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3 years ago

Give the slope perpendicular to the equation 2x-y=5

Mathematics

1 answer:

nikklg [1K]3 years ago

4 0

Answer:

- $\frac{1}{2}$

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

2x - y = 5 ( subtract 2x from both sides )

- y = - 2x + 5 ( multiply through by - 1 )

y = 2x - 5 ← in slope- intercept form

with slope m = 2

Given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{2}$

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<h3> What is trigonometry?</h3>

Trigonometry is the branch of geometry that gives the relationship between the angles and sides of a triangle.

c) proof. if cos∅ = $\frac{x}{\sqrt{x^{2} + y^{2} } }$ <em>, </em>then x = adjacent, $\sqrt{x^{2} + y^{2} }$ = hypotenuse

To solve for opposite, since it is a right angle triangle

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ycos∅ = $\frac{xy}{\sqrt{x^{2} } + y^{2} }$

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d) 41 sin∅ = 40

sin∅ = 40/41

solving for adjacent = 9

tan ∅ = opp/adj = 40/9

$\frac{tan \alpha }{tan^{2}\alpha - 1 }$ = 40/9 ÷ $(\frac{40}{9} )^{2}$ - 1 = 40/9 ÷ 1600/81 - 1

= 40/9 ÷ 1519/81 = 40/9 x 81/1519 = 360/1519

e) 1 - cos ∅ = 1/2

cos ∅ = 1 - 1/2 = 1/2

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tan 60 = $\sqrt{3}$ $tan^{2} 60$ = 3, sin 60 = $\frac{\sqrt{3} }{2}$ $sin^{2} 60$ = 3/4

$tan^{2} 60$ + 4 $sin^{2} 60$ = 3 + 4(3/4) = 3 + 3 = 6

Learn more about trigonometry: brainly.com/question/24349828

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