CPhill

2(-1)^2 -8 (-1) + 1   =  11

2(2)^2 - 8(2)+ 1 =  -7

Range =  [ -7, 11 ]

\(f(x) = \frac{1}{1+\frac{2}{1+\frac 3x}}. \)

x cannot  = 0

1 + 3/x  =  [ x + 3] / x

2 /  [ ( x + 3) /x ] =   2x / [x + 3]

x cannot = -3

1 + 2x  / [ x + 3]  =   [ x + 3 + 2x ] / [ x + 3] = [ 3x + 3] / [x + 3]

1/ ([3x + 3] /[x + 3]) =    [x + 3] / [ 3x + 3]

x cannot = -1

(c) Does there exist a combination of red potion and blue potion that can produce a potion that is 75% magical syrup by volume?

Not possible since neither potion exceeds 60 %  magical syrup

(b) Find the amounts of red potion and blue potion (in mL) that can be combined in order to produce 100 mL of a potion that is 54 magical syrup by volume.

Let x be the amount of red potion and 100-x be the amount of blue

.60x + .30 (100-x)  = .54 * 100

.60x + 30  - .30x  = 54

.30x + 30 = 54

.30x = 24

x = 24/.30  = 80

80ml of red must be added

20 ml of blue must be added

(a) Find the amount of red potion (in mL) that must be added to 500 mL of blue potion in order to produce potion that is 40% magical syrup by volume.

Let x be the amount of red potion to  be added

.60x  + 500 * .30  =   (500 + x) *.40

.60x + 150  = .200 + .40x

,60x -.40x = 200 - 150

.20x = 50

x = 50 /.20  = 250ml  of red must be added

https://web2.0calc.com/questions/algebra_55993

Simplify as

x^2 + 12x - 4   = 0

Let the roots = a,b

Sum of the  roots  =  a + b =  -12

Product of the roots = -4  = ab  →   -8 = 2ab

a + b  = -12        square both sides

a^2 + 2ab + b^2 = 144

a^2  - 8 + b^2 =144

a^2 + b^2  = 152

Slope of line = -1

Slope of any tangent line to the circle=   -x/y

So

-x/y = -1

x/y =1

x = y

But 

x + y =3 + k

x^2 + y^2 = k

So

x + x  =  3+k

x^2 + x^2 = k

So

2x = 3+k    →  x = (3 + k) / 2

So

2x^2  = k

2 [ (3 + k) / 2] ^2  = k

(3 + k)^2  = 2k

k^2 + 6k + 9 = 2k

k^2 + 4k+ 9  = 0

The discriminant =  16 - 4 * 1 * 9  =   -20

So....k is  not a real value

So....no real solution

                A

            30   30

B                D              C

Impossible

AD is a bisector

Angle CAD  =  60/2 =  30

So...it can't also  = 45

 v

The circles have the same radius....set their equations equal

x^2 + (y -1)^2  = (x-1)^2 + y^2

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

-2y =-2x

y = x

So

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

2x^2 - 2x + 1 = 1

2x^2 - 2x =0

x^2 - x =0

x ( x -1)  = 0

Solving this produces x =0  and x =1

Since x = y

x = 0    y  = 0

x =1    y =1

Points of intersection are (0,0)  and (1,1)

PQ =  sqrt [ (1-0)^2 + (1-0)^2 ]  = sqrt 2