CPhill

1.  Definitely geometric....the common ratio  is 2

We can see this because

12/ 6   =  24 / 12  = 48/ 24   =  2  {the ratio of each successive term to the prior one is the same }

2. Fifth term of the geometric  sequence.....the common ratio is 3

So...the fith term  is  just.....  5 * (3)^5-1   =  5 * 3⁴  =   5 * 81   =  405

3. 

a₁  = -4      a₂  =  8     a₃  =  -16

Note, NSS...it appears that we are multiplying each term by -2  to get the next term

So.....the "formula"  appears  to be

-4 * (-2)^n-1   

The  5th term  is     -4 *(-2)^{(5 - 1)}  =  -4 * (-2)⁴   =  -4 * 16   =   -64

4 .   This is a geometric series in disguise

The side of the the nth triangle  is given by

(50) * (.80)^{n - 1}

We could use some fancy math here...but let's just do this

Side  of  2nd trangle  =  50 * (.80)^{(2 - 1)}  = 50 * (.80)   =  45

Side of third triangle  =  45*.80  =  36

Side of 4th triangle  =  36 * .80  = 28.8 inches

5.

Note, NSS  ....that we multiplied 50  by the common ratio, r, twice to  get 450

So

50 * r^2   =  450    divide bott sides by  50

r^2  =  9        take the positive root of both sides

r  = 3

So.....the missing term is

50 * r   =    50 * 3   =    150 

Let  BC  =  x

So BD   =  (2/7)x

And CD  =  (5/7)x

So

AD  =  √ (AB^2  - BD^2]   = √   ( 10^2  - [ (2/7)x]^2 )   and

AD  =  √  (AC^2  - CD^2 ) = √ ( 17^2 - [ (5/7)x ]^2 )

Since   AD   =  AD......we have that

√   ( 10^2  - [ (2/7)x]^2 )  =  √ ( 17^2 - [ (5/7)x ]^2 )     square  both sides

 ( 10^2  - [ (2/7)x]^2 )  =   ( 17^2 - [ (5/7)x ]^2 )     simplify

100 -  (4/49)x^2  =  289 - (25/49)x^2     rearrange

(25/49)x^2  - (4/49)x^2  =  289 - 100

(21/49)x^2  =  189   

(3/7)x^2 = 189    multiply both sides by  7/3

x^2  =  (7/3)189

x^2  =  63 * 7    

x^2  =  441    

So....AD  =

√  (100  -  (4/49)*441 )  =

√ ( 100 -  (4*9) )  =

√ (100 - 36)  =

√64   =

8

4py  =  x^2      p =  -1.5   ....so....

4 (-1.5)y  = x^2

-6y  = x^2

y  =  (-1/6)x^2

y  =  x^2 / 4

4y  = x^2           the vertex  is (0,0)

To find the focal length, we have

4p  =  4   ⇒   p   =  1

So....the focus  is  (0,  0+ p)   =  (0 , 1)

The directrix  is   y  = -1

A parabola is set of all points in a plane which are an equal distance away from a given point and given line. The point is called the focus of the parabola and the line is called the directrix.

The longer the focus, the wider the parabola.....

1. What is the value of x?

sin(x+22)°=cos(2x−7)°

Remember that the  sine and cosine  are co-functions

So.....sin (theta)   =  cos (90 - theta)

This implies that  the angles sum to 90°

So

( x + 22)  + (2x - 7)  =  90    simplify

3x + 15  = 90     subtract  15 from both sides

3x  =  75      divide both sides by 3

x  = 25

2. Given cosx=12/13and sinx=5/13

What is ratio for ​ tanx ​ ?

tan x   =   sinx / cos x

tan x  =   (5/13) / (12/13)   =  (5/13) * (13/12)  =  5/12

This is known as a cyclic quadrilateral.....its opposite angle sum to 180°

So.....angle A  + angle C  = 180  ......so we have

(3x + 53)  + (2x - 68)   = 180   simplify

5x  - 15  =  180      add 15 to both sides

5x  =  195      divide both sides by 5

x  = 39

So   A  =   3(39) - 15 =  102°

D  =   2(39) + 30   =  108°....and B  is supplemental to this  = 180- 108 =  72°

And C   =  2(39) - 68   =  10°

Draw BC

We have triangle  BPC

Angle PBC  = PCB  because they are inscribed angles intercepting the same arc

So....the sides opposite thes angles are equal....so... BP = PC  = 8m

Draw AD

And we have triangle APD

For the same reasonas above angle  PAD = angle PDA

So.....AP  =  PD = 6m

So....BD  = BP + PD   =  8 +  6     =   14 m

log ( .6)   =   - .2218

9)In cyclic quadrilateral ABCD, AB = 2, BC = 3, CD = 10, and DA = 6. Let P be the intersection of lines AB and CD. Find the length BP.

Draw AC

AC^2  =  2^2  + 3^2   -  2(3)(2)cos (ABC)  

AC^2  = 13  - 12cos(ABC)   (1)

And since ABC is obtuse and ADC  is supplemental to ABC, cos(ADC)  = -cos(ABC)

So

AC^2   = 6^2  + 10^2 - 2(6)(10)cos(ADC)

AC^2  = 6^2 + 10^2  - 120 [ -cos(ABC) ]

AC^2  = 6^2 + 10^2 + 120cos(ABC)   

AC^2  = 136 + 120cos(ABC)   (2)

Subtract (1) from (2)

0  = 123  + (120 + 12) cos(ABC)

0  =  123  + 132cos(ABC)

-123 / 132  =  cos (ABC)

-41/44 = cos(ABC)   

So  sin (ABC)  =  √ [ 1 -  (41/44)^2 ]  =   √ (255) / 44   =  sin PBC

Likewise.....draw  DB

DB^2  = 6^2  + 2^2  - 2(6)(2)cos(DAB)

DB^2  = 40 - 24cos(DAB)

And sincw DAB is obtuse and DCB is supplemental to DCB, cos(DCB) = -cos(DAB)

So

DB^2  = 10^2 + 3^2  - 2(10)(3)cos(DCB)

DB^2 =  10^2 + 3^2  - 60 [-cos(DAB) ]  

DB^2   =  109 + 60cos(DAB)

Subtract (3) from (4)

0  = 69 + (60 + 24) cos(DAB)

-69/84  = cos(DAB)

-23/28  = cos(DAB)

So sin (DAB)  = √  [ 1 - (23/28)^2 ]  = √ (255)/28  =  sin PCB

sin PBC / sin PCB   = PC/ PB

(√ (255) / 44)  / ( √ (255)/28)  = PC / PB

28 / 44 = PC /PB

7/11 = PC /PB

PC   =  (7/11)PB

And we have that

(PB + AB) * PB   = PC(PC + CD)

(PB + 2) * PB  = (7/11)PB [ (7/11)PB + 10)

PB^2 + 2PB  =  (49/121)PB^2 + (70/11)PB

Let PB  = x

x^2 + 2x =  (49/121)x^2 + (70/11)x

(121 - 49) x^2 / 121   +  (2 - 70/11)x  = 0

(72/121)x^2  -  (48/11)x  = 0

(1/11)x [ (72/11)x - 48 ] = 0

So..... either x  =  0   {reject}    or

(72/11)x  - 48  = 0

(72/11)x  =  48

  x =  48(11)/72  =  (2/3) * 11  =   22/3   =  PB  = BP

Proof   {PC  = (7/11)PB  =  (7/11)(22/3)  = 14/3 }

(PB + AB) * PB   = PC(PC + CD)

(22/3 + 2) * (22/3)  =  (14/3) * ( 14/3 + 10)

616 / 9    =   616 / 9