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3 years ago

Graph a direct variation function that goes through the point (-2, -6). what is the slope

Mathematics

2 answers:

Effectus [21]3 years ago

6 0

Answer:

The slope is 3.

Step-by-step explanation:

A direct variation function will always go through the origin point (0, 0).

Therefore, we have the two points (0, 0) and (-2, -6).

We can use the slope formula to find the slope:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Substitute and evaluate:

$\displaystyle m=\frac{0-(-6)}{0-(-2)}=\frac{6}{2}=3$

Therefore, the slope of the function is 3.

To graph, simply plot (0, 0) and (-2, -6) and draw a line connecting them. This is shown below.

denis23 [38]3 years ago

3 0

Answer:

y = 2/3x - 3

Slope: 2/3x

Step-by-step explanation:

Slope: 2/3x

y- intercept: (0,-3)

once it goes down it'll be -2/-3x which will go to through (-2,-6) as soon as it goes down from the y-intercept.

equation: y = 2/3x - 3

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3 years ago

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3 years ago

CAN SOMEONE PLS ANSWER-If W(- 10, 4), X(- 3, - 1) , and Y(- 5, 11) classify AEXY by its sides . Show all work to justify your an

AnnZ [28]

Answer:

WX = $\sqrt{74} \approx 8.6023253\\\\$
XY = $2\sqrt{37} \approx 12.1655251\\\\$
WY = $\sqrt{74} \approx 8.6023253\\\\$
Classify: Isosceles

============================================================

Explanation:

Apply the distance formula to find the length of segment WX

W = (x1,y1) = (-10,4)

X = (x2,y2) = (-3, -1)

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-10-(-3))^2 + (4-(-1))^2}\\\\d = \sqrt{(-10+3)^2 + (4+1)^2}\\\\d = \sqrt{(-7)^2 + (5)^2}\\\\d = \sqrt{49 + 25}\\\\d = \sqrt{74}\\\\d \approx 8.6023253\\\\$

Segment WX is exactly $\sqrt{74}$ units long which approximates to roughly 8.6023253

-------------------

Now let's find the length of segment XY

X = (x1,y1) = (-3, -1)

Y = (x2,y2) = (-5, 11)

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-3-(-5))^2 + (-1-11)^2}\\\\d = \sqrt{(-3+5)^2 + (-1-11)^2}\\\\d = \sqrt{(2)^2 + (-12)^2}\\\\d = \sqrt{4 + 144}\\\\d = \sqrt{148}\\\\d = \sqrt{4*37}\\\\d = \sqrt{4}*\sqrt{37}\\\\d = 2\sqrt{37}\\\\d \approx 12.1655251\\\\$

Segment XY is exactly $2\sqrt{37}$ units long which approximates to 12.1655251

-------------------

Lastly, let's find the length of segment WY

W = (x1,y1) = (-10,4)

Y = (x2,y2) = (-5, 11)

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-10-(-5))^2 + (4-11)^2}\\\\d = \sqrt{(-10+5)^2 + (4-11)^2}\\\\d = \sqrt{(-5)^2 + (-7)^2}\\\\d = \sqrt{25 + 49}\\\\d = \sqrt{74}\\\\d \approx 8.6023253\\\\$