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2 years ago

Find the equation of the line passing through the points (2, -1) and (-6, -5)

Mathematics

1 answer:

blagie [28]2 years ago

7 0

Step 1: Find the slope

Point 1 = (2, -1)

Point 2 = (-6, -5)

m = (-5 - - 1) / (-6 - 2) = (-5 + 1) / (-6 - 2) = -4 / -8 = 1/2

m = 1/2

Step 2: Find the y-intercept

Slope-Intercept Form: y = mx + b

-m is the slope

-b is the y-intercept

---Plug in whichever point you want as well as the slope and solve for b

Point = (2, -1)

-1 = (1/2)(2) + b

-1 = 1 + b

b = -2

Step 3: Form the equation

Slope-Intercept Form: y = mx + b

m = 1/2

b = -2

y = 1/2x - 2

Hope this helps!! :)

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The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by $t$ , then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

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2 years ago

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stiks02 [169]

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Step-by-step explanation:

The given formula seems to represent the volume of right circular cone. The correct formula is:

$V=\frac{1}{3} \pi r^{2} h$

We have to rearrange the formula for h. This means we have to move h on one side of the formula and all the other variables and constants on the other side of the equation. This can be done as shown below:

$V=\frac{1}{3} \pi r^{2} h\\\\3V=\pi r^{2} h\\\\h=\frac{3V}{\pi r^{2} }$

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