CPhill

625x = 9

x = 9/625

So

125 (9 / 625) = (125 * 9) / 625  =    (5 *25 * 9) / (25 * 25 )  =   45 / 25 =  9 / 5

2 / (2 - 3x)

The denominator cannot = 0   so x cannot =  (2/3)

So  domain =  (-inf, 2/3) U (2/3 , inf)

We have a lower power polynomial over a higher power polynomial....in such a case we will have a horizontal  asymptote at y  = 0

So....the range =  (-inf, 0) U (0, inf)

f(x)  = e^x

Proof  

(x, f(x) )  = (0,1)  means that  e⁰   =1

(x, f(x)) = (1,e)  means that e¹ =  e

Let the Blue circle = Blue ribbons

Orange circle = Green ribbons

Purple circle = White ribbons

____ =  (81 * 5) / 3 =  (81/3) * 5  =   27 * 5 =   135

Let A =  (0,1)

Let B = (1,1)

Let X = ( 1, 1/2)

Let D = (0,0)

The beam will reflect from  X to D

The path length =  2 * AX  = 2 sqrt [ ( 0 -1)^2 + (1 -1/2)^2 ]  = 2 sqrt [ 1 + 1/4] =  2 sqrt (5) / 2  = sqrt 5

Angle BDX = 90 = angle ADX

DA = BD

XD = XD

So, by LL, triangles BXD and AXD are congruent right triangles

And by the same reasoning triangles, AXE and CXE  are congruent right triangles

And angle DXE  = the angle formed by the intersection of  lines p,q = 45 = the sum of  angles AXD and   AXE

But angles  BXD and CXE  = angles AXD and AXE....so their sum is  also = 45

So angle BXC = AXD + AXE + BXD + CXE    =    45 + 45  =  90

A  will have an area of (-1/5)^2   =1/25  of XYZ

So         (1/25) [ XYZ ] / [ XYZ ] =  1/25

Triangle ADC is right with angle ADC = 90

AC = 2

So DC = 2 / sqrt 2 =  sqrt 2 =  AB  = x

(2x^3 + x^2 - 3x + 5)  ( x^4 - 7x^3 + tx^2 - 11x + 15)

2x^7 - 14x^6 + 2tx^5 - 22x^4 + 30x^3

              x^6  -  7x^5  +tx^4   -   11x^3 + 15x^2

                      -  3x^5  +21x^4 -  3tx^3 +  33x^2 - 45x

                                     5x^4  +  35x^3 + 5tx^2 - 55x + 75

To  have no x^2 term

15 + 33 + 5t  =  0

48 + 5t =  0

-5t = 48

t = -48/5