geno3141

Every line that is parallel to the line  x + 3y  =  -5  has the form  x + 3y  =  k

where diffierent values of k will give you different (but parallel) lines.

Placing the point  (-7, 2)  into  x + 3y  =  k     --->   (-7) + 3(2)  =  -7 + 6  =  -1   --->   k  =  -1.

So, the equation of the line that passes through  (-7, 2)  that is parallel to  x + 3y  =  -5

      is  x + 3y  =  -1.

To get this equation to have the form  Ax + By  =  3

                                             instead of  x + 3y  =  -1

we'll need to multiply the bottom equation by  -3:

     -3 [ x + 3y  =  -1 ]     --->     -3x + -9y  =  3

This makes  A = -3  and  B = -9.

Your instinct is correct!

Call that whose expression on the left side of the equal sign "expression" so you have

   [expression]  =  9

Square both sides  --  you'll get:

  x + [expression]  =  81

Since  [expression]  =  9  --  by substituting  '9'  for  [expression]  --  you'll now have:

  x + 9  =  81

Therefore,  x  =  72

You can do this just by listing the ways:

H-H-H-T  =  (½)·(½)·(½)·(½)  =  1/16

H-H-T-H  =  (½)·(½)·(½)·(½)  =  1/16

H-T-H-H  =  (½)·(½)·(½)·(½)  =  1/16

T-H-H-H  =  (½)·(½)·(½)·(½)  =  1/16

T-T-T-H  =  (½)·(½)·(½)·(½)  =  1/16

T-T-H-T  =  (½)·(½)·(½)·(½)  =  1/16

T-H-T-T  =  (½)·(½)·(½)·(½)  =  1/16

H-T-T-T  =  (½)·(½)·(½)·(½)  =  1/16

Adding the individual results give ½

There is another way, using combinations ...

p - 2q  =  0     --->                       p - 2q       =  0

q - 2r  =  0     --->   x 2   --->            2q - 4r  =  0

Add down:                                 p         - 4r  =  0

p - 4r  =  0     --->                      p - 4r  =  0

p + r  =  5     --->   x -1   --->    -p  -  r  =  -5

Add down:                                   - 5r  =  -5

                                                        r  =  1

p + r  =  5   --->   p + 1  =  5     --->     p  =  4

q - 2r  =  0   --->   q - 2(1)  =  0     --->   q  =  2

arc(FD)  =  2·76^o  =  152^o 

arc(GF)  =  2·40^o  =  80^o 

arc(GSD)  =  2·64^o  =  128^o 

arc(SD)  =  ½·arc(FD)  =  76^o 

arc(SG)  =  arc(GSD) - arc(SD)  =   ...

angle(SDG)  =  ½·arc(SG)  =  ...

1)  Find the midpoint of the line segment:  mdpt  =  (  (1 + 5)/2,  (3 + 7)/2  )  =  ( 3, 5 )

2)  Find the slope of the line segment:  m  =  (y₂ - y₁) / (x₂ - x₁)  =  (7 - 3) / (5 - 1)  =  4/4  =  1

3)  The slope of the perpendicular bisector will be the negative reciprocal of the slope found in 2).

     m  =  -1

4)  We have a point  (3, 5)  and a slope  -1,  so we can use the point-slope form:

      y - y₁  =  m(x - x₁)     --->     y - 5  =  -1(x - 3)

5)  Rewrite this equation:     y - 5  =  -1(x - 3)

                                             y - 5  =  -x + 3

                                                  y  =  -x + 8

                                             x + y  =  8 

Call the center of the half-circle  'O'

       the right-hand base corner of the triangle  'A'

       the point of tangency on the right side  'X'.

Because OX is perpendicular to AX, the angle(AXO) is a right angle.

Because the triangle is equilateral, the angle(XAO) is a 60^o angle.

This means that angle(XOA) is a 30^o angle.

OA has length 2, so side OX has length sqrt(3).

OA is a radius of the half-circle, so the diameter is  2·sqrt(3).

Step 1:  Find the slope of  3x + 4y  =  -14

                                                  4y  =  -3x - 14

                                                    y  =  (-3/4)x - 14/4

             The slope is  -3/4

Step 2:  Find the slope of the perpendicular line; it will be the negative reciprocal:  4/3

Step 3:  Use the point slope form to find the equation:  y - y₁  =  m(x - x₁)

                                                                                       y - 7    =  (4/3)(x + 5)

Step 4:  Write this equation in slope-intercept form:

                          y - 7  =  (4/3)(x + 5)

                          y - 7  =  (4/3)x + 20/3

                               y  =  (4/3)x + 41/3

Step 5:  Identify  m  and  b.

Step 6:  Find  m + b.

The easiest way would be to draw an equilateral triangle 3 units on each side.

Call the original triangle ABC. 

Divide each side into thirds.

Label the two points of trisection of AB as U and V.

Label the two points of trisection of BC as W and X.

Label the two points of trisection of CA as Y and Z.

Draw VW, XY and ZU to get hexagon UVWXYZ.

Draw UX, VY, and WZ.

They will meet at O, the center of the triangle.

You now have 9 congruent trianges, 6 in the hexagon.

The ratio of the hexagon to the area of the original triangle:  6 : 9  =  2 : 3.

You can also find the areas of the triangles and get the same answer that way.

Start listing them (have the second digit larger than the first and the third digit larger than the second):

123     126     129     135     138     147     156     159     168

234     237     246     249     258     267     279

345     348     357     369     378    

456     459     468    489

567     579

678

789

But each of these can be chosen in 6 ways:   123  =  132  =  213  =  231  =  312  =  321

So take the number of ways listed and multiply by 6.

It would also be wise to check to see if I have missed any.