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3 years ago

(2,1),(-3,1) how do you solve this I’m looking for the slope I also have to show work

Mathematics

1 answer:

shutvik [7]3 years ago

7 0

Answer: 0

Step-by-step explanation:

To find the slope, you would need to use the formula $m=\frac{y_2-y_1}{x_2-x_1}$ .

$m=\frac{1-1}{-3-2} =\frac{0}{-5} =0$

The slope is 0, meaning the line is horizontal.

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3 years ago

arlik [135]

Given:

In a right angle triangle θ is an acute angle and $\tan\theta =\dfrac{3}{5}$ .

To find:

The value of $\cos \theta$ .

Solution:

In a right angle triangle,

$\tan \theta=\dfrac{Perpendicular}{Base}$

We have,

$\tan\theta =\dfrac{3}{5}$

It means the ratio of perpendicular to base is 3:5. Let 3x be the perpendicular and 5x be the base.

By using Pythagoras theorem,

$Hypotenuse=\sqrt{Perpendicular^2+base^2}$

$Hypotenuse=\sqrt{(3x)^2+(5x)^2}$

$Hypotenuse=\sqrt{9x^2+25x^2}$

$Hypotenuse=\sqrt{34x^2}$

$Hypotenuse=x\sqrt{34}$

In a right angle triangle,

$\cos \theta=\dfrac{Base}{Hypotenuse}$

$\cos \theta=\dfrac{5x}{x\sqrt{34}}$

$\cos \theta=\dfrac{5}{\sqrt{34}}$

Therefore, the value of $\cos \theta$ is $\dfrac{5}{\sqrt{34}}$ .

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