CPhill

4a + 6b = 5

        (2 - m) b  =  k

This only has infinite solutions if  both equations  are the same......not possible

2(x + 8)(x - 4p)

2 (x^2 + (8 -4p)x - 32p)

2x^2 + (16 - 8p)x - 64p      (1)

The x coordinate of the vertex is  [ - (16 - 8p) / (2*2) ] =  [8p - 16) /4 = 2p - 4

So sub this into this equation

2(2p-4)^2 + (16 - 8p)(2p -4) - 64(2p -4)  =  -32

Simplifying

-8(p^2 + 12p - 28)  = -32

p^2 + 12p - 28  = 4

p^2 + 12p - 32  =  0

p^2 + 12p  =  32

p^2 + 12p + 36  = 32 + 36

(p + 6)^2  = 68       take both roots

p + 6 = sqrt (68)                    p+ 6  = -sqrt (68)

p =  2sqrt (17) - 6                  p = -2sqrt(17)  - 6

To have a double root, the discriminant must  =  0

So

11^2  - 4(2)k  = 0

121 - 8k  =  0

121  = 8k

k = 121 /  8

Simplify  as

x^2 -3x  -  p

{ p must be negative }

To find p

take (1/2) of (-3)  = -3/2.....square it =  9/4

p = -9/4

Check

x^2  -3x  - (-9/4)  =

x^2 -3x + 9/4  =

(x -3/2)^2

Here :https://web2.0calc.com/questions/parabola_31828

a11 =  a10 + a9

4 =  1 + a9

a9 = 3

a10 = a9 + a8

1 =  3 + a8

a8 = -2

a9 = a8 + a7

3 = -2 + a7

a7 = 5

a8 = a7 + a6

-2 = 7 + a6

a6 = -9

Simplify as

x^2 -21x + y^2 + 17y + 8

Take the derivative with respect to x and set to  0

2x -21  =  0

x = 21/2

Take the derivative with respect to y and set to 0

2y + 17 = 0

y = -17/2

The coldest temp  is

(21/2)^2 - 21 (21/2) + (-17/2)^2 + 17(-17/2) + 8    =  -174.5

Just a slight error by EnormousBighead....

BC has to be >  5 because angle A > angle B

AC/ sin 34  = BC / sin 108

5 / sin 34 = BC / sin 108

BC  = 5*sin 108 / sin 34  ≈ 8.50

Nice solution, EnormousBighead   !!!!

Simplify as

-t^2 + 23t +  25

This is a  parabola that turns downward

The t that produces the max value is    (-23) / (2 * -1)  =  23/2

The max value is   - (23/2)^2 + 23(23/2) + 25    = 629/4  =  157.25