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YAYYY. :DDDD

That is such an intresting solution. 

I have not yet learned about AM-GM/AM-HM, but I'm planning to do so soon through alcumus. 

I think the mistake I made was forgetting that x, y, and z were all positive, meaning that (x+y)/(xyz) was also positive. 

=^._.^=

Oh, i've also had this burning question, why do all your answers have hundreds of spaces XD

Thank you!

(x +  y)^2   =   x^2  +  y^2

x^2  +  2xy  +  y^2 =  x^2  + y^2                   (subtract   x^2, y^2   from both sides)

2xy     =  0

xy =  0                which  implies  that  x =  0   or y =  0     or  both  =   0

The roots  are  1 + sqrt 2  ,1 - sqrt 2  and  r

Sum  of the  products of the roots taken two  at  a time  =  2r  - 1  =   b

Product of  the roots  =    -r =  -c    ⇒    r =  c

So     

x^3   +  (2r - 1) x  + r =   0        since  1 + sqrt 2   is a root....then

(1 + sqrt 2)^3  +  (2r - 1)(1 + sqrt 2)   +  r   =   0

1  + 3sqrt 2  + 3*2 + 2sqrt 2  + 2r  - 1 + 2rsqrt 2 - sqrt 2  +  r  = 0

6 +  4sqrt 2  + 2r sqrt 2   +  3r  = 0

6 + 3r   +  2sqrt 2 ( 2 + r)  =  0

3 ( 2 + r)   + 2sqrt 2  ( 2 + r)  =  0

(2 + r)  ( 3  + 2sqrt 2)   =   0

r  =  -2

So   b =  2(-2)  -  1   =   - 5

c  =  -2  =    r

The polynomial is

x^3  -  5x   -  2      and  the integer root   =    -2

Ay!  Turns out, the answer was 16, occurring when x=y=1/4 and z=1/2.  

We use a combination of the AM-HM and the AM-GM in that order.  

Starting, we have by the AM-HM $\frac{x + y}{2} \ge \frac{2}{\frac{1}{x} + \frac{1}{y}} = \frac{2xy}{x + y}$

So we have $\frac{x + y}{xy} \ge \frac{4}{x + y}$.

Multiplying by 1/z, we get $\frac{x + y}{xyz} \ge \frac{4}{(x + y)z}$.

By the AM-GM inequality,
 

$\sqrt{(x + y)z} \le \frac{x + y + z}{2} = \frac{1}{2}$, 

meaning $(x + y)z \le \frac{1}{4}$.

 Hence,
 

$\frac{4}{(x + y)z} \ge 16$.

Using what we had before, we can write 

$\frac{x + y}{xy} \ge \frac{4}{x + y} \ge 16$.

Equality occurs when x=y, and the minimum value is 16.  

If you ask me, this solution was quite interesting.  I thought of using the QM-AM-GM-HM relationship, as well as the Cauchy-Schwarz Inequality, but not in this way.  Thanks for your help! 

Your questions are always so hard but I'll try my best. 

Since x and y seem to be used the same in the equation.

I'm going to assume that x = y. *This could be totally incorect.

(2x)/(x^2z) is our new equation. 

z = 1 - 2x

Let a be (x+y)/(xyz).

(2x)/(x^2(1-2x)) = a

2x = a(x^2 - 2x^3)

Just quickly graphing this out in desmos by setting a = y (not the y in the original equation, but for coordinate purposes), it seems like the sum can be any value other than from the range (0, 16]. 

Not sure what went wrong, but I hope this helped a bit. 

Good luck. 

=^._.^=

(x  -  sqrt 125)   ( x - sqrt 125)   =  x^2  - 2sqrt (125)x   +  125

( x +  sqrt 125)  ( x +  sqrt 125)   =  x^2  + 2sqrt (125)x  +  125

a =  -2 sqrt (125)     and   2 sqrt (125)

Simplified

a  =  -10sqrt 5        and  10sqrt 5

Using coordinate  geometry

Let A  = (0,0)

B =   (7,0)

C = (7  +  4cos 45  , 0  + 4sin 45)  =  ( 7  + 2sqrt 2 ,  2sqrt 2) 

D =  (7 + 2sqrt 2  ,  2 + 2sqrt 2)

E  = (7 + 2sqrt 2 -  (5/2)sqrt 2  ,  2  + 2sqrt 2 + (5/2)sqrt 2)  =  ( 7 - (1/2)sqrt 2 , 2 +(9/2)sqrt 2  )

F =   ( 7  -  (1/2)sqrt 2  - 6   ,  2  + (9/2) sqrt 2)   =   ( 1  - (1/2)sqrt 2 ,  2 + (9/2)sqrt 2 )

G =    ( 1  - (1/2)sqrt 2  - sqrt 2    ,  2  + (9/2)sqrt 2  -  sqrt 2)  =   (1 - (3/2)sqrt 2 ,2 + (7/2)sqrt 2  )

The  line   through   HA  has  the  equation    y =  -x

The line  through  GH   has  the  equation  x = 1 - (3/2)sqrt 2

So  the    coordinates of  H  =     ( 1 - (3/2) sqrt 2  , (3/2) sqrt (2)  -  1)

The  distance  from  A  to H  =    sqrt  ( 11  - 6sqrt 2)  =  3  -  sqrt 2     (simplified using WolframAlpha)

Distance  from   G to H   =  3  +  2sqrt 2

Total   perimeter   =    7 +  4  +   2  +   5   +  6  + 2   +   3 - sqrt 2  +  3  +  2sqrt 2  =  

32   +  sqrt 2

w  +  x  +  y  =   -2        (1)  

w + x +   z  =   -4           (2)

w  +  y  +  z  =   19        (3)

x + y +   z  = 12             (4)

Add  the  first  three  equations

3w  +  2 ( x  +  y  + z)  =  13

3w  +  2 ( 12)  =    13

3w  +  24   =  13

3w   =  -11

w   =   -11/3

Manipualting  the  first two equations

x +  y =  -2  + 11/3    ⇒   x +  y  =   5/3            (5)

x + z =  -4  +   11/3   ⇒   x  +  z  =  -1/3  ⇒    -x  - z  =  1/3       (6)

Add  (5)  and (6)

y - z  =  2          (7)

And  manipulating (3)

y +  z  =  19  +  11/3    ⇒       y + z =   68/3         (8)

Add (7)  and (8)

2y   =  74/3

y  = 74/6   =  37/3

And   

z    =  68/3 - 37/3  =  31/3

And

x =  5/3 - 37/3   =   -32/3

wx  + yz   =

(-11/3)(-32/3)   +  (37/3)  ( 31/3)     = 

1499  /  9

|2x| = 9

-4.5 < x < 4.5

|3y| = 20

-20/3 < y < 20/3

There are 9 options for x, 13 options for y. 

9*13 = 117

=^._.^=

using "x" this should be

x < 5

so as an interval: $\left(-\infty \:,\:5\right)$

f(x) degree = 7           x^7 in f(x)  when multiplied by the (x-1) will result in a x^8 term

The other x-axis intercept will be the same distance from the vertex as the given intercept..EXCEPT it will be on the OTHER SIDE of the axis of symmetry .

(6,0) 

Yea... wonder who made this problem!

$\frac{j+k}{jk} = \frac{2}{3}$

$3j + 3k = 2jk$

$3j - 2jk + 3k = 0.$

Here is the graph: https://www.desmos.com/calculator/u6uyo6fqj3

Consider asymptotes and reflections. LMK if you have any more questions.

Seems too complex of an answer?

Edit: You misplaced commas: https://www.wolframalpha.com/input/?i=w%2Bx%2By+%3D+-2%2C++w%2Bx%2Bz+%3D+-4%2C++w%2By%2Bz+%3D+19%2C+x%2By%2Bz+%3D+12%2C

Well, let's see.

We can expand and simplify the product: $\left(\frac{a + b}{ab}\right)\left(\frac{a - b}{ab}\right) = \frac{a^2 - b^2}{a^2 b^2}$

We could try and make a negative with them, but $b$ is always positive, so that doesn't work.

Because of the squares, the numerator will always be nonnegative(the least value is $(-2)^2 - 2^2 = 0$) and the denominator always positive. So the least positive value is when the numerator is 0, which gives 0 as the solution.

Equation of a parabola is $y = a(x-h)^2 + k$

The vertex is $(h,k)$

We plug in and get $y = a(x-12)^2 - 4$

Plug in the coordinates: $0 = a(6^2) - 4$

$0 = 36a - 4$

$a = \frac{1}{9}$

The equation is now $y = \frac{1}{9}(x - 12)^2 - 4$

The other x-intercept is when $y=0,$ so we have $4 = \frac{1}{9}(x-12)^2$

$36 = x^2 - 24x + 144$

$x^2 - 24x + 108 = 0$

$r_1 + r_2 = 24$

$r_1 r_2 = 108$

We know that one root is $18,$ so we have $18 + r_2 = 24$ giving the answer to be $(24,0).$ The x-coordinate of that is $24.$

Your answer is correct!!! But, it would be more interesting if they asked for this ratio [ACD]/[ABC].

The possible values of a are 2, 3, 5, 7, for a total of 4 possible values.

Assuming that  a can equal b   and  c can equal  d, then  K + L + M  = 

(K   +    L)     +   M   = 

2(10C2  + 10)  +  [ 10  + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 ]  =  

2  (45  +  10)   +  55 = 

2 (55 )  +  55  =

3 (55)   =

165