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19, 19, 20, 20, 21, 21, 22, 24, 25, 25, 26, 27, 28, 28, 30, 30, 30, 31, 32, 32, 34, 34, 35, 38, 39, 45, 48

Mean   = 29  = x

Median   =  28    =   y

Mode  = 30   = z

x ( y - z)    =   29  ( 28 - 30)  =   29  (-2)   =   -58

Sorry i read the problem wrong 

Trying a few values, I'm getting f(x) = x + 1.

That's not possible if x+y=3 means that x+y = 5 cant happen.

You might have made a typo

Since the question is just asking about the center, ignore the fact that it's a circle. You won't need it.

So you're essentially reflecting the point (6,-5) over the line y=1.

(1-(-5))=6. (difference of y values (-5 and 1))

So the point is 6 units from the line. You just need to move the point (6*2)=12 units up. (The *2 comes from the reflection.)

So it becomes (6,-5+12)=$\boxed{(6,7)}$.

-1620/9=-180

w = -1620

Your Welcome 

oh my

Hello ss333!

Background

The line on top of ABCD means to read it as a number instead of A*B*C*D. 

Equation

line(ABCD) + line(BCD) + line(CD) + line(D) = 2012

(1000A + 100B + 10C + D) + (100B + 10C + D) + (10C + D) + (D) = 2012

1000A + 200B + 30C + 4D = 2012. 

Letter A

First, let's try to find A. 

A has to be 1 or 2, since if A was 3, then 1000*A = 3000, which is bigger than 2012. 

However, if we try setting A as 2, we find out that B would be 0 since 200B + 30C + 4D = 12. 

B can't be 0 since B has to be a positive digit, making A 1. 

Equation

A = 1

1000(1) + 200B + 30C + 4D = 2012. 

200B + 30C + 4D = 1012

100B + 15C + 2D = 506

Letter B

Following the same logic as letter A, B has to be in the range 1-5 since if B was 6, 100*B = 600, which is greater than 506. 

However, if B were 5, then C would be 0, and if B were less than or equal to 3, then C would be greater than 10, which isn't a digit number, making B 4. 

Equation

B = 4

100(4) + 15C + 2D = 506. 

15C + 2D = 106. 

Letter C

Following the same letter as A and B, C has to be in the range 1-7 since if C were greater than 7, then 15C > 106. 

However, if C were less than 6, then D would be greater than 10 and not a digit. 

Similarly, if C were greater than 6, then D would be a decimal number.

Thus C has to be 6. 

Equation

15(6) + 2D = 106. 

2D = 16

D = 8

Answer

A = 1

B = 4

C = 6

D = 8

1+4+6+8 = 19.

Our final answer is 19. 

I hope this helped. :)))))))

=^._.^=

Δ ADB  ~ Δ AEC      

Δ ADB  ~ Δ PEB

ΔAEC  ~   Δ PEB

AE / EC  = PE / EB

AE / [CP + EP ]    = PE / EB

AE  / [ 10 + 20 ]  = 20 / 30

AE / 30  = 20 / 30  ⇒   AE   = 20

20!

Like so

To check if x = 12 is correct, substitute this value back into the original equation and see if the left-hand side equals the right hand side.

Can someone please check me?

I think the answer is x=12

Let's solve your equation step-by-step.

12−x=−3x+36

Simplify both sides of the equation.

12−x=−3x+36

12+−x=−3x+36

−x+12=−3x+36

Add 3x to both sides.

−x+12+3x=−3x+36+3x

2x+12=36

Subtract 12 from both sides.

2x+12−12=36−12

2x=24

 Divide both sides by 2.

2x=24

2     2

x=12

Answer:

x=12

(this may not be right but it was kinda confusing...)

-Grace

As follows:

The following satisfies the conditions

P = 124           P/2 = 62            ∠B = ∠D  => x

7 / x + 24 / x = 62         x = 1/2     (this is sine of angles B and D)

AB = 1/2 * 7 = 14               AD = 1/2 * 24 = 48

Area             [ABCD] = 48 * 7 = 336     or      14 * 24 = 336

You "borrowed" that answer from jugoslav, didn't you? 

ab + a + b  =  95

ab +  a  =  95 - b

a ( b + 1)  = 95 -  b

a  =   (95 - b) / ( b + 1)

When   b = 7  a =  11    relatively prime

When  b = 15   a  =  5     not relatively prime

a +  b =   20

See the following  :

Let C  =(-10,20)     C = (0,0)

The slope of  this line =  -2

And the  eqation of this  line is   y =   -2x    (1)

The equation of the  circle  with its ceenter at  A  is   ( x+10)^2  + ( y - 10)^2  =  100    (2)

Sub (1)  into (2)  and we have that

(x + 10)^2   + ( -2x - 10)^2  = 100

x^2  + 20x + 100  +  4x^2  + 40x + 100    = 100

5x^2 + 60x + 100   = 0

x^2   + 12x  +  20  =   0     factor as

( x  + 10) ( x + 2)   =  0

The second factor set to 0   and solved for x  gives  x  = -2 =  x coordinate of F

And  the y coordinate is   -2 (-2)    =  4  = HF  =    side of the  square

The 500th term in the sequence = 522

An equilateral triangle of side length 2 units is inscribed in a circle. Find the length of a chord of this circle which passes through the midpoints of two sides of the triangle.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Find the diameter of a circle      D = 2(1 / cos30º) = 2.309401076

Height of a triangle           h = √3 = 1.732050808

Subtract height from diameter            M = D - h = 0.577350268

The length of a chord            L = sqrt( D² - M² ) = √5

       p   n   d

1  = 30 0 0
2  = 25 1 0
3  = 20 2 0
4  = 15 3 0
5  = 10 4 0
6  = 5 5 0
7  = 0 6 0
8  = 20 0 1
9  = 15 1 1
10  = 10 2 1
11  = 5 3 1
12  = 0 4 1
13  = 10 0 2
14  = 5 1 2
15  = 0 2 2
16  = 0 0 3
 

 "The integers are not relatively prime, and each is less than 20".

[15  x  5]  +  [15 + 5] ==95

[15  +  5] ==20

The height of the  larger equilateral  triangle  is  sqrt ( 3)

So....the  height of the  smaller equilateral  triangle  formed  by the chord is sqrt (3)/ 2  = (1/2)sqrt (3)

The  radius of  the cirlcle  can be found as

2^2  =  2R^2  - 2R^2  cos 120°

4  = 2R^2 + R^2

4  = 3R^2

R^2 = 4/3

R = 2/sqrt (3)  =  (2/3)sqrt (3)

So  the distance from the center of the  circle to  the  chord is   (2/3)sqrt (3)  -(1/2) sqrt(3)   =  (1/6)sqrt (3)= sqrt (3)/6

Using  the Pythagorean  Theorem, 1/2 the chord length can be found as

sqrt  [ ( 4/3)  -  ( sqrt (3)/6 )^2  ] =     sqrt [ 48/36  -  3/36 ]  =  (1/6) sqrt (45)  = (1/6) sqrt (9 * 5)  =

(3/6) sqrt (5)  =    sqrt ( 5)/2 

So....the chord length is twice this =  sqrt (5)

( (1*14)+(2*5)+(3*9)+(4*7)+(5*7)+(6*8) ) / 50

( 14 + 10 + 27 + 28 + 35 + 48 ) / 50

= 3.24

In rectangle ABCD, AB = 6 and BC = 4.  EF is parallel to BC, and EC = 1.  Rectangle GHIJ is inscribed in rectangle AFED, such that it is similar to rectangle ABCD.  Find the ratio of the area of GHIJ to the area of ABCD.

The process is really long but here is the answer:

The ratio of the area of GHIJ to the area of ABCD is 36/65