What is the equation of the line shown on the graph?

Mathematics

1 answer:

Olegator [25]3 years ago

4 0

Answer:

B. y = 4x

Step-by-step explanation:

** 4 blocks = 1 unit [vertically]

Starting from the origin, you do rise\run by moving four blocks north over one block east. This graph is an example of what is known as direct variation.

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Three vertices of the trapezoid are A(4d,4e), B(4f,4e), and C(4g,0). The fourth vertex lies on the origin. Find the midpoint of

satela [25.4K]

We have that

point C and point D have y = 0-----------> (the bottom of the trapezoid).

point A and point B have y = 4e ---------- > (the top of the trapezoid)

the y component of midpoint would be halfway between these lines
y = (4e+ 0)/2 = 2e.

the x component of the midpoint of the midsegment would be halfway between the midpoint of AB and the midpoint of CD.

x component of midpoint of AB is (4d + 4f)/2.
x component of midpoint of CD is (4g + 0)/2 = 4g/2.
x component of a point between the two we just found is
[(4d + 4f)/2 + 4g/2]/2 = [(4d + 4f + 4g)/2]/2 = (4d + 4f + 4g)/4 = d + f + g.

therefore

the midpoint of the midsegment is (d + f + g, 2e)

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Answer:

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Step-by-step explanation:

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Calculate the straight line distance between the points (-10, -7) and (-8,1). G.2(B)

tia_tia [17]

Answer:

2√(17) or about 8.2462 units.

Step-by-step explanation:

We want to determine the distance between the two points (-10, -7) and (-8, 1).

We can use the distance formula. Recall that:

$\displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substitute and evaluate:

$\displaystyle \begin{aligned} d &= \sqrt{((-10)-(-8))^2 + ((-7) - (1))^2} \\ \\ &=\sqrt{(-2)^2+(-8)^2} \\ \\ &= \sqrt{(4) + (64)} \\ \\ &= \sqrt{68}\\ \\ &= 2\sqrt{17}\end{aligned}$