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The distance  between  the  centers of the  circles  = 6  +  R   where R is the radius of  the larger circle

So  we  can  form a right triangle  with   the hypotenuse =   (6 + R)  and  one leg =  R - 6

So.....the  width of the rectangle is    sqrt   [ (6 + R)^2   - ( R - 6)^2 ]   =   sqrt  (24R)

And the height of the rectangle  = R

So

sqrt ( 24R) * R  = 243        square both sides

24R * R ^2   =   243^2

24R^3   = 243^2

R^3  =  243^2 / 24

R  =    (243^2 / 24)^(1/3)  =  27/2  =   13.5

Let the first term be a. Because the sum of the series is 60, we have 

\(60= \frac{a}{1-(1/8)} = \frac{a}{7/8} = \frac{8a}{7}.\)

Therefore, \(a=\frac{7}{8}\cdot60=\boxed{\frac{105}{2}}\).

There are five different figures that can be constructed.

using law of cosines
8^2 = 7^2+9^2 - 2*7*9 cos A
64 = 49 + 81 - 126 cos A
cos A = 66/126

x^2 = 130 - 126 cos(180-A)
but cos(180-A) = -cos A
so
x^2 = 130 + 126(66/126)
x^2 = 196
x = 14

Thx CPhill I got it this way 

This is a geometric sequence with first term 1/4 and common ratio 1/2. Thus the sum of the first n terms is: 

\(\frac{63}{128}=\frac{1}{4}\left(\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}}\right)=\frac{2^n-1}{2^{n+1}}\).

We see that \(\frac{63}{128}=\frac{2^6-1}{2^7}\), so \(n=\boxed{6}\).

We can work backwards here....the common ratio  is  1/8  / 1/4   =   4/8 =  1/2

63/ 128  =   (1/4)  ( 1 - (1/2)^n )  / ( 1 - 1/2)

63/128  =  (1/4) ( 1 - (1/2)^n) / (1/2)

63/128 =  (1/4) (2/1)  ( 1 - (1/2)^n)

63/128 =  (1/2) ( 1 - (1/2)^n)

(2/1) (63/128) =   1  - (1/2)^n

63/ 64  =  1 - (1/2)^n

(1/2)^n  =   1 - 63/64

(1/2)^n =   64/64 - 63/64

1/2^n =  1/64

2^n =   64

n =   6

This answer is going to assume that the short side of the rectange is equal to the radius of the large circle.

Our goal is to find the short side of the rectangle. The short side must be between 6 and 12(radius and diameter of small circle). Let's start by finding some factor pairs for 243.

Pair 1: 1*243 (doesn't work)

Pair 2: 3*81(doesn't work)

Pair 3: 9*27(works)

As those are the only three pairs, pair 3 will work.

Now that we know the short side is 9, we can see that the radius of the large circle is 9

I assume that the 31.5° is relative to the horizontal.

Then if the total force is F = 125N, then the force parallel to the water is

Fp = FCos31.5 = 125(0.85) = 106.625N

The resisting force by the water is 84.8N, so the net force is 106.625 – 84.8 = 21.825N

Then from F = ma

21.825 = 63.9a

So a = 0.342m/s^2 in the x-direction

Then the force normal to the water is

Fn = FSin31.5 = 125(0.5225) = 63.31N

Then again from F = ma

63.31N = 63.9a

So a = 63.31/63.9 = 0.991

So a = 0.991m/s^2 in the y-direction

This ok

OOOPS!!!...misread the problem....I just did the first three terms....your answer is  correct  !!!

THX for the  correction  !!

THX, GR  !!!!

Here's another way

100   -  100  (.90)  (.80) (.70) 

100  - 50.4  =   49.6

Found the answer after working on the problem for a bit:

Pick 5 states out of the 50 possible, 50 choose 5 (2118760)

Choose 1 out of 2 competitors out of each state. 

2⁵ because there are 5 states and 2 choices for each.

2118760*32=67800320

Found it!

Pick 5 states out of the 50 possible, 50 choose 5 (2118760)

Choose 1 out of 2 competitors out of each state. 

2⁵ because there are 5 states and 2 choices for each.

2118760*32=67800320

CPhill I got something different 

This is a finite geometric series with first term \(1/3\) and common ratio \(1/3\). There are five terms, so the sum of this series is \(\frac{\frac13\left(1-\left(\frac13\right)^5\right)}{1-\frac13} = \boxed{\frac{121}{243}} \).

That's okay, I'll keep working to see how to arrive at 67,800,320. Thanks anyways though!

Sum of a geometric series

S =  first term  (  1 -  common ratio ^n)   / ( 1  - common ratio)     where n =  the  number of terms we are  summing

The common ratio  is    (1/9) / (1/3)  =  3/9  = 1/3

So

S =   ( 1/3)  ( 1 - (1/3)^3 ) / ( 1 - 1/3)   =

(1/3)  ( 1 - 1/27) / ( 2/3)   =

(1/3) (3/2)  * ( 26/27 )   =   

(1/2) ( 26/27)  = 

26/54  =

13/27

hihihi's solution is good, but this one may help you solve the problem quicker and easier

(x+y)² = x²+2xy+y²

(x-y)²=x²-2xy+y²

As such, when you add them together, you get 2x²+2y²

When you plug in x and y, this equals to 2(2006)+2(2007)=4012+4014=8026

Yes, but some of them contain 2 zero's

Don't know where I made an error....maybe someone else can  give you the  correct answer....sorry!!!

Answer: 

\((x+y)^2+(x-y)^2\)

\((√2006+√2007)+(√2006-√2007)\)

\(\left(3\sqrt{223}+\sqrt{2006}\right)^2+\left(\sqrt{2006}-\sqrt{2007}\right)^2\)

\(\left(3\sqrt{223}+\sqrt{2006}\right)^2+\left(\sqrt{2006}-3\sqrt{223}\right)^2\)

\(4013+6\sqrt{447338}+\left(\sqrt{2006}-3\sqrt{223}\right)^2\)

\(4013+6\sqrt{447338}+4013-6\sqrt{447338}\)

8026

Area of  base =   QP * QR   = 6       (1)

Area of one side  POUT  = QP * QU   = 15   ⇒   QP =  15/QU     (2)

Area of the other side  QUVR  =   QU * QR  =  10  ⇒  QR =  10/QU    (3)

Sub (2) , (3) into (1)

15/ QU  *  10/QU  =  6

150 / QU^2  =   6      rearrange  as

QU^2  =  150/6

QU^2 =  25          take the square root  

QU   =   5

So  QP   =  15/5   =  3

And QR =   10/5  =  2

So....the volume is   QU  * QP * QR   =   5 * 3   *  2    =     30  units^3

Wait, but according to the official national MATHCOUNTS solutions, the answer is 67,800,320 ways, was there something wrong in your solution?

(x+y)² = x²+2xy+y²

(x-y)²=x²-2xy+y²

As such, when you add them together, you get 2x²+2y²

When you plug in x and y, this equals to 2(2006)+2(2007)=4012+4014=8026

Any  1  of the 100  can be chosen first

Next.....we  can  choose  any 1  of the other (100 - 2)  =   98   ( we can't choose  the other  one from the  first state)

Next.....we can choose any of the remaining  (100 - 4)  = 96  (we have chosen two  from two states  and we  can't choose the  remaining two from either state)...so...we only have 96 to choose from

Continuing with this pattern : 

Then.... one from the remaining  (100 - 6)  = 94

Then... one from the remaining (100 - 8)  = 92

So

100 * 98 * 96 * 94 * 92    =   8,136,038,400   ways  !!!

THIS IS INCORRECT   !!!