CPhill

3) 

Because transversal EF cuts parallels AB, CD.....then angle  BEF  = angle GFE

And since  EF = FG.....then angles FEG  and EGF are equal

So.....angle FEG  =  [  180  - (100 +x) ] / 2   =  (80 - x)/2

And angles  BEF +  FEG + AEG  = 180

Sustituting....we have   that

(100 + x)  +  (80 - x) / 2  +  x   =   180      multiply through by 2

2(100 + x)  + 80 - x  + 2x  =  360     simplify

200 +2x  + 80 - x + 2x  =  360

280  + 3x  =  360     subtract 280  from both sides

3x  =  80       divide both sides by 3

x  = 80/3 °   ≈  26.66°

Mmmmm....let's see..

1) Since BAC  = 23, then angle DAQ  will be complementary to this  = 90 - 23  = 67

And QDA  =  90 - DAQ  =  23

And since DP = PC, then in triangle DPC.....angle  DCP = angle PDC

But......since tranversal AC  cuts parallels DC and AB....then angle BAC  = angle DCP  = 23

So....PDC  also = 23

So....

QDA  + PDC   =  23 + 23   =  46°

The new area  =   

Original Area *  Dilation Factor^2  =

58 * 8^2   =   

3712 ft^2

Let's convert "bearing"  into standard degrees....we can convert back at the end.

A bearing of  130°  is  320°

A wind from the SW means that the wind is blowing at a 45° angle

a)  Velocity of the bat  =   

[ 10cos (320°) i, 10 sin (320°) j  ]  =

[ 7.66 i, - 6.43 j ]

 Wind  velocity  =  [  5cos (45°) i , 5 sin (45°) j  ]  =

[ 3.54 i, 3.54 j ]

b)   Velocity of the bat =   

[ (7.66 + 3.54) i ,  ( -6.43 + 3.54 j ]  =

[ 11.2 i ,- 2.89 j ]

c)  Resultant velocity of the bat  = 

sqrt  [  11.2^2  + 2.89^2 ]  = 

11.57  m/s

Final direction  =    arctan  (-2.89/ 11.2)  = -14.47° = -14°  [to the nearest degree ]

Converting this to bearing, we have   90  + 14   =    104°

I liked the way you did this, hectictar....the "congruent" triangles approach was a nice touch...!!!

Angle ADP  = 30°   and angle APD  = 60°

So.....since we have a 30-60-90 triangle......AP  =  1/√3   = √3/3

And PD  =   2 / √3  =  2√3 / 3

And angle DPB = 120°  and angle PDB = 30° ..... so   angle  PBD = 30°

So...triangle DPB  isosceles with  PD  = PB  =  2√3 / 3

And using the Law of Sines

DB/ sin(120)  = PB / sin (30)

DB  =  sin (120) * [ 2√3 / 3 ]  /  (1/2)  =    √3 * 2√3 / 3   =  2

So...the perimeter of triangle BDP  =

PD + PB + DB  =     4√3 / 3  +  2   =   

2  +  4√3 / 3

So   w  = 2,  x = 4 , y  = 3  and z  = 3 

And their sum is  12

-2r^2  - 6r   =  0

Factor out the -2

-2 ( r^2  + 3r )  =  0

We are going to complete the square on  "r"

Take   1/2 of 3   =  3/2    .....square it  =9/4.....add and subtract it inside the parentheses....so we have...

-2 ( r^2  + 3r  +  9/4   -  9/4)        factor the first three terms

-2 [  (  r + 3/2)^2  - 9/4)  ]      distribute the -2 over each term in the parentheses

-2(r + 3/2)^2  +  18/4

-2(r + 3/2)^2  +  9/2

This is vertex form

The vertex  is  (-3/2,  9/2)

-

2.

4p+5c=6.71

5p+3c=7.12

Multiply the first equation by 3    and the second by - 5

12p  + 15c   =  20.13

-25p  - 5c   =  -35.60            add these

-13p   =  -15.47      divide both sides by -13

p  =  1.19

Using  any equation to find c, we have

4(1.19)  + 5c  = 6.71  

4.76  + 5c  = 6.71     subtract 4.76 from both sides

5c  = 1.95     divide both sides by 5

c  = 0.39

So   .... { p, c }  =  { 1.19, 0.39}

1. 

12o+7a=5.36

 8o+5a=3.65 

Multiply the first equation through by 5    and the second through by -7

So we have

60o  + 35a   =  26.8

-56o -  35a  =  - 25.55          add these

4o   =  1.25  =  5/4    divide both sides by  4

o =  5/16  =  .3125

We can find a with any of the equations

8(5/16) + 5a  =  3.65

5/2 +  5a  = 3.65

2.5  + 5a  = 3.65      subtract  2.5 from each side

5a  =  1.15       divide both sides by 5

a  =  .23

So... {o, a}    =   { .3125, .23 }

p(n)  =  6 (.98)ⁿ