CPhill

Law of Cosines

BC^2  = AC^2 + AB^2  - 2 (AC * AB) cos (BAC)

7^2  = 5^2 + 3^2 - 2 ( 5 * 3) cos (BAC)

[ 7^2 -5^2 -3^2] / [ -2 * 5 * 3] =  cos (BAC)   =  -1/2

arccos (-1/2)  =  BAC =   120°

Since AD is a bisector ofthis angle, then BAD  = 60°

cos 60°  = cos (BAD)  =  1/2

side AB  is contained on the line

y = (3/5)x

side AC is  contained on the line

y = -(1/4)(x -32)  = (-1/4)x + 8

The x intersection of these lines is

(3/5)x  = -(1/4)x + 8

[ 3/5 + 1/4] x  = 8

[17/20]x =8

x = 8 *20 / 17  = 160 /17

And the y value at this  point =  (3/5)(160 /17)  = 96/17  = the altitude...call this AD

[ABC]  =  (1/2) BC * AD   =  (1/2)  32 * 96/17 =  1536  /  17  

Complete the square on x, y

x^2 -10x + 25 + y^2 + 10y + 25  =  25 + 25 - c

(x -5)^2  + ( y+ 5)^2  =  50   - c

We want 50 - c  =1

c =  49

We have 6 equilatera ltriangles with sides  =3

The area =    6 (1/2) *3^2 * sqrt 3  /2 =  (27/2)sqrt 3

https://web2.0calc.com/questions/geometry_74033

When Dirk finishes, Edith has run 14 km and Foley has run 9 km

So ....  his rate to  her rate =  9/14

When  she finishes, Foley has run   (9/14 )* 24 km

So he is   24 -  (9/14) * 24  =  24 ( 1 - 9/14)  =  24 * ( 5/14)  ≈  8.57 km behind

AP = sqrt [5^2 + 4^2]  = sqrt 41

PQ = sqrt [2^2 + 1^1] =  sqrt 5

AQ = sqrt [ 5^2 + 2^2] = sqrt 29

Law of Cosines

PQ^2  = AP^2 + AQ^2  - 2 (AP * AQ) cos (PAQ)

5  = 41 + 29 - 2 [sqrt [41] *sqrt[29] cos (PAQ)

cos (PAQ)  =  [ 5 - 41 - 29 ] / [ - 2sqrt [41]sqrt[29] ]  = 65  / sqrt [ 4756 ]

sin PAQ =   sqrt [ 1 - 65^2 / 4756 ]  =  3sqrt [ 70151] / 2378

https://web2.0calc.com/questions/help-pls_105

https://web2.0calc.com/questions/algebra_78048

\( x \mathbin{\spadesuit} y = \frac{x^2}{y} \)

\($a \mathbin{\spadesuit} (a + 1) = 9 \)

So, using substitution

(a)^2 / (a + 1)  = 9

a^2  = 9a + 9

a^2 -9a   =  9         complete the square

a^2 - 9a + 81/4  = 9 + 81/4

(a - 9/2)^2  =  117 /4

a -9/2 =  +/- sqrt [117] /2

a = [ 9 + sqrt (117) ] / 2         and       a = [ 9 - sqrt (117) ] / 2

We need to  make f(2x/3)  = f(3)

So...   2x / 3  = 3 →  2x = 9 →  x = 9/2  = 4.5

So....remembering that f(3)  =  4 we have

5 + f( 2(4.5) / 3) / 11  =   5 + f(9/3) / 11  =   5 + f(3)/11  =  5 + 4 / 11  =  59 / 11

The point is  (4.5 , 59/11)

x + y =  13 →  y = 13 - x        (1)

Construct a circle with a radius of 5 centered at (2,7)

(x - 2)^2  + (y -7)^2 =  25       (2)

Sub (1)  into (2)

(x -2)^2  + ( 13 -x -7)^2  =25

( x -2)^2 + ( 6 - x)^2  = 25

x^2 -4x + 4 + x^2 -12x + 36  =25

2x^2 -16x + 15  = 0

x = [ 16  +  sqrt [ 256 - 120] ] / 4 =   [16 + 2sqrt 34] / 4  =  4 + sqrt (34) / 2

y = 13- (4 + sqrt (34)/2)  =   9 - sqrt (34) /2

x = 4 -sqrt (34)/2

y = 13 - (4 -sqrt (34) /2)  =  9 + sqrt (34)/2

The points are 

( 4 + sqrt (34)/2 , 9 - sqrt (34)/2)   and ( 4 -sqrt (34)/2, 9 + sqrt (34) / 2)

Rewrite as

-15x^2 -kx + 53 = 0 

This has no real solutions when the discriminant is < 0

k^2 - (-15)(4)(53) < 0

k^2 + 3180 < 0

The discriminant is never < 0  so no k  produces non-real solutions

Since M is the midpoint of TU  and P  is the midpoint of ST, then MP divides sides TU and TS of triangle TUS proportionally...so  ....MP is parallel to US

So  angle TMP  = angle TUS  = 89

Angle TZU = 90   Angle UTS  = 29

So angle TUZ =  90 - 29  =  61

So angles TMP  + TUZ  =  89 + 61 =  150

The given line  has a slope of   1

A perpendicular line to  this  passing through  (7,1)  has the equation

y = - ( x -7) + 1

y = -x + 8

The x intersection of these two lines is

x + 3 =  -x + 8

2x = 5

x =5/2

And y = (5/2) + 3 =  11/2

From  (7,1)  to ( 5/2, 11/2)  we went  back to the left  7 -5/2  = 9/2  unirs

And  11/2 - 1  =  9/2  units up

So

B =  [ 5/2 - 9/2,  11/2 + 9/2]  =  ( -2, 10 )

Simplify as

x^2  - mx  + 14  =  0  

Possible linear factors

( x - 2) ( x -7)      m =  9

(x + 2) (x + 7)     m =  -9

(x - 1) ( x -14)    m  = 15

(x + 1) ( x + 14)  m = -15