CPhill

I'll  do  one.....the calculation of the  other two is similar

(The median to BC is AF , not CF)

Mid point of BC  = [ (-9 + 1)  /2 ,  (6 + -4) /2)  =  (-4, 1) =  F     (5,4) (-4,1)

Slope of a line through AF  =  ( 1 - 4)  / (-4 - 5)  =  -3/-9  = 1/3

Equation of line containing  median AF   using the  slope and (5,4)  =

y = (1/3) ( x - 5) + 4

y = (1/3)x - 5/3 + 4

y = (1/3) x + 7/3

If the mean  =  0......then the sum of all the oods must =  0  

The 500th positive odd   =  2 (500) - 1  =  999

The 500th negative odd  =  2 (-500) + 1   = -999  =  the least number

We can use  somethng called "Pick's Theorem"

Lattice point =  a point defined by the intersection of  grid  lines

Interior  lattice points   =  17

Boundary lattice points =   8

Area =  interior lattice points  + boundary lattice points / 2     - 1   =

17 + 8/2  -  1    =

14 + 4   - 1     =

20  

THX, Imcool.....very observant !!!!

Best to check the logic of some of these questions  before attempting an answer 

We want to find the intersection points of

(x -2)^2  + (y -7)^2   = 25       and  y = 5x -28

So

(x -2)^2  + (5x -28 - 7)^2   =   25

(x -2)^2  + (5(x -7))^2    = 25

(x -2)^2 + 25(x - 7)^2  = 25

x^2 - 4x + 4 + 25 (x^2 - 14x + 49)  = 25

x^2  - 4x + 4  + 25x^2 - 350x + 1225 - 25 =  0

26x^2 - 354x +  1204   =  0

x =  (354 + sqrt [ 354^2  - 4*26*1204 ])   / 52    =  7    and

x =   (354 - sqrt [ 354^2  - 4*26*1204 ])   / 52 =   86/13

So   y = 5(7) - 28 =  7

And

y = 5(86/13) - 28  =  66 /13

So  the points  are   (7,7)   and (86/13  , 66 / 13)

        W   60                     120   X

                                        -          

                        -      

          -                         

Z                                            120     Y

This is impossible  .....triangle WXZ's   interior angle sum would exceed 180°

Draw altitude CN  to AB =  h

We will have a right triangle CNB

Let BN = x

And angle NCB = 90 - 55 = 35

Using the Law of Sines

h /sin (55) = x / sin 35

h = xsin 55 / sin 35       (1)

Next, draw altitude DO  to AB  = h

We have another right triangle

Let AO = 96 - (44 + x) =  52 - x

Angle DAO = 180 - 110  = 70 

Angle ADO  = 90 - 70  = 20

Using the Law of Sines again

h / sin 70  = (52 - x)/ sin 20     

h =  (52 - x)sin 70 / sin 20    (2)

Equating (1) , (2)

xsin 55  / sin 35  =  (52 - x)sin (70) / sin 20

x (sin 55 / sin 35)  =  52(sin 70 / sin 20)  - x ( sin70 /sin20)

x [ sin 55 / sin 35 + sin 70 /sin20 ]  = 52 (sin70 /sin20)

x =  (52)(sin 70 / sin20) / ( sin55 / sin35 + sin 70/sin 20)  

h = (52)(sin 70 / sin20) / ( sin55 / sin35 + sin 70/sin 20)  * (sin 70 / sin20) ≈  94

Area =   (1/2) (94) (44 + 96)  = 47 * 140  ≈  6580 

This isn't always true, Imcool....see here :

However....it is true that the sum of the interior angles =  360

This one is pretty easy

The triangle is equilateral

The altitude of an equilateral triangle  is a median

And since M is the midpoint of ST, then UM is also a median

Their intersection, X, is the centroid of STU

SN = sqrt [ST^2 - TN^2]  = sqrt [ 8^2 - 4^2]  = sqrt [ 48]  = 4sqrt (3)

By a property of centroids ....SX = (2/3)(SN)

So

SX = (2/3) * 4 (sqrt3)  = 8sqrt (3)  / 3  =  8 / sqrt 3   ≈  4.618

Oscar and Felix leave the same house driving separately to the same conference, along the same route, at a speed of $x$ mi/h. If Oscar had driven $10$ mi/h faster, he would have driven for $15$ fewer minutes. If Felix had driven $30$ mi/h faster, he would have driven for $20$ fewer minutes. What is the distance from their house to the conference?

Let D be the distance , x be the normal rate  and  T  be the normal time (in hours)

15min  =1/4 hr     20 min = 1/3 hr

We have this system

D / x    =  T     →   D = Tx  (1)

D / ( x + 10)  = T - 1/4     (2)

D / ( x + 30)  = T - 1/3     (3)

Sub (1) into (2) , (3)

Tx / (x + 10) = T -1/4

Tx / ( x + 30) = T -1/3

Tx  = (T -1/4)(x + 10)

Tx  = (T -1/3)(x + 30)

Tx = Tx - (1/4)x + 10T - 5/2

Tx = Tx -  (1/3) x  + 30T -10

(1/4)x - 10T  = -5/2  →   x - 40T   =  -10      (4)

(1/3)x - 30T = -10  →    x - 90 T  =  -30       (5)     subtract (5) from (4)

50T = 20

T = 2/5 hrs

x =  40T - 10  =  40 (2/5)   - 10   =   6 miles/hr

The distance is   (2/5 - 1/4) (6 + 10)  =  (3/20)(16)  = 48/20  = 24/10  = 2.4 miles