CPhill

Simplify as

-5x^2 - 6x + 10

We have th form   ax^2 + bx + c

The x coordiante of the vertex =  -(b) / (2a) =   - (-6) / (2 * -5 )  =  6 / -10    =  -,6

The y coordinate is   -5(-.6)^2 - 6(-.6) + 10  =   11.8

Simplify as

x^2 + 15x + k   

This will have a double root when the discriminant =  0

15^2 - 4 (1) k  = 0

225 = 4k

k =  225 /  4

Since both of the radiuses = 1, set these circles =

x^2  + (y-1)^2  = (x -1)^2 + y^2

x^2  + y^2 - 2y + 1   = x^2 -2x + 1 + y^2

-2y =-2x

x = y

So

x^2 + (x -1)^2 = 1

x^2 + x^2 -2x + 1  =1

2x^2 - 2x  = 0

x^2 -x =0

x ( x -1)  = 0

x=0     x=1

The intersection points  are (0,0) and (1,1)

PQ =   sqrt [1 + 1]  =  sqrt 2

Find the area with Heron's Formula

semi-perimeter = 17

Area =  sqrt [ 17 * 2 * 7 * 8 ]  = 4sqrt 119

The shortest altitude is drawn to the longest side....so....

4sqrt119 = (1/2)  15  * altitude

altitude  =  4/ 7.5  * sqrt 119  =   4 / (15/2) * sqrt 119  =  (8/15)sqrt 119

a^3 - 1/a^3  =  (a -1/a) (a^2 + 1 + 1/a^2)

a^3 - 1/a^3 =  (0)  ( a^2 + 1 + 1/a^2)  =  0

a11 = a10 + a9

1  = 4  + a9

a9 = -3

a10 = a9 + a8

4 = -3 + a8

a8 = 7

a9 = a8 + a7

-3 = 7 + a7

a7 = -10

a8 = a7 + a6

7 = -10 + a6

a6 = 17

              A

        15       45                   

               24

B               D                C

If AD is a  bisector  and angle BAC  = 60

Then angle CAD should be 30 , not 45

You are correct, NTS !!!

This can't be a quadratric  because of the Conjugate Property

If a + bi is a  root so is a - bi

If c + di is a root so is c - di

We would have 4 complex roots, not  just two

$\left(\sin(a)\right)^7 + \left(\cos ( a)\right)^7< 2*sin(a)^4*cos(a)^4.$

$0 < a < \frac{\pi}{2}$

This doesn't appear to be true

Simplify as

8x^2 + 8x + 28 = 0

2x^2 +  2x + 7 = 0

x =  [-2 + sqrt (-52) ] / 4  =  [-2 + 2sqrt (13) i ] / 4  =  -1/2 + (1/2) sqrt (13) i

"a"  is not positive

a * b =  (-1/2) (1/2)sqrt 13  =   -sqrt (13)  /  4