geno3141

                                     x² - 2x  =  0

Factor out the x-term:  x(x - 2)  =  0

Now, either                   x = 0          or          x - 2  =  0   --->   x  =  2

To win the tournament you will need to win two more games and not lose more than two games.

The various ways that you can win are:

WW             (½)(½)            =

WLW           (½)(½)(½)       =  

WLLW         (½)(½)(½)(½)  =

LWW           (½)(½)(½)       =

LWLW         (½)(½)(½)(½)  =

LLWW         (½)(½)(½)(½)  =

Calculate each of the above situations and add them together for your final answer.

I would suggest drawing this on a sheet of graph paper but make the distance from A to B to be 12, not 1.

This will make finding the points of reflection, and the distances, much easier.

However, the last step will be to divide the answer by 12.

Place A at the origin:  A = (0, 0)     B = (12, 0)     C = (12, 12)     D = (0, 12)

Also, since there will be many points of reflecttion, rename point X as point X₁.

Since X₁B = three-fourths of AB,  X₁ = (12, 9).

The point X₂ will be on side CD.  

The "angle going in" will equal the "angle coming out", so CX₂ / CX₁  =  AB / X₁B.

This makes  X₂ = (8, 12).

Point X₃ will be on side DA   --->   DX₃ / DX₂  =  CX₁ / CX₂   --->   X₃ = (0, 6).

Point X₄ will be on side AB   --->   X₄ = (8, 0).

Point X₅ will be on side BC   --->   X₅ = (12, 3).

Point X₆ will be point D   --->   X₆ = (0, 12).

Now, use the distance formula to find all the distances:

AX₁   =  sqrt( (12 - 0)² + (9 - 0)² )  =  15

X₁X₂  = sqrt( (8 - 12)² + (12 - 9)² )  =  5

X₂X₃  =  ...

X₃X₄  =  ...

X₄X₅  =  ...

X₅D   =  ...

Add these distances together and then divide by 12.

Angle(D) cuts off arc(ABC)   --->   angle(D)  =  ½·arc(ABC)

Angle(B) cuts off arc(ADC)   --->   angle(B)  =  ½·arc(ADC)  

Since arc(ABC) + arc(ADC) creates the whole circle:  arc(ABC) + arc(ADC)  =  360^o 

Therefore:  angle(D) + angle(B)  =  ½·arc(ABC) + ½·arc(ADC) 

                                                    =  ½·( arc(ABC) + arc(ADC) )

                                                    =   ½·( 360^o )  

                                                     =  180^o 

--->   (x + 24)^o + (x + 10)^o  =  180^o

--->                    2x^o + 34^o  =  180^o

--->                              2x^o  =  146^o

--->                                x^o  =  73^o 

Now, find the sizes of angle(A), angle(B), and angle(D).

Add these values together and subtract from 360^o to find the size of angle(C).

A 5 cm by 5 cm by 5 cm box contains 5 x 5 x 5 = 125 1-cm by 1-cm by 1-cm cubes.

None of the interior cubes touch the box  -  there are 4 x 4 x 4 = 64 interior cubes.

Also, none of the interior cubes on the top layer touch the box; there are 4 x 4 = 16 of these.

The total number that touch the cardboard is  125 - 64 - 16  =  45  1-cm by 1-cm by 1-cm cubes.

15 miles = 15 miles x ( 5280 feet / mile   =  79 200 feet

1 hour  =  1 hour x ( 60 minutes / hour ) x ( 60 seconds / minute )  =  3600 seconds

15 miles per hour  =  79 200 feet per 3600 seconds  =  22 feet per second

"per" means "divided by"

The top angle in he triangle is 180^o - 121^o  =  59^o.

The angle in the right corner of the triangle is 180^o - 144^o  =  36^o.

This means that the angle in the left corner of the triangle is 180^o - (59^o + 36^o)  =  85^o                 

To find the value of n^o:  n^o  =  180^o - 85^o  =  95^o.

First, I went into algebra mode and set up equations for both area and perimeter -- this got very difficult,

very quickly.

So, I began looking at Pythagorean Triples and found  15 - 20 - 25.

To center something that is 2 1/2 inches wide in an area that is 3 3/4 inches wide:

1)  subtract  2 1/2 from 3 3/4   --->   3 3/4 -  2 1/2  =  1 1/4

2)  divide this answer by 2 so that there is an equal amount on each side:

               1 1/4  /  2  =   5/4  /  2  =  5/4 x 1/2  =  5/8

60 out of 100  =  60/100  =  0.60  =  60%

36%

0.589  =  58.9%

Which percentage is greatest?

Originally, there is a right triangle:  call the length of the pole "p"

At first, the distance from the bottom of the pole to the wall is "g"

and the distance from the top of the pole to the ground is 8.

--->   p²  =  g² + 8²

If the pole is moved another 2 feet from the wall, it will lie on the ground; now the distance from the bottom of the 

pole to the wall is  g + 2; so  p  =  g + 2.

Substituting  "g + 2" for "p" in the above equation:  (g + 2)²  =  g² + 8² 

                                                                           g² + 4g + 4  =  g² + 64

                                                                                   4g + 4  =  64

                                                                                         4g  =  60

                                                                                           g  =  15

The pole is  g + 2  =  17