CPhill

See the image

AB = 12       BC  = 15       so     AC =   9

Since BD bisects   angle ABC,  then

AD / DC =  AB/ BC

AD / DC = 12/15

AD / DC =  4/5

So   AD =  (4/9)(9) =  4

So    D  =(4,0)    and  DC =  AC - AD  =  9 - 4    =  5

And the distance from B to D  =  sqrt  (12^2  + 4^2 )  =  sqrt  (160)

So....using the Law of Cosines....

BC^2  =  DC^2  + BD^2   -2(DC* BD)cos(BDC)

15^2  = 5^2  +  160  -  2 ( 5sqrt(160)) cos(BDC)

40 =  -10sqrt (160) cos (BDC)

-40 /  (10 sqrt (160) )  =  cos (BDC)

-4 / [ sqrt (160) ]  =cos (BDC)

-4sqrt (160)  / 160  = cos (BDC)

-sqrt (160) / 40  = cos (BDC)

- sqrt (10)  / 10  =  cos (BDC)

Last one

By  the intersecting secants theorem

SU * ST   =   SW * SV

18 * 3  =  SW * SV

54  =  SW * SV

Let SV  =  x      so    SW = x + 3   .....so.....

54 = ( x + 3) * x     simplify

54  = x^2 +3x      rearrange as

x^2 + 3x  - 54   =  0       factor

(x - 6) ( x - 9)  =  0

Set each factor to  0  and solve for x and we have that   x  = 6  or x  = 9

Only  x = 6  = SV

So..SW =  6 + 3   = 9

Chords AB and CD of a circle meet at Q inside the circle. If CD = 38, AQ = 6, and BQ = 12, then what is the minimum length of CQ?

By the intersecting chord theorem

AQ * BQ   =  CQ  * DQ

6 * 12  = CQ * DQ

72  = CQ * DQ

Let CQ  = x    Let  DQ  =  38 - x    ....so we have

72 =  x  ( 38 - x)      simplify

x^2 - 38x  + 72  = 0       factor

(x -36) ( x- 2)  =  0

Setting each factor to 0   and solving for x we have that  x  = 36    or  x  = 2

So.....the minimum length  of  CQ   = 2

First one

By the tangent-secant theorem, we have that

WZ^2   = WY * WX

8^2  =  WY *  4

64  =  WY * 4

WY  = 16

And using the Pythagorean Theorem

YZ  = sqrt  [ WY^2  + WZ^2]   = sqrt  [ 16^2  + 8^2]  =sqrt [ 320 ] =  8sqrt(5)

Call the radius  of the small circles, R

Call the radius of the large circle, 2R

(1/2)  of one of the  yellow areas = [ (1/4)pi R^2  -  (1/2)R^2  ]

So   the total of the yellow area  = 8 [ (1/4)pi R^2   - (1/2) R^2]  = R^2 [ 2pi - 4] 

And  one of the pink areas  =   pi R^2  - [ pi R^2  - 2 R^2]   =  2R^2

So....the total of the pink areas =  8R^2

And the  area  of the large circle is  pi (2R)^2 =  4 pi R^2

So....the  blue area  =

Area of large circle -  pink areas  - yellow area

   4 pi R^2 - 8R^2  - R^2 [ 2pi - 4]  =

4pi R^2  - 8R^2  - 2pi R^2  + 4R^2  =

2 pi R^2  - 4R^2

R^2  [ 2pi  - 4 ] 

So....the areas are equal   !!!

Let   a be the longer side of the  original rectangle and  b be  the shorter side

So

 a*b  =12

b  =  12/a   (1)

When folded in half, the  longer side  =  b   and the shorter side  is  a/2

a/ b  =  b / (a/2)

a/b  = 2b / a

a^2  = 2b^2

a = √2b

So

a * b   = 12

√2 b* b  = 12

b^2  =  12/√2

b^2 =  6√2

b^2 = √72

b = ⁴√72    

And a  = √2 ⁴√72  

So

a / b =  √2 ⁴√72  / [  ⁴√72  ] =  √2

And   b/ [a/2]  =   2b / a  =  2  ⁴√72  / [√2 ⁴√72 ] =  2/√2  = √2

So

a  = √2 ⁴√72 

b =  ⁴√72 

And

a₁  = b =   ⁴√72 

b₁  =  a/2  =   [√2 ⁴√72] / 2

BA =  (1, 0, -1)

CA  = ( 1, -1, 2)

Find the  cross-product   BA x CA

i     j      k        i         j

1    0     -1      1       0

1   -1      2      1       -1

[ 0i  - 1j  - 1k ]  - [ 0k + 1i + 2j ]  =   -1i - 3j - 1k   =   ( -1, -3 , -1)  = the  normal vector

Note that if we multiply this  vector by -1  we get

( 1, 3 , 1)

And adding these components gives us  5

Nice, Melody.....!!!

Because  angle  ADE  = angle B  and   angle EAD  = angle BAC....then

Triangle ADE  is similar to triangle ABC

So

DE / AD  = BC / AB

DE / 6.8  =  5.5 / (8.6 + 2.4)

DE / 6.8  =  5.5 / 11 

DE / 6.8  = 1 / 2       multiply both sides by 6.8

DE =  6.8  / 2   =   3.4

f^-1 (-3)     means that  f ( x)  =  - 3

This happens when x  =  3

So  f^-1 ( -3)   = 3

f^-1 (0)  means that   f(x)  =  0

This happens when x = 2

So  f^-1(0)  =  2

f^-1(3)   means  that  f(x)  = 3

This happens when  x = -1

So   f^-1(3)  = -1

So

f^-1 (-3)   + f^-1(0) +  f^-1(3)  =

     3       +     2       +  (-1)  =

4