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3 years ago

12

Which of the following equations does not represent a linear relationship between x and y

1 answer:

Ksenya-84 [330]3 years ago

7 0

A linear relationship is an equation that has a constant slope

The equation that does not represent a linear relationship is (b) y = 2x^2 + 5x

<h3>How to determine the non-linear relationship</h3>

A linear equation can take any of the following forms:

y = mx + b

y - y1 = m(x - x1)

Ax + By = c

Any form different from the above forms is not a linear equation

Using the above format as a guide, the equation that does not represent a linear relationship is (b) y = 2x^2 + 5x

Read more about linear relationships at:

brainly.com/question/15602982

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tigry1 [53]

Answer:

Yes, Her father is correct.

Step-by-step explanation:

75/15=5 which means that there will be fifteen rows, with five pennies in each row. Since one side has 5 pennies and the other has fifteen, the sides are different and form a rectangle. To check your work, you can do 15*5 which equals 75 the rectangle of pennies should look like this:

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3 years ago

HELP i need it on my final

ANTONII [103]

Answer:

-2

Step-by-step explanation:

when x = 5, f(x) = -2

hope this helps. good luck

8 0

3 years ago

The area of a rectangle is 4n+16. What could the length and the width be?

zysi [14]

Answer:

length can be 4

width can be n+4

IT CAN ALSO BE THE OTHER WAY ROUND

Step-by-step explanation:

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3 years ago

The price for 7 lb of potatoes is $5 what is the price per pound.

Thepotemich [5.8K]

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The answer is $1.4.

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3 years ago

the slope of a line passing through h (-2, 5) is -3/4. Which ordered pair represents a point on this line?

rosijanka [135]

Answer:

A

Step-by-step explanation:

Calculate the slope of the given points with the point (- 2, 5 )

If the slope is - $\frac{3}{4}$ then the point is on the line

Calculate slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (6, - 1)

m = $\frac{-1-5}{6+2}$ = $\frac{-6}{8}$ = - $\frac{3}{4}$ ← point (6, - 1) is on the line

Repeat with (x₁, y₁ ) = (- 2, 5 ) and (x₂, y₂ ) = (2, 8)

m = $\frac{8-5}{2+2}$ = $\frac{3}{4}$ ← point (2, 8) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (- 5, 1)

m = $\frac{1-5}{-5+2}$ = $\frac{-4}{-3}$ = $\frac{4}{3}$ ← point (- 5, 1) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (1, 1)

m = $\frac{1-5}{1+2}$ = - $\frac{4}{3}$ ← point (1, 1) is not on the line

Thus the point on the line is (6, - 1 ) → A

4 0

3 years ago