CPhill

x^3 - x^2 -21x + 45  factors as   (x-3)^2 (x + 5)

Multiply through by this factorization  and  we  get

1  =  A(x-3)^2 + B (x-3) ( x + 5)  + C(x + 5)

Let x = -5

Then  A(-8)^2  = 1  →  A = 1/64

f(a)  =  (13 + a) / (3a + 7)

There is  no max value

At  a = 0   f(a)  =  13/7

As a →  inf,    f(a) →  1/3

See the graph here : https://www.desmos.com/calculator/reaplfkcny

The first function will  be <  0  on [-4,  -3]

x^2 + 5x + 6  = ( x + 3) ( x + 2)

At  x = -3 ....   x^2 + 5x + 6  =  (-3 + 3) (-3 + 2)  =  0

At x = -4........x^2 + 5x + 6  =   (-4 + 3) (-4 + 2)  = 2

So  x = -4  is  correct

See the graph here : https://www.desmos.com/calculator/nf9utyhcye

https://web2.0calc.com/questions/quadratic_39762

PT = PU  )tangents form a point outside a circle are equal_

PO = PO

TO = UO  =   radii of the same  circle

So, by SSS  triangle TOP   is congruent to triangle UOP

So angleTOP  = angle UOP  = 45

So angle TOU = 90

This is a  central angle intercepting minor arc TU  ...so minor arc TU  = 90

https://web2.0calc.com/questions/triangle_4602

https://web2.0calc.com/questions/geometry-question_42617

Here's what I think......

https://web2.0calc.com/questions/coordinates_87801#r1

\( \frac{a}{3} + 1 = \frac{a + 3}{a} + 1 \)

a/3 = (a + 3) / a       cross-multiply

a^2  = 3(a + 3)

a^2  = 3a + 9

a^2 - 3a  = 9

a^2 -3a + 9/4 = 9 + 9/4

(a - 3/2)^2  = 45 / 4      take both roots

a - 3/2  =  + / -  3sqrt 5 / 2

a = [ 3 +/- 3sqrt 5 ] / 2

\(xy + x = \frac{3x + 2y + z + y + 2z}{3} \)

3x (y + 1) =  3x + 2y + z + y + 2z

3x (y+ 1) - 3x = 3y + 3z

3x (y + 1 - 1)  = 3 (y + z)

x (y )  = (y + z)

         ( y + z)

x =     _________

               y