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Vesnalui [34]
3 years ago
8

Please show work

Mathematics
1 answer:
Anit [1.1K]3 years ago
6 0

(1)

since both equations express y in terms of x, equate the right sides

- 2x = - 4x + 10 ( subtract 4x from both sides )

2x = 10 ( divide both sides by 2 )

x = 5

substitute x = 5 into y = - 2x → y = - 10

solution is : (x, y ) → (5, - 10 )

(2)

equate the right sides of both equations

3x = 2x - 7 ( subtract 2x from both sides )

x = - 7

substitute x = - 7 into y = 3x → y = - 21

solution is : (x, y) → (- 7, - 21)

(3)

substitute y = - 8 into the other equation

- 8 = 6x + 22 ( subtract 22 from both sides )

- 30 = 6x ( divide both sides by 6 )

x = - 5

solution is : (x, y ) → (- 5, - 8 )



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forsale [732]

See below for the proof that the areas of the lune and the isosceles triangle are equal

<h3>How to prove the areas?</h3>

The area of the isosceles triangle is:

A_1 = \frac 12r^2\sin(\theta)

Where r represents the radius.

From the figure, we have:

\theta = 90

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A_1 = \frac 12r^2\sin(90)

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Multiply by 2

L = 2r\sin(45)

This gives

L = 2r\times \frac{\sqrt 2}{2}

L = r\sqrt 2

The area of the semicircle is then calculated as:

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Next, calculate the area of the chord using

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Recall that:

\theta = 90

Convert to radians

\theta = \frac{\pi}{2}

So, we have:

A_3 = \frac 12r^2(\frac{\pi}{2} - \sin(\frac{\pi}{2}))

This gives

A_3 = \frac 12r^2(\frac{\pi}{2} - 1)

The area of the lune is then calculated as:

A = A_2 - A_3

This gives

A = \frac{\pi r^2}{4} -  \frac 12r^2(\frac{\pi}{2} - 1)

Expand

A = \frac{\pi r^2}{4} -  \frac{\pi r^2}{4} + \frac 12r^2

Evaluate the difference

A =  \frac 12r^2

Recall that the area of the isosceles triangle is

A_1 = \frac 12r^2

By comparison, we have:

A = A_1 = \frac 12r^2

This means that the areas of the lune and the isosceles triangle are equal

Read more about areas at:

brainly.com/question/27683633

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5 0
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