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2 years ago

Find the slope of a line PERPENDICULAR TO LM: L(3,-5), M(-4, -1)

Mathematics

1 answer:

Ede4ka [16]2 years ago

7 0

Answer:

$\frac{7}4$

Step-by-step explanation:

First, we need to find the slope, which is defined as the change in y over the change in x:

$\frac{-5-(-1)}{3-(-4)} = \frac{-4}{7}$

To find a slope that's perpendicular to that, we just take the negative reciprocal of that slope, which is $\frac{7}4$ .

I hope this helped.

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Can someone please help me with this? I'll give brainliest :)

vlabodo [156]

The slope of y = 3x - 4 on the interval [2, 5] is 3 and the slope of y = 2x^2-4x - 2 on the interval [2, 5] is 10

<h3>How to determine the slope?</h3>

The interval is given as:

x = 2 to x = 5

The slope is calculated as:

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

Substitute 2 and 5 for x

y = 3*2 - 4 = 2

y = 3*5 - 4 = 11

So, we have:

$m = \frac{11 - 2}{5 - 2}$

$m = \frac{9}{3}$

Divide

m = 3

Hence, the slope of y = 3x - 4 on the interval [2, 5] is 3

Substitute 2 and 5 for x

y = 2 * 2^2 - 4 * 2 - 2 = -2

y = 2 * 5^2 - 4 * 5 - 2 = 28

So, we have:

$m = \frac{28 + 2}{5 - 2}$

$m = \frac{30}{3}$

Divide

m = 10

Hence, the slope of y = 2x^2-4x - 2 on the interval [2, 5] is 10