CPhill

2x+3y=6 (for x) 

Subtract 3 from both sides

2x  = 6 -  3y      divide both sides by 2

x  =  [ 6 -  3y ]  / 2

Fm-3=Q (for F)   →      F  =   [Q + 3] / m

This will have vertical asymptotes as well as a "slant" asymptote....

The  y  ...and x....   intercept is at x  = 0

It will be defined everywhere except at the x values that make the denominator  = 0

We can find these values as follows :

x^2 -  3  = 0

x^2  = 3    take both roots

x  = ± √3     the graph will have vertical asymptotes at these values

To find the asymptote ......    divide x^3  b x^2 - 3 and disregard any remainder

                      x

x^2 - 3     [    x^3           ]

                     x^3 - 3x

                             3x 

So.....the asymptote is  y  = x

There are other Calculus techniques involved in graphing this...but, let's cut to the chase.....you can see it here :    https://www.desmos.com/calculator/gedjottav7

I can't upload a pic....so use your imagination ....!!!!

Draw DH  perpendicular to AE

By SAS.....ΔABE  = ΔBEF

<BAE = <CBF 

<ABF  + <CBF  = 90

So <BAE + <ABF  =90

So <AGB  =90

<BAE + <DAH  = 90

<BAE + <ABF  = 90

So < DAH  = < ABF

And < AHD =< AGB  =90

So by  AAS .......ΔHAD  = ΔGBA

And by AA congruency

ΔABE ≈ ΔAGB   →  ΔBEA ≈ ΔGBA

Since AB  = the side of the square and BE  = (1/2) the side of the square

Then...... 2BE  = AB   and, by similar reasoning.....  2GB = AG

Now AG  = HD   and by the same reasoning as before 2AH  = HD  →  2AH = AG

But

AG  =AH + GH

2AH = AH + GH

So   AH  = GH

By SAS

ΔAHD = ΔGHD

And <HAD  = < HGD

But in Δ AGD, <GAD = <AGD

Therefore.......DG  = AD

But      AD  = AB

Therefore.......DG  = AB

If this stupid thing will let me upload a pic.....I'll do so....!!!!

3 / 5  = x / 2    where x = cans of blue  ..........  cross-multiply

3 * 2   =  5 * x

6  = 5x     dide b 5

6/5   =  x  =   1 + 1/5   cans of blue

246 / 574  =

[ 123 * 2 ]  /  [ 287 *2 ] =

123 / 287  =

[41 * 3 ] / [ 41 * 7] =

3 / 7

I assume we have :

9^x=27^y     (1)

64^(xy) = 512^(x+1)   (2)

(1)  can be written as

(3^2)^x  = (3^3)^y    → (3)^(2x)  = (3)^(3y)  ... equate exponents  →   2x = 3y  →  y  = (2/3)x

Sub this into (2)

64^(x* (2/3)x )  =  512^(x + 1)

(2^6)^[ (2/3)x^2]  = (2^9)^(x + 1)

(2)^[4x^2]  = (2)^[9x + 9]     equate exponents again

4x^2  = 9x + 9

4x^2 - 9x - 9  = 0      factor

(4x + 3)(x - 3)  = 0   set each factor to 0   

And

x = -3 / 4    → y  = (2/3)(-3/4)  =  -1/2 

Or

x  = 3    →  y  = (2/3)(3)   =  2

So....the solutions are   (x, y )  =    ( - 3/4, -1/2)     or  (3, 2)

3(x+y) - 5(x-y)  =

3x + 3y  - 5x + 5y  =

-2x + 8y

3x+23=54       subtract 23 from both sides

3x =  31         divide by 3 

x   = 31 / 3

The had a total of  $28/.20 = 140 coins...so...

A + B + C + D  = 140   ( 1)

A + D  = B + C   →  A + D - B - C  = 0  (2)

A - C - D = 2    →  A - 2  = C + D   (3)

B + D - 8  = A + C      (4)

Add (1)  and (2)   →  2A + 2D  = 140  →  A + D  = 70 →  D  = 70 - A   (5)

Put (5)  into (3)  →   A - 2  = C + ( 70 - A)  →   2A  - 72  = C   (6)

Put  (5)  and (6)  into (4)

B + [70 - A] - 8  =  A + [2A - 72]   →   B  =  4A - 134

So we have

A + B + C + D  = 140

A + [4A - 134] + [2A - 72] + [70  - A]  = 140

6A =   276  →  A  = 46  →   $9.20    =  $2.20 profit

D = 70 - 46  =   24   →  $4.80   =  $2.20  loss

C  = 2(46) - 72  = 20  →  $4.00 → $3.00 loss

B = 4(46) - 134  = 50   →  $10.00 → $3.00 profit

Notice that this is a "zero sum" game with regard to profit and losses...just as we would expect !!!!

The constant  willl occur at the 5th  term in the binomial expansion of this  =

C(6,4) * (2x^2)^2 * (1/x)^4  =

15 (4x^4) (1/x^4)  =

15* 4   =

60