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The diameter of this smaller circle is  20/3  for this reason.

Take out any round coin. 

Trace around it on a sheet of paper and find the center of this circle.

You can then trace another six coins (of the same size) around the original coin so you have six outer circles

which are the same size an the innter circle.

Now, draw a circle whose center is the center of the original coin and is externally tangent to the six outer coins.

Draw radii of this larger circle which are tangent to each of the six ourter circles.

If you look at the center of the original circle and one of the outer circles, you have the picture to this problem.

From the center of the original circle to the edge of an outer circle is one adn one-half diameters.

Therefore, the diameter of each of the coins is two-thirds of 10.

NVM....I messed up   !!!

3y - 49  =0

3y = 49

y =  49/3

Subbing this into the second equation we have that

x^2 +  (49/3)^2  -  281  =  0

x^2  + 2401/ 9  - 281  = 0     multiply through by   9

9x^2  + 2401  - 2529  = 0

9x^2  - 128  = 0

9x^2  = 128       divide both sides by 9

x^2  =  128 / 9

x^2  =  2  (64/9)     take  both roots

x = ± (8/3)√2

Of course,  the sum of these x values =   0

Angle  ADC =  180 - 45  =  135

So angle  DAC =  180 - 30  - 135  =  15

Let angle BAD  = x

Using the Law of Sines

sin 30 /  AB  = sin ( 15 + x) / (2BD)   ⇒  (2BD) / AB =   sin (15 + x) / sin 30     (1)

Also

sin 45 / AB  =  sin x  / BD

sin 45/ AB =  2 sin x / (2BD)   ⇒  (2BD) / AB = 2 sin x / sin 45    (2)

Equate (1) , (2)

sin (15 + x)  / sin 30  =  2 sin x / sin 45

sin (x + 15) / ( 2sin 30)  = sin x / sin 45            { 2sin 30  = 1 }

sin (x + 15)  = sin x / sin 45

sin x cos 15 + cos x sin 15  =  sin x / sin 45

sin x  ( 1/sin 45  - cos 15)  = cosx sin 15

sin x / cos x  =  sin 15  / ( 1/sin 45 - cos 15)

tan x  =  sin 15  / ( √2 - cos 15)

tan x  =  1/ √3

arctan (1/ √3)  =   x =   30° =   BAC

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Since you know that the roots are  a  and  b, the quadratic can be written as  (x - a)(x - b)  =  0

(x - a)(x - b)  =  0     --->     x² - ax - bx + ab  =  0     --->     x² + (-a -b)x + ab  =  0

We are given that the equation is:  x² + ax + b  =  0

Comparing these two equations, we see that  a  =  -a - b     and     b  =  ab

If b  =  ab,  a  must be 1.

If  a  =  -a - b     --->     2a  =  - b     --->     b =  -2a

Since  a = 1     --->     b  =  -2

I believe that there is only one ordered pair,  (1, -2)

a = 14-b

b^3 + (14-b)^3 = b^3+      (196 -28b+b^2)(14-b) =

       b^3 +       2744-196b-392b+ 28b^2 +14b^2 -b^3 = 812

      42b^2 -588b +1932 = 0            Quadratic Formula      a and b = 7+- sqrt 3          or    b = 5.268   a=8.732

                                                                                                                                            b= 8.732   a = 5.268

For the first two hours, he drove at 50 mph: 

   Distance = Rate x Time   --->  Distance  =  50 mph x 2 hrs  =  100 miles

For the next three hours, he drove at 60 mph:

    Distance  =  60 mph x 3 hours  =  180 miles

Total distance  =  280 miles                           Total time  =  5 hours

To find the average speed:

    Distance  =  Rate x Time

     280 mi    =  Rate x 5 hours                 (divide both sides by 5 hours)

      56 mph  =  Rate  

mmmmmbhhkkknnn,,,,,,,,,,,,,,,kbhgvcfdxxxxfzbdxgnfhcgjjjjjkkkkkknn1112345678890

Chris, I'm not understanding one part of your explanation. 

Each [triangle] will have legs  of (1/2)S   where S is the side of the square 

Are you saying that the length of each leg of the triangles equals half the length of the side of the square? 

_.

THis is how I did it(or similar, this is from a few days ago, doing it now)

So we first convert all 25^whatevers into 5^somethings

so

5^-4 = 5^48/x OVER 5^26/x * 5*34/x = 5 ^(48-26-34)x = 5^-12x

so -12x = -4, x = 3

essentially the same as what you did

And also I'd nominate your answer for "Best Answer"

     5^(48/x)                             1

_________________ =        ____

5^(26/x) * 25^(17/x)               25^2

5^(48/x - 26/x)                     1

_____________ =          ______ 

    (5^2)^(17/x)                   25^2

5^( 22/x)                       1

_______   =             ________              

5^(34/x)                     (5^2)^2

   1                             1

_______   =          _____

5^(12/x)                   5^4

Which implies that    12/x = 4     ⇒   x  = 3

Let  the smaller  number  =  x  and  the greater  = y

We know that

x / y   = 3/5     ⇒ x = (3/5)y       (1)

And  we know that

[ x - 4]              2

______  =     ___

[y + 8]              7

Sub (1) into this for  x

[ (3/5)y - 4 ]              2

_________  =       ___

 [ y + 8]                   7

Cross-multiply  and you have

7 [ (3/5)y - 4 ]  =  2 [ y + 8]

You should be able to solve it now for the larger number, y

......

okie

So you see under any of my questions/answers there's a blue username bubble with my username.

Click that.

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cos θ  = (8/10)

Take  the  arccos  to  find the  angles

arcos (8/10) = θ  ≈  36.9°

The other  value is   360 - 36.9  =  323.1°

The sum of the coefficients  is just  the  sum of the coefficients in the 11th row of Pascal's Triangle

This is given  by  2^11  =   2048

Well if you only need the bottom row tell us what you have done for the top row.

The formula for each interior angle of a regular polygon of n sides is:  (n - 2)(180^o)/n

For a regular hexagon:  (6 - 2)(180^o)/6  =  120^o.

For a regular pentagon:  (5 - 2)(180^o)/5  =  108^o.

For the red triangle that is outside the blue hexagon, the two base angles are each: 

    (180^o - 108^o) / 2  =  36^o.

This means that each base angle of the dark-red trapezoid is 108^o - 36^o  =  72^o.

Subtracting this from the hexagon angle, we get  120^o - 72^o  =  48^o

ok but how do I do that?

Ok glad to hear you understand

Tomorrow you can send me a private message with your question in case I don't spot ti on the thread :)

that will be tmrw but really thank u!

np 

Is there anything you need me to explain or something you have any confusions, even if a smidget of confusion lingers, feel free to ask me.