}}{13}$

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3 years ago

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Zolol [24]

Answer:

$x=0,x=-2$

Step-by-step explanation:

From the question we are told that:

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$y'=10x^4+20x^3-0$

Where

y'=0

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Factorizing,We have

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Therefore

The critical points are

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3 years ago

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nlexa [21]

Answer:

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Step-by-step explanation:

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4 years ago

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4 years ago

Which value of k will make this relationship NOT a function? {(−7, 2), (−1, 0), (4, 8), (k, 0)} ​

Which value of k will make this relationship NOT a function? {(−7, 2), (−1, 0), (4, 8), (k, 0)}