CPhill

3.   Note that  triangle ABC is a 6-8-10 right triangle

And angle BCA  =  90

And angle PCA  =  45

So....angle BCP  = 45

Then.....PC  bisects  angle BCA

Let the intersection of PC and AB  =  D

So.....in triangle ABC, since angle BCA  is bisected, we have the following relationship

BC/CA   = BD/AD

6/8  = BD/AD

3/4  = BD/AD

So....since BA  = 10.....there  are 7 equal parts of BA

BD  =  (3/7)*10   =  30/7

AD = (4/7)* 10  = 40/7

Note....sin BAC  =  6/10  =  3/5

And  sin PCA = sin 45  =  1/√2

So.....using the Law of Sines

AD/ sin PCA  = CD/ sin BAC

(40/7)/ (1/√2)  =  CD / (3/5)

CD  =  (40/7)(3/5)*√2   =  (120/35)√2  =  (24/7)√2

And using the intersecting chord theorem

BD * AD  = CD * PD

(30/7) (40/7)  =  24√2 / 7  * PD

1200/7  =  24√2 * PD

PD  =  (1200) / ( 7 * 24 * √2) 

PD  =  (25/7)√2

So....CP  = 

PD + CD  =

(25/7)√2  + (24/7)√2  =

(49/7)√2 =

7√2

1.  In the form  (x - h)^2  + (y - k)^2  = r^2, the center is  (h, k)

So we have

(x + 7)^2  +  (y - 13)^2  =  196

(x  -  (-7) )^2  + (y - 13)^2  = 196

So.....the center  is   (-7, 13)

2. x^2 + y^2  + 14x + 10y - 7  = 0        re-write as

x^2 + 14x  +  y^2  + 10y     =   7        complete the square on x and y

x^2 + 14x + 49  + y^2  + 10y + 25  =  7 + 49 + 25     factor and simplify

(x + 7)^2  + (y + 5)^2  =  81

So....the center is  (-7, -5)

3.  Hyperbola

Let the height of the plane above the helicopter be "h"

We have a right triangle  where  the side opposite the 71 degree angle  =  h   and the side adjacent to this angle =  3055

So   we can use the tangent here

tan (71)  =  h / 3055       multiply both sides by 3055

3055 * tan (71)  =  h ≈   8872.4  ft

So.......the total altitude of the plane is

1600 + 8872.4  =

10472. 4 ft

1. The 12 ft   can be considered the hypotenuse of a right triangle

And the 8 ft  can be considered the leg opposite the angle we are considering

So....to  find out if this angle is at least 45°, we have take thw sine inverse of  8/12

So  we have

asin (8/12)  =  the angle  ≈ 41.8°.....so....it won't pour out smoothly

2. Area  =  (3.14) r^2 *  (40/ 360)  =

(3.14) * 4^2 * 1/9   =   

16 * 3.14 / 9    =

5.6 cm^2

3. We have  a 30-60-90  right triangle.....the side opposite the 30 degree angle  = 10 = base

And the side oposite the 60 degree angle  = 10√3 cm = height

So....the area  is   (1/2) base * height  =

(1/2) 10 * 10√3  = 

50√3   cm^2  ≈    86.6 cm^2

1.  We can find AE  with the secant product theorem

CA * AB   =  AE * DE

18 * 3   =  AE * AD

54  =  AE * AD

Let  AD  =  x

Then AE  =  3 + x        and we have that

54  = (3 + x)* x

54  = 3x + x^2       rearrange as

x^2 + 3x - 54   =  0        factor

(x + 9) (x - 6)  = 0

Setting each factor to 0 and solving for x and we have that

x  = -9    (reject)          and      x  = 6  (accept)

So

x = AD  =  6

And AE  =  3 + x   =   3 + 6   =  9

Note....that the diagonals of a rhombus are perpendicular.....they  form  a  90 degree angle

So...triangle  CEB  is a 30-60-90 right triangle

The hypotenuse of this triangle is 12

Angle ECB  = 30 degrees.....and the side opposite this angle   = 1/2 of the hypotenuse  =  6

So EB  = 6......and since the diagonals bisect each other....then  DB  = 12

Also..in triangle CEB....the side opposite the 60 degree angle  = √3  times the side  opposite the 30 degree angle  = 6√3

So.....CE  =  6√3    and CA  = 2 * 6√3  =  12√3

So...the area  =  (1/2) product of the diagonal lengths  =

(1/2) (12) (12√3)  =

72√3   units^2

Task 4

x + a      =    b

____           ___

  ax               x

We can "engineer"  an extraneous solution, as follows :

Cross-multiply ;

x (x + a)  =  bax    simplify

x^2  + ax  = bax

x^2  + ax - bax  = 0              let  a  =  1, let  b  = 5....so we have

x^2  + 1x  - (5)(1)(x)  = 0

x^2 + 1x  - 5x  = 0

x^2  - 4x  = 0      factor

x(x  - 4)  = 0

Set both factors to 0   and solve for  x

x  = 0           x  - 4  =0

                      x  = 4

So.....note that the original problem becomes ;

x + 1            5

____   =      __

   1x              x

Note that if  x  = 4

4 + 1             5

_____   =    ___       which is true

   1*4             4     

However...if x  =  0   ....both denominators are undefined because we cannot divide  by 0....thus....x  = 0  is an extraneous solution....!!!!!

Task 6

Let  N  = 37

We can write this as  (x + y)  =  (49-  12)

So....note that

(49 -  12) ^2  =    49^2  - 2(49*12) + 12^2  =   2401  - 1176  + 144    = 1369  =  37² 

Task 5

Let  the polynomial  be    x^2  - 4x  + 4

Let the  linear binomial be    ( x - 2)

Part 1

               x  -  2

x - 2   [   x^2  -  4x  + 4  ]

              x^2   -2x

             _____________

                       -2x   + 4

                       -2x   + 4

                       _______

                                 0

Part 2

f(a)   = f(2)  =   2^2  - 4(2)  + 4   =    4 - 8  + 4   =  0

Part 3 

Since  we obtained a "0"  with polynomial long division......then x  = 2   is  a  root. The Remainder Theorem says that if f(a)  = 0, then  "a" is a root.  This is borne out by the fact that f(2) = 0  where  "a"   = 2.

tangent P  =   opp/adj =  24/18

tangent Q  = adj/opp = 18/24

tan 26/10

tan B  = 10/26

Note...on this silly  question.......one leg of the right triangle is LONGER than the hypotenuse...IMPOSSIBLE!!!!

Last one   .....we can use the tangent here 

tan 19  =   height  / 4100     multiply both sides by  4100

4100 * tan 19   = height  ≈   1412 ft