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3 years ago

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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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:

$\theta = 90$

So, the equation becomes

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

Evaluate

$A_1 = \frac 12r^2$

Next, we calculate the length (L) of the chord as follows:

$\sin(45) = \frac{\frac 12L}{r}$

Multiply both sides by r

$r\sin(45) = \frac 12L$

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:

$A_2 = \frac 12 \pi (\frac{L}{2})^2$

This gives

$A_2 = \frac 12 \pi (\frac{r\sqrt 2}{2})^2$

Evaluate the square

$A_2 = \frac 12 \pi (\frac{2r^2}{4})$

Divide

$A_2 = \frac{\pi r^2}{4}$

Next, calculate the area of the chord using

$A_3 = \frac 12r^2(\theta - \sin(\theta))$

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