CPhill

Here's the solution ....

Note that we should have this equation

9x + 17y  =  541           which we can manipulate as follows

9x  =     541  - 17y

x  =        541  - 17y

             ________

                     9

x  =  541   -  (18 - 1)y

       _____________

                   9

x  =  541 + y                18y

       _______     -        ____

             9                      9

x  =  540  + 1 +  y                18y

        ___________   -         ____

                   9                         9

x =  540           1 + y             18y

      ___   +       _____  -       ___

        9                  9                 9

Note that the first and last terms will always be divisible by 9

And.....the middle term will be divisible by 9  when y  =  8, 17  or 26

So   we have the following possibilities for x

x  =  541 - 17(8)        =      45

        _________   

               9

x =   541 - 17*17    =    28

        __________

                9

x =   541  -   17*26      =   11

        ____________

                  9

Note   if y =  35,we get the next multiple of 9...but  x  = -6 if this is so....    so....these are the only possible solutions

So....the solutions  (x, y)   = ( 11,26)  (28, 17)  (45, 8)

Just add these coordinates and you will get your answer, CU

Here, Logic  :   https://web2.0calc.com/questions/geometry_64537

(c)         x^2

          ______

            1  -  x

Unfortunately....other than a graph, we will need to use some Calculus to find the range..this is a little involved, but not that difficult....hence.....the graph might be the best way to go :

https://www.desmos.com/calculator/vdr0itmcy5

The graph shows that the range is  (-inf, 4]   and [ 0, inf )

(b)         3x + 1

             ______

               x + 8

We  have     a   same degree polynomial  / same degree polynomial......this means that we will have a horizontal asymptote    at   y   =    ratio of  the coefficients on x  in the numerator/denominator  

So....the horizontal asymptote occurs  at     y  =  3 / 1  =  3

So...the range will be   (-infinity, 3)   and  (3, infinity )

Check the graph, here :   https://www.desmos.com/calculator/umaqjrwwtr

The lines y = (5/12)x and y = (4/3)x are drawn in the coordinate plane. Find the slope of the line that bisects the acute angle between these lines.

Here's one way, Logic.....but I don't know if it's the most elegant

Construct a circle centered at the origin with a radius of 1

The  equation  is

x^2 + y^2  = 1

We can find the x coordinate  of the intersection of the first line and this circle,thusly :

x^2  + [ (5/12)x ]^2  = 1

x^2 + (25/144)x^2   =1

[144 + 25] / 144  x^2  = 1

[169] /144 x^2  =1

x^2 = 144/169

x = 12/13

And y = (5/12)(12/13) = 5/13

So   ( 12/13, 5/13 )  is on the circle

Likewise...we can find the intersection of this circle with the  second line :

x^2  +[ (4/3)x\^2  =1

x^2  + (16/9)x^2  = 1

[9 + 16 ] / 9 * x^2  =1

25/9 x^2  =1

x^2  = 9/25

x = 3/5

And y = (4/3)(3/5)  = 4/5

So  (3/5, 4/5)  is on the circle

And a chord can be drawn on this circle with endpoints   ( 12/13, 5/13)  and (3/5, 4/5 )

And the midpoint of this chord is

  (  [12/13 + 3/5]  / 2,  [ 5/13 + 4/5 ] / 2 )  =

(99/130 , 77/130  )

And  the line drawn  from the origin to this point is the line we seek and it will have the slope

[77/130]  / [99/130]  =    77/99   =  7/9

2.     [  log (a / b) ]^2  +  [ (log ab) ] ^2  =  20         and    (log a - log b) log(ab)  = 8

Note that

log (a / b)  = log a - log b

log (ab)  =   log a + log b

So we have

(log a - log b)^2  + (log a + log b)^2  =  20         and   (log a - log b) ( log a + log b )  = 8

Let log a  = u       Let log b = v

So

(u - v)^2   + ( u + v)^2  = 20                                 and        (u - v) (u + v)  = 8

u^2  - 2uv + v^2   + u^2 + 2uv + v^2  = 20                         u^2  - v^2  = 8     (2)

2u^2 + 2v^2  = 20

u^2 + v^2  = 10    (1)

Add (1)  and (2)    and we have that

2u^2  = 18

u^2  = 9

u = ±3  

So 

u =  log  a  = ± 3

So

l log a  l     =   l ± 3 l  =      3

Using the change-of base theorem  we have

log (3x + 4)             log (3x + 4)         log (3x + 4)               (1/2)log(3x + 4)

_________    =      __________  =   ______________=    ____________    (1)

log 4                          log 2^2                2 log 2                         log 2

log ( 7x + 8)        log(7x + 8)           log (7x + 8)                (1/3)log(7x + 8)

_________   =  __________  =  _____________ =        _____________   (2)

log 8                    log 2^3                  3 log 2                           log 2

The arithmetic difference must  be   (2) - (1)  =

(1/3) log (7x + 8) - (1/2)log(3x + 4)

__________________________

                  log 2

This implies that

log (x + 2)               (1/3)log(7x + 8) - (1/2)log (3x + 4)              (1/2)log(3x + 4)

________    +        ____________________________   =    ______________

  log   2                                  log 2                                                log 2

;log (x + 2)  +  (1/3)log(7x + 8) - (1/2)log(3x + 4)  =  (1/2)log(3x + 4)

log (x + 2) + (1/3)log(7x + 8)    =   log(3x + 4)

3 log( x + 2)  + log(7x + 8)  = 3log(3x + 4)

log ( x + 2)^3 + log(7x + 8)    =  log (3x + 4)^3

log  [ (x + 2)^3 * (7x + 8)]  = log(3x + 4)^3

Which implies that

(x + 2)^3 (7x + 8)  = (3x + 4)^3

(x^3 + 3x^2*2 + 3x*4 + 2^3) (7x + 8)   =  (3x)^3  + 3* (3x)^2*4 + 3* 3x*4^2 + 4^3

( x^3 + 6x^2 + 12x + 8)(7x + 8)  =  27x^3 + 108x^2 + 144x + 64

7 x^4 + 50 x^3 + 132 x^2 + 152 x + 64  =  27x^3 + 108x^2 + 144x + 64

7x^4 + 23x^3 + 24x^2 + 8x  = 0

x (7x^3 + 23x^2 + 24x + 8)  = 0

x = 0 is one solution

7x^3 + 23x^2 + 24x + 8  = 0

Using the Rational zeroes theorem, I see that x = -1 is also another solution

So....we can use synthetic division to find the residual polynomial

-1  [  7     23     24      8   ]

                -7     -16    -8

       ________________

        7     16       8       0

The   remaining polynomial is     7x^2  + 16x + 8

Using the Quadratic Formula  

x =    -16 ±√ [ 16^2  - 4(7)(8) ]      =          -16  ±√ 32      =      -16 ± 4√2    =     -8 ±2√2

        ____________________                __________          _________        _______

                   2 *7                                            14                          14                       7

The solution  -8 - 2√2

                     _______  ≈  -1.55       which makes two of the original logs undefined  

                           7

So

x  =   -8 + 2 √2

        _________  ≈  -.74

               7

a =  8   b =  2    c  = 2   d  = 7

And their sum  =    19

We can  choose any 2 of 5 postions for the "9s" to occupy.....so

5C2  =   10   five-digit integers

We can place all three candies in any box  =  4 ways

We can place  the same candies in any one of the 4  boxes and the different one in any of the remaining 3 boxes

= 4 * 3  = 12 ways

We can place  the different candy and either piece of the same candy in any of the 4 boxes and the other piece of the same candy in any of the remaining 3 boxes = 4 * 3  = 12 ways

[ The same candies are indistinguishable, so choosing either doesn't matter]

We can place the different candy in one of the four boxes and choose any 2 of the remaining 3 boxes to put one piece of each of the same candies in   =  4 * 3C2  =  4 * 3  =  12 ways

So......

4 + 3(12)   = 40 ways

√3 *∛2  =

(3)^1/2  * (2)^(1/3)  =

(3)^(3/6) * 2^(2/6)  =

(3^3)^(1/6)  *  (2^2)^(1/6)  =

(27)^(1/6) * (4)^(1/6)  =

(27 * 4)^(1/6)    =

(108)^(1/6)   =

⁶√108