CPhill

2 sqrt (x^2 + y^2))  = sqrt [ (x-12)^2 + y^2]

4 [ x^2 + y^2]  = [ x^2 - 24x + 144 + y^2]

4x^2 + 4y^2  = x^2 - 24x + 144 + y^2

3x^2 + 3y^2  = 144 - 24x

x^2 + y^2  = 48 - 8x

x^2 + 8x + y^2  = 48   ........complete the square on x

x^2 + 8x + 16 + y^2 = 48 + 16

(x + 4)^2 + y^2  = 64

The set of points lie on a circle with center (-4,0) and radius = 8

f(x) = 2x + 8 if 1 \le x \le 2

f(x) = 13 - 5x if 2 < x \le 3

f(x) = 20 - 14x if 3 < x \le 4

f(x) = 40 + 5x if 4 < x \le 5

x= 2x + 8                 x = 13 - 5x                   x = 20 - 14x                 x  = 40 + 5x

-x = 8                      6x  = 13                       15x = 20                       -4x  =  40

x = -8                      x = 13/6                         x = 20 / 15  = 4/3        x = -10 

-8 not in interval      13/6 in the interval        4/3 not in interval       -10 not in interval

Answer :   13 / 6

Top rectangle = 6 * 2  =12

Rectangle directly below this = 6 * 1  = 6

Right triangle on the left = (1/2)(3)(3)  = 4.5

Middle rectangle = 6 * 2 = 12

Rectangle below this = 6 * 2 = 12

Half-circle with radius of 5  =  (1/2) pi * (5^2)  = 12.5 pi

Total area =  [ 12 + 6 + 4.5+ 12 + 12 + 12.5 pi ]  cm^2  = 85.77 cm^2

Hey, NTS !!!

Law of Cosines

AC^2  = BA^2 + BC^2  - 2 (BA * BC) cos B

7^2  = 5^2 + 32  - 2 (5 * sqrt 32) cos B

(-8) / (-10 sqrt 32)  = cos B

(-8) / (- 5 * 2 * 4 sqrt 2)  = cos B

1 / (5 sqrt 2)  = cos B

General formula :   arctan (a) + arctan (1/a)  =  90°  if  a is a positive integer

Slope of first line =  2 ....angle formed by first line with x axis = arctan(2)

Slope of  second line = 1/2  ....angle formed by second line with x axis

[ arctan (2) + arctan (1/2) ] / 2  =  = 90 ° / 2 =  45°

m =  slope 45° angle  =  1

sin2A  =  2sinAcosA

sin3A = sin (2A + A)  =  sin2AcosA + sinAcos2A  = 2sinAcos^2A + sinA (2cos^2A -1)

sin4A = sin (2A + 2A)  = 2sin2Acos2A = 4sinAcosA *(2cos^2A - 1)

2sinAcosA  = sinA [ 2cos^2A + 2cos^2A - 1] + 4sinAcosA *(2cos^2A  -1)

(divide out sinA since sinA  = 0 so angle A = 0 , but A is acute)

2cos A  = 4cos^2A -1 + 8cos^3A - 4cosA

8cos^3A + 4cos^2A - 6cosA - 1 =  0

Let cos A  =x

8x^3 + 4x^2 - 6x - 1  = 0

x ≈ .7406

cos A = .7406

arcos (.7406) = A  ≈ 42.217°

tan A =  y /x = - 4 /-1  =  4

cot A =  x /y  = -1 /-4 =  1/4

Nice, Bosco !!!!

41409 / 9  =  4601

Factorization of 4601 = 43 * 107

sum = AB + CDE =   43 + 107  =   150

E = (10, 10sqrt 3)  F = (10sqrt 3, 10)

The area is compsed of a  square whose  area = EF^2  = [ 2(10 - 10sqrt3)]^2  =  400 (2 -sqrt 3) =

800 - 400sqrt 3

And  4  equal sectors  =  4 [ pi /6 * 20^2  - (1/2) (20^2) sqrt (3) /2)]  = 4 (200pi/3 - 100) =

(800/3)pi  - 400

Total area =  400 - 400sqrt 3 + (800/3)pi =    400 [ 1 - sqrt 3 + 2 pi / 3 ]

Here's what I get

1,2,3,4,5,5,5,5,5,5,5,5,6,7,8,9

Mean = 5

Median = 5

Mode = 5

Largest value  =  9

surface area =  2 (LW + LH + WH)  =  24    ⇒    LW + LH + WH  =  12

sum of  edge lengths = 4 [ L + W + H)  = 24   ⇒  L + W + H =   6

Using a little logic and assuming integer values for  each dimensinon, each edge measures 2

Proof

L + W + H = 6                 LW + LH + WH  =12

2 + 2 + 2  =  6                 2*2 + 2*2 + 2*2 =  12

The volume =  LWH =  2*2*2  =   8 cubic units

          P

     5           8

Q         M           R

     5.5       5.5

Law of Cosines  (twice)

[QP^2 - QR^2 - PR^2] /[-2* QR * PR] =  cos PRQ

[ 25 - 121 - 64] / [-2 * 11 *8 ] = cos PRQ  = 10/11

PM^2 = MR^2 + PR^2  - 2 (MR * PR) *cos (PRQ)

PM^2  = 30.25 + 64 - 2[5.5 * 8]* (10/11)

PM ^2  =14.25

PM =sqrt [14.25] ≈ 3.77