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3 years ago

What is the steepness of this line?

Mathematics

1 answer:

andrezito [222]3 years ago

8 0

<em>The steepness of the line on which the plane is descending is 1/3</em>

<h2>Explanation:</h2>

The steepness of a lines tells us <em>how steep</em> this straight line is. In other words, this is the slope of the line. This can be calculated as:

$steepness=\frac{Change \ in \ y}{Change \ in \ x}=m$

From the figure, the plane is descending through a line whose Change in y and x are as follows:

$Change \ in \ y:1000ft \\ \\ Change \ in \ x:3000ft$

So:

$steepness=\frac{1000}{3000}=\frac{1}{3}$

<em>So the steepness of the line on which the plane is descending is 1/3</em>

<h2>Learn more:</h2>

Inverse variation: brainly.com/question/13881905#

#LearnWithBrainly

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Find the perimeter of the triangle with vertices A(negative 2,2,3), B(4,negative 4,3), and C(7,6,4).

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$P\approx 28.872\,\text{units}$

Step-by-step explanation:

We are given coordinates of each of the vertices of the triangle ABC, hence we can use the distance formula to find the lengths of each of the side.

the distance formula is generally written as:

$r = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

for the side AB: A(-2,2,3) and B(4,-4,3)

$AB = \sqrt{(-2-4)^2+(2-(-4))^2+(3-3)^2}$

$AB = \sqrt{(-6)^2+(6)^2+(3-3)^2}$

$AB = \sqrt{72}$

for the side BC: B(4,-4,3) and C(7,6,4)

$BC = \sqrt{(4-7)^2+((-4)-6)^2+(3-4)^2}$

$BC = \sqrt{110}$

for the side CA: C(7,6,4) and A(-2,2,3)

$CA = \sqrt{(7-(-2))^2+(6-2)^2+(4-3)^2}$

$CA = \sqrt{98}$

Now we all the sidelengths:

to find the perimeter we need to just sum the three side lengths:

$P = AB + BC + CA$

$P = \sqrt{72} + \sqrt{110} + \sqrt{98}$

$P\approx 28.872\,\text{units}$

this is the perimeter of triangle ABC

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