hectictar

Also....

There are  180°  in triangle ACD, so..

m∠DAC  =  180° - 90° - 50°

m∠DAC =  40°

An inscribed angle is half the measure of its intercepted arc, so....

(1/2)m∠AOC  =  m∠ABC

(1/2)m∠AOC  =  70°

m∠AOC  =  140°

OA  and  OC  are radii of circle  O , so they are the same length, and triangle AOC  is isosceles. So...

m∠OAC  =  m∠OCA

m∠OAC + m∠OCA + 140°  =  180°

m∠OAC + m∠OAC + 140°  =  180°

2m∠OAC + 140°  =  180°

2m∠OAC  =  40°

m∠OAC  =  20°

m∠DAE  =  m∠DAC - m∠OAC

m∠DAE  =  40° - 20°

m∠DAE  =  20°

\(\sin(x +\frac{11\pi}{6})\,-\,\sin(x -\frac{11\pi}{6})\,=\,1\)

We can use the sum of angles formula for sine:

\(\sin(\alpha \pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\~\\ \ \\ {\color{BlueViolet} \sin(x +\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}+\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\\~\\ {\color{Fuchsia} \sin(x -\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}-\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}\)

So if...

\({\color{BlueViolet}\sin(x +\frac{11\pi}{6})}\,-\,{\color{Fuchsia}\sin(x -\frac{11\pi}{6})}\,=\,1\\~\\ [\,{\color{BlueViolet}(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\,]\,-\,[\,{\color{Fuchsia}(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}\,]\,=\,1\\~\\ (\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)-(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)\,=\,1\\~\\ 2(\cos x)(-\frac12)\,=\,1\\~\\ - \cos x\,=\,1\\~\\ \cos x\,=\,-1\\~\\ x\,=\,\pi+2\pi n\quad \text{, where }\ n\ \text{ is an integer.}\)      then...

(XY / PQ)²   =   area of UVWXY / area of LMNPQ

(XY / 10)²   =   198 / 88

(XY / 10)²   =   9 / 4

XY / 10   =   √[ 9 / 4 ]

XY / 10   =   3 / 2

XY   =   10 * 3 / 2

XY   =   30 / 2

XY   =   15

diameter of circle  M  =  CB  =  36

radius of circle  M  =  36 / 2  =  18

area of semicircle  M  =  (1/2)pi( 18² )  =  162pi

radius of circle  O  =  OC + CB  =  32 + 36  =  68

diameter of circle  O  =  AB  =  2 * 68  =  136

area of semicircle  O  =  (1/2)pi( 68² )  =  2312pi

diameter of circle  K  =  AB - CB  =  136 - 36  =  100

radius of circle  K  =  100 / 2  =  50

area of semicircle  K  =  (1/2)pi( 50² )  =  1250pi

area of shaded region   =   area of semicircle O  -  area of semicircle K  -  area of semicircle M

area of shaded region   =   2312pi  -  1250pi  -  162pi

area of shaded region   =   900pi

In point slope form, the equation is:

y - 2  =  7(x - 1)      Now get this equation into standard form  (   Ax + By  =  C   ).

                            Distribute  7  to the terms in parenthesees.

y - 2  =  7x - 7

                            Add  7  to both sides of the equation.

y + 5  =  7x

                            Subtract  y  from both sides.

5  =  7x - y

7x - y  =  5

If we drew a line connecting the two vertices that are furthest from each other, we would create two triangles, and by SSS, these triangles would be congruent.

And  n°  would be the same as the measure of its corresponding angle, which is  142° .   So...

n  =  142

In every quadrilateral, the sum of the interior angles  =  360° ,   so....

m  =  360 - n - 142 - 20  =  360 - 142 - 142 - 20  =  56

m  =  56

Assuming that this is a regular octagon.......

We can break the octagon into 8 congruent triangles like this:

The sum of the 8 central angles  =  360° , and each central angle has the same measure,

so the measure of one central angle  =  360° / 8  =  45°

Draw a height of one of the triangles perpendicular to a side of the octagon. I labeled the height "a".

By the HL theorem, the two smaller triangles that are formed are congruent.

So the measure of the angle formed by the height and the hypotenuse  =  45° / 2  =  22.5°

tan 22.5°  =  1 / a  cm

a tan 22.5°  =  1  cm

a  =  1 / tan 22.5°  cm

area of one of the eight triangles  =  (1/2)(base)(height)

area of one of the eight triangles  =  (1/2)(2)(1/tan 22.5°)   sq cm

area of one of the eight triangles  =  1 / tan 22.5°   sq cm

area of octagon  =  8 * (area of one of the eight triangles)

area of octagon  =  8 * (1 / tan 22.5° )   sq cm

area of octagon  =  8 / tan 22.5°   sq cm

area of octagon  ≈  19.3   sq cm

x² + 30x + 180  =  -36

                                         Add  36  to both sides of the equation.

x² + 30x + 216  =  0

                                         What two numbers add to  30  and multiply to 216 ?  +12  and  +18

                                         So the left side of the equation factors like this...

(x + 12)(x + 18)  =  0

                                         Set each factor equal to zero and solve for  x .

x + 12  =  0          or          x + 18  =  0

  x  =  -12                            x  =  -18

The roots of the equation are  -12  and  -18 .

The nonnegative difference between  -12  and  -18  is   (-12) - (-18)   =   -12 + 18   =   6

Any value of  y  in the interval  (10, 11)  will make the equation true, for example....

\(\lceil{10.1}\rceil\cdot\lfloor{10.1}\rfloor\,=\,11\cdot10\,=\,110 \)

But also, any value of  y  in the interval  (-11, -10)  will make the equation true, for example...

\( \lceil{-10.9}\rceil\cdot\lfloor{-10.9}\rfloor\,=\,-10\cdot-11\,=\,110\)

Since the problem says  y < 0 ,  the only values of  y  that work are those in the interval  (-11, -10) .

These price values are in dollars:

price after  0  years   =   200

price after  1  year    =   200  +  8% of 200   =   200  +  0.08(200)  =  200(1 + 0.08)  =  200(1.08)

price after  2  years   =   200(1.08)(1.08)

price after  3  years   =   200(1.08)(1.08)(1.08)  =  200(1.08)³

price after  x  years   =   200(1.08)^x

price after  20  years  =  200(1.08)²⁰   ≈   932.19

The price after 20 years is about  $932.19 .