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The function is  not  defined  if the  denominator = 0

Note that     l x l  = sqrt (x^2)

So

sqrt [ (x - 3)^2 ]  - sqrt [ (x + 1)^2 ]  = 0

sqrt [ (x - 3)^2 ]  =   sqrt [ ( x + 1)^2 ]        square both sides

(x - 3)^2  = (x + 1)^2

x^2 - 6x + 9  = x^2 + 2x + 1

-6x + 9  = 2x + 1

9 - 1 = 6x + 2x

8 = 8x

x = 8/8  = 1

So  x =1 makes the  denominator = 0

So the  domain  is (-inf, 1) U ( 1, inf)

(8 times 2) - - 5  =

16 + 5  =  21

We have these equations

-11 = (-1)^2 + b(-1) + c

-17 = (3)^2 + b(3) +  c   simplify

-11 = 1 -b + c

-17=  9 + 3b + c

-12 = -b + c

-26 = 3b + c       subtract  these

14 = -4b

14/ -4 = b =  -7/2

And we can  find  c as

-12 = - (-7/2) + c

-12  = 7/2 + c

-12 - 7/2  = c = -24/2 - 7/2 =    -31/2

I do not know how to solve this problem, as there is i in the problem, but I'll try my best in solving it in terms of z. Firstly, I would put the two intergers together, which turns into -1 = 2z - 3iz. Then, we can turn it into 2z = -3iz - 1, and divide two from both sides. This give us z = (-3iz - 1)/2. 

Given that triangle XYZ is equilateral and X is the circumcenter of triangle OYZ, it implies that angle OXY is 120 degrees. This makes angle OXZ also 120 degrees, making triangle OXZ isosceles.

Since X is the circumcenter of triangle OYZ, XY = XZ. Therefore, angle XYZ = angle XZY = (180 - 120) / 2 = 30 degrees.

Now , we can use the area of the circle

           A = πr²

where r is the radius of the circle centered at o.

Given that the area of the circle is 48π

 we can find the radius

     48π = πr²

        r² = 48

        r = √48

        r = 4√3

Since triangle OXY is isosceles, we can find the height of the triangle (which is also the radius of the circle):

Height = r = 4√3

Now, we can use the forumala for the area of an equilateral triangle A = (side² * √3)/4

Area of XYZ = (XY² * √3)/4

                     = (4√3)² * √3)/4 

                     = 48√3/4

Area of XYZ = 12√3

So, the area of triangle XYZ is 12√3 square units.

I am editing this answer in order to make it more readable.

It does look computer generated but I personally don't really care, except that where an answer comes from should ALWAYS be cited.

The answer appears to be a good one.

Melody 

(all my changes are clearly marked)

Let’s assume the length of the rectangle is L and the width is W.
The area of the rectangle is given by A = L * W. The perimeter of the rectangle is given by P = 2L + 2W.
According to the problem, the number of square units in the area is four times the number of units in the perimeter. Mathe- matically, this can be represented as:
A = 4P
Substituting the expressions for A and P, we get:
L * W = 4(2L + 2W)

L * W = 8L + 8W

Rearranging the equation, we get:
L * W - 8L - 8W = 0
We can rewrite this equation as:
LW - 8L - 8W = 0
To make it easier to solve, let’s add 64 to both sides of the equation:
LW - 8L - 8W + 64 = 64

LW - 8L - 8W + 64 = 64
Now, we can factor the left side of the equation:
(L - 8)(W - 8) = 64

We are looking for the smallest possible perimeter, which means we want to minimize the sum of the length and width.

Therefore, we need to find the smallest possible values for L and W that satisfy the equation.

The factors of 64 are: 1, 2, 4, 8, 16, 32, 64.
We can try different combinations of     L - 8    and    W - 8    to see which ones give us integer values for L and W.

If L - 8 = 1 and W - 8 = 64, we get L = 9 and W = 72, which are not integer values.  (I think you mean that they are integer values)

If L - 8 = 2 and W - 8 = 32, we get L = 10 and W = 40, which are integer values.
If L - 8 = 4 and W - 8 = 16, we get L = 12 and W = 24, which are integer values.
If L - 8 = 8 and W - 8 = 8, we get L = 16 and W = 16, which are integer values.
 

So far what you have said appears to make sense  -  But it also looks like it has been computer generated.

Therefore, the smallest possible perimeter of the rectangle is 2L + 2W = 2(10) + 2(40) = 20 + 80 = 100 units. You lose me here - Melody

Added by Melody.

I think this gives possible perimeters of    

2(9+72)      = 162

2(10+40)     = 100

2(12+24)     = 72

2(16+16)     = 64  ( but this one is a square which is contrary to the conditions. )

So the smallest perimeter appears to be 72. 

This happens when the length is 24 and the width is 12

Area = 288 u^2

Perimeter = 72

72* 4 = 288      excellent        

There are 16 odd cards and 12 face cards

The probability is  [ 16 + 12 ] / 52 =   28 / 52  =   7 / 13

There are 15 line segments to be colored:
(1) 6 edges
(2) 3 diagonals that connect opposite vertices
(3) 6 other diagonals

For the appearance of the figure to be unchanaged when it is rotated by 60 degrees, all segments of the same type must be the same one of the two colors.

So the number of different colorings is 2*2*2 = 8.

Using the 'vertical line rule'   ---->   a and b are functions because any vertical line crosses the graphs at only one point......not so for (c) 

 8 loaves  *   8 ounces / loaf     / ( 16 ounces/ pound )  =  8 * 8 / 16 pounds = 4 pounds 

...ooops !    Remember when dealing with inequalities, when you divide or multiply both sides by a negative the  <  or   >  symbol reverses ! 

3x-7 ≤ 5x+9 

-16 ≤ 2x

-8 ≤ x                         solution is then  [-8, + ∞ )    

Vertex (3,-2)  form of the parabola is then :

 y =  n ( x-3)^2 -2    and it has point (2,0) ....sub in this point  to calculate the value of 'n'

0 = a(2-3)^2 -2    shows n = 2 

so the parabola is    y = 2 ( x-3)^2 -2      expand R side to get 

                               y = 2x^2 -12x + 16      I think you can see what 'a'  'b' and 'c' are now......

Breaking into parts

1/3 < x / 5   cross-multiply

5 / 3 < x     so x >  5/3

Also

x / 5 < 7/8   cross-multiply

8x < 35

x < 35 / 8

The solution  is     5 / 3 < x < 35 / 8

The largest integral value of  x  is   32 / 8  =   4

The smallest multiple of 12 that is greater than 200 is 17x12, or 204.

204's prime factorization is 2^2 times 3^1 times 17^1.

36's prime factorization is 2^2 times 3^2. The GCD is 2^2 times 3^1, which is 12. 

The answer is 204.

Plug the points that are given in.

-2=9a+3b+c

0=4a+2b+c

We need a third equation to solve. We are given the vertex; so the symmetrical line of this parabola is x=3. Reflect the other given point, (2,0), over this line to get your third point (4,0).

0=16a+4b+c. 3rd equation minus second equation is 12a+2b=0, second equation minus the first is -5a-b=2. -10a-2b=4. 2a=4, a=2, b=-12, c=16. (2,-12,16)

Note that, area of semicircle = πr²/2, where

                          r - radius of semicircle.

The area of semicircle APC = 7

Let r₁ be radius of semicircle APC.

By above note,

Area of semicircle APC = πr₁²/2

           => πr₁²/2 = 7

           => r₁² = 14/π

                r₁ = √14/π.

 length AC = 2r₁ = 2√14/√π

Again, the area of semicircle CQB = 13

Let r₂ be radius of semicircle CQB.

       By above rule,

Area of semicircle CQB = πr₂²/2

           => πr₂²/2 = 13

           => r₂² = 26/π

                r₂ = √26/π.

length CB = 2r₂ = 2√26/√π

--------------------------------------

Now, in △ACB, MLC = 90°

     By Pythagoras theorem,

          AB² = AC² + CB² 

                  =  4(14)/π + 4(26)/π

        => AB² = 160/π

            AB = √160/π

Therefore, radius of third circle = r₃ = 1/2 √160/π.

Area of third circle = πr₃²/2 

                              = π(1/4 * 160/π)/2

                              = 40/2 = 20 

Therefore area of third circle is 20 square unit.