CPhill

Note that triangle QVR is similar to triangle PVS

And QR /PS = 5/7

Therefore QV / PV = (5/7)

QV = (5/7)PV 

PQ^2  = PV^2 + QV^2

PQ^2 = PV^2 +[ (5/7)PV]^2

3^2 =  PV^2 + (25/49)PV^2

9 = [74/49]PV^2

[441] / 74 = PV^2

21 / sqrt [74 ] = PV

QV = (5/7)* 21 /[sqrt [74]

QV = 15 / sqrt [74]

PS^2 = PV^2  + VS^2                     QR^2 = QV^2 + VR^2                               

7^2  = [ 441/74] + VS^2                     5^2 =  [225/74] + VR^2

49 - [441/74 ]  =VS^2                        25  - [225/74] = VR^2

3185/74 = VS^2                                 1625/74 = VR^2

RS   =sqrt [  VS^2 + VR^2]

RS  = sqrt [  [ 3185 + 1625 ] / 74 ] 

RS = sqrt [ 4810 / 74 ] 

RS =  sqrt [ 65 ] 

Simplify as

x^2 + 15x + k   

Double Root means discriminant = 0

15^2  - 4 (1)(k) = 0

225 - 4k  = 0

225 = 4k

k = 225 / 4

Impossible

Tangent - Secant Theorem

XY^2 =  XB ( XB + BA)

8^2  = 10 (10 + BA)

64  = 100 + 10BA

-36 = 10BA

BA   -36/10 =  =  -3.6       (BA is a  distance and  must be positive)

https://web2.0calc.com/questions/sequence-question_5

Write as

-(1/4)x^2 + (5/2)x  =  0

Find the roots

Factor as

-x [ (1/4)x - (5/2)  ]  = 0

Set the factors  to  0 and  solve  for  x

-x  =  0            (1/4)x - (5/2)   =0

x  = 0              (1/4)x = (5/2)

                              x = (5/2)(4/1)  = 10

The roots are

x = 0   and  x  = 10

The horizontal dstance =  10 -  0    =    10 (meters)

Simplify as

x^2 +10x +  p

Take 1/2 of 10 = 5

Square it =  25  =   p

x^2 + 10x + 25 =   (x + 5) (x + 5)  = (x + 5)^2

                        20 A

                                      E                    

                     144 H        

        B                  D               C

ABH forms a triangle

Angle AHB = 144

Angle BAH = 20

So angle ABH =  180 - 144 - 20  =  16° 

1,2,3,4,5,6,6

Number of  identifiable sequences = 

(Number of rolls)   ! / ( Number of Repeated 6's ) ! =   7! / 2! =    2520

Semi-perimieter  = (15 + 9 + 10)  / 2 =    17  = s

Area = sqrt [ s ( s -15) (s- 10) (17-9) ] =  sqrt [ 17 * 2 * 7 * 8 ]  = sqrt (1904)

Shortest altitude, sa,  is  always drawn  to  the  longest base

Area = (1/2) (AB) (sa)

sqrt (1904) = (1/2(15) ( sa)

sqrt (1904) / 7.5  = sa ≈ 5.82

The point  (0,3)  is  on the line = the y intercept

The equation of the  line is

y= 2x + 3

The x intercept is

0 = 2x + 3

-3 =2x

x = -3/2

Area of triangle   =   (1/2) (3/2) (3)  = 9/4