From the center of  the small  circle on the bottom left draw a radius to the tangent  formed by the side of the equilateral triangle and  draw a segment to the bottom left vertex of the equilateral triangle....call this x

This will form a 30-60-90 right triangle

The vertex angle  of the equilateral triangle will be bisected = 30°

The side opposite of this = radius of a small circle  = 1

So....x  =  twice the radius  =  2

So.....the distance  from the bottom left vertex to the center of the middle  circle  = x + 2 = 2 + 2  = 4

And we can form another triangle  with two sides of 4  and an included angle of 120°

And the side of the equilateral triangle ,S,  wil be  opposite this angle

So.....using the Law of Cosines 

S^2  =  4^2  + 4^2  - 2 (16) cos (120°)

S^2  = 32 - 32 (-1/2)

S^2   = 48

S  =  sqrt (48)

So....the area of the equilateral triangle  = 

(1/2)  (sqrt (48))^2  sin (60°)  = 

(1/2) (48)  sqrt (3)  /2   =

12 sqrt (3)  units ^2

We can split the base into two parts....m  and n

tan 30° =  3 /m

sqrt (3) / 3 =  3/m

m = 3 / ( sqrt (3) / 3)

m = 9 / sqrt (3)  =   3 sqrt (3)

tan 60° = 3 / n

sqrt (3)  = 3 / n

n  = 3 /sqrt (3)  =  sqrt (3)

So....area  =

(1/2) base length * height  =

(1/2) (m + n) * height  =

(1/2)  (3sqrt (3)  + sqrt (3) )* 3  =

(1/2) ( 4 sqrt (3) )  * 3   =

2sqrt (3) * 3  = 

6sqrt (3)  (square units)

Factors  of 2008   = 

1  2  4   8     251   502    1004    2008

The smallest  value  for  b  is

( x + 8)  ( x +251)   =    x^2  +  ( 251 + 8) x  + 2008   =   259

THX, guest.....my previous "solution"  was totally wrong !!!

The  vertical asymptotes  will  occur at the  x values   produced by setting the denominator = 0   :

x^2  + 6x   - 7     =  0    factor

( x - 7) ( x + 1) = 0     set each factor to 0  and solve for x

x  -7   =  0                x  +  1   =   0

x  = 7                      x  = -1

These are the equations of  the vertical asymptotes

Here I wanted to correct answer #1. I copied and pasted the whole answer and have corrected what was wrong.

I apologize if I've done something wrong, teacher.

Using Chinese Remainder Theorem + Modular Multiplicative Inverse, we have:

K mod 3 = 1,   

K mod 4 = 1,   

K mod 5 = 1,   

K mod 6 = 1,   

K mod 7 = 1

K =420m  +  1, where m=0, 1, 2, 3........etc.

So, the smallest K under 1000=420 + 1 =421

r = sqrt(-3² + 2²)     or    r = sqrt(-2² + 3²)  ==>  r = sqrt(13)

We have that

1/3  =  7/9  +  d (13 - 1)

1/3  = 7/9  +  12d

(1/3 - 7/9)  =12 d

(3/9 - 7/9)  = 12 d

-4/9  = 12 d      divide both sides  by  12

-4/ 9  *  1/12   = d

-4/108  =d

-1/27    = d

To find the seventh term we have

7/9  +  (-1/27) (7 -1)

7/9 + (-1/27) (6)

7/9  -6/27  = 

7/9 - 2/9  = 

   5/9

Thank you so much. 

Nice explanation too. 

If the center lies on the x axis, call the   center  (a, 0)

So.....the distance from this point to each of the given points  is equal

So....equating distances, we have that

(-3-a)^2  + ( 2 -0)^2  = (-2-a)^2 + (3-0)^2        simplify

((-1) (3 + a))^2  + 2^2  =  ((-1)(2 + a))^2  + 3^2

(3 + a)^2  + 4  = (2 + a)^2  +  9

9+ 6a + a^2  + 4  = 4 + 4a + a^2 + 9        subtract a^2  from both sides and simplify

6a +13 = 4a + 13     subtract 13 from both sides

6a  = 4a

This is true when a  = 0

So  the center is    (a, 0)   = (0, 0)

And the radius  is  sqrt  [ ( 0 - -3)^2 +  (0 -2)^2 ] =   sqrt  [ 3^2 +2^2]  = sqrt  ( 13)

Here's a graph :   https://www.desmos.com/calculator/dd6nbqzagb

The points (-3, 2) and (-2, 3) lie on a circle whose center is on the x-axis. What is the radius of the circle?

r = √13

Here's a link:    http://mathforum.org/library/drmath/view/54103.html

That is not correct.

Total salt of ingredients    = total salt in final solution

  .40 (40)                =  .25 (40 + x)

    16                       = 10 + .25x

      6                       =   .25 x

     24 liters  of water = x

We are not here to do your homework for you. We are here to support you.

f(g(x) = sqrt(5(x-2)+20) = sqrt(5x-10+20) = sqrt(5x+10) 5x+10>=0 since sqrt cannot contain negative numbers so 5x>-10 x>-2 so domain is [-2,infinite) and range [0,infinte)

but so hard

Hi dqwadszda! 

You just have to follow the steps as follows=

Question #1 How were Sparta and Athens alike? Question #2 How were they different? Be sure to include details about how the city you are visiting and how it compares to your city.

Step 1: Choose Your Point of View
First, you must choose one of the following types of persons and write your letter from their point of view:

young teenage girl of the citizen class, slave,
boy of the citizen class ,young soldier , very wealthy person of the citizen class
Step 2: Choose Your Destination
Second, you must choose one of the following destinations:

an Athenian visiting Sparta
a Spartan visiting Athens
Step 3: Write Your Letter To Home
Your letter must:

be written from the perspective of a slave, young teenage boy or girl citizen, solider or wealthy citizen
be at least two paragraphs
compare how the city you are visiting is the same or different as your home city for each of the following categories: geography, social structure, and government

This cannot be a rectangle....see the graph :

For this to be a rectangle, angle ABC would have to be  = 90°

LOL!!!