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hjlf
3 years ago
9

Is the relationship shown by the data linear? If so, model the data with an equation. x y –7 5 –5 9 –3 13 –1 17

Mathematics
1 answer:
diamong [38]3 years ago
8 0

Given:

The table of values is:

x :     -7          -5         -3           -1

y :      5           9         13          17

To find:

Whether the given data is linear and then find the equation.

Solution:

The given data is linear if the slope and rate of change is constant.

Slope formula:

m=\dfrac{y_2-y_1}{x_2-x_1}

Using the slope formula. we get

\dfrac{9-5}{-5-(-7)}=2

\dfrac{13-9}{-3-(-5)}=2

\dfrac{17-13}{-1-(-3)}=2

Since the rate of change is constant, i.e., 2, therefore the given data is linear.

The slope of a linear equation is 2 and it passes through the point (-7,5). So, the equation of the line is:

y-y_1=m(x-x_1)

y-5=2(x-(-7))

y-5=2(x+7)

y-5=2x+14

Adding 5 on both sides, we get

y-5+5=2x+14+5

y=2x+19

Therefore, the equation for the given data is y=2x+19.

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Trapezoid MNPQ is similar to Trapezoid RSTU. What is the length of the "x" on the misside side NP ? Also, what is the length of
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Answer:

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Step-by-step explanation:

Given

See attachment for MNPQ and RSTU

Required

Find x and y

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The ratio of corresponding sides are:

MN : RS

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PQ : TU

MQ : RU

From the attachment, we have:

MN : RS \to 18 : 30

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To solve for x, we equate MN : RS and NP : ST

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Express as fraction

\frac{18 }{ 30 }= \frac{x }{ 25}

Make x the subject

x =  25 * \frac{18 }{ 30 }

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To solve for y, we equate MN : RS and PQ : TU

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Express as fraction

\frac{18 }{ 30 }= \frac{15 }{ y}

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