hectictar

Hey...here's another way!  

∠KRM  =  130° / 2  =  65°

∠RKM  =  130° / 2  =  65°

∠QKM  =  28° / 2  =  14°

∠RKQ  =  65° - 14°  =  51°

And since there are 180° in triangle KPR....     ∠RPK  =  180° - 65° - 51°  =  64°

*edit*

Wait a minute...this isn't any different than the second one!!! Ah..sorry about that!

Hahah!! Well, maybe it's a little more efficient, but it isn't as explanatory  :)

Since   \(MR = MK\)   and   \(\overset{\frown}{MK} = 130°\)   ,   \(\overset{\frown}{MR} = 130°\)

\(\overset{\frown}{MK} +\overset{\frown}{KR} +\overset{\frown}{MR} =360°\)

Replace  \(\overset{\frown}{MK}\)  with  130°  and  \(\overset{\frown}{MR}\)  with  130°

\(130°+\overset{\frown}{KR}+130°=360° \\~\\ \overset{\frown}{KR}=360°-130°-130° \\~\\ \overset{\frown}{KR}=100°\)

And..

\(m\angle RPK=\frac{\overset{\frown}{KR}+\overset{\frown}{MQ}}{2}\\~\\ m\angle RPK=\frac{100°+28°}{2} \\~\\ m\angle RPK=64°\)

That's a really good way of explaining it Gh0sty !! 

First let's find the slope.

slope  =  \(\frac{\text{difference of y values}}{\text{difference of x values}}=\frac{-6--6}{-7-3}=\frac{0}{-10}=0\)

So, in point-slope form, using the point  (-7, -6) ,  the equation is....

y - -6  =  0(x - -7)

y + 6  =  0

y  =  -6

And..we could have noticed from the beginning that the  y  value of both points is  -6  . That means it must be a horizontal line at  y = -6  . 

You can click the  " 2^nd "  button and then click  " acos " .

Or you can click and hold the  " cos "  button.

Or you can type  " acos(  ) "  into the bar.

2m + 12  =  3m - 31

                                  Subtract  2m  from both sides of the equation.

12  =  3m - 2m - 31

12  =  m - 31

                                  Add  31  to both sides of the equation.

43  =  m

On the home page, you can either press the button that looks like this:     x^y

Or, you can type this symbol into the bar:     ^

For example, to do:     5³

Type:                           5^3

I think the base is the number that is being raised to an exponent.

So....the base in   2x + 1²   is   1   .

The base in   (2x + 1)²   is   2x + 1   .

Here's another way...but it looks a bit longer than CPhill's!

First let's extend line  AB .

Then....

Draw a line perpendicular to  AB  that goes through point  C  .

Draw a line perpendicular to AB  that goes through point  E .

Draw a line parallel to  AB  (and perpendicular to the previous two lines) that goes through point  D .

Now let's label these new points of intersection W, X, Y, and Z , like the picture,

and put the information from the problem.

∠EAZ  =  ∠CBY  =  180° - 120°  =  60°

Since triangles AEZ and BCY are 30-60-90 triangles where the side across from the 90° angle is 2....

the side across from the 30° angle  =  2/2  =  1

the side across from the 60° angle  =  √3

And...

WD + DX  =  2 + 1 + 1             And  WD = DX , so we can substitute  WD  in for  DX .

WD + WD  =  2 + 1 + 1

2WD  =  4

WD  =  2

Now we can use the Pythagorean theorem to find  CX .

2² + CX²  =  4²

4 + CX²  =  16             Subtract  4  from both sides of the equation.

CX²  =  12                   Take the positive square root of both sides.

CX  =  2√3

And.....

area of ABCDE   =   area of WXYZ  -  2(area of △EDW)  -  2(area of △AEZ)

                           =   (width)(length)  -  2(1/2)(base)(height)  -  2(1/2)(base)(height)

                           =   (2 + 2)(2√3 + √3)  -  2(1/2)(2)(2√3)  -  2(1/2)(1)(√3)

                           =   12√3  -  4√3  -  √3

                           =   7√3  square units