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3 years ago

Need help number 8 it’s decreasing or increasing

Mathematics

1 answer:

andrew-mc [135]3 years ago

5 0

Increasing I think dryyhiiouiiiiuyutff

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Which relationships have the same constant of proportionality between y and x as in the equation y=1/2x Choose 3 answers:

Andrew [12]

Answer:

A and B has the same constant of proportionality

Step-by-step explanation:

$y \propto x$

$y = kx ----1$

Where k is the constant of proportionality

We are supposed to find Which relationships have the same constant of proportionality between y and x as in the equation $y=\frac{1}{2}x$

On comparing with 1

$k = \frac{1}{2}$

A)6y = 3x

$y = \frac{3}{6}x\\y = \frac{1}{2}x$

So, this equation has the same constant of proportionality

B) $(x_1,y_1)=(2,1)\\(x_2,y_2)=(4,2)$

To find the equation :

Formula : $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

So, $y - 1=\frac{2-1}{4-2}(x-2)\\y-1=\frac{1}{2}(x-2)\\y-1=\frac{1}{2}x-1\\y=\frac{1}{2}x$

So, this equation has the same constant of proportionality

$(x_1,y_1)=(1,2)\\(x_2,y_2)=(2,4)$

To find the equation :

Formula : $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

So, $y - 2=\frac{4-2}{2-1}(x-1)\\y - 2=2(x-1)\\y - 2=2x-2\\y=2x$

So, this equation do not has the same constant of proportionality

$(x_1,y_1)=(2,1)\\(x_2,y_2)=(3,2.5)$

To find the equation :

Formula : $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

So, $y - 1=\frac{2.5-1}{3-2}(x-2)$

$y-1=1.5(x-2)\\y-1=1.5x-3\\y=1.5x-2$

So, this equation do not has the same constant of proportionality

Hence A and B has the same constant of proportionality

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Step-by-step explanation:

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Which of the following points lie in the solution set to the following system of inequalities?

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