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Very nice, geno....

I happened to notice another pattern, too

1st ring = (3^2 - 1^2)  = 8

2nd ring (5^2 - 3^2)  = 16

3rd ring (7^2 - 5^2) = 24

4th ring (9^2 - 7^2)  = 32

So the nth ring is just [(2n + 1)^2 - (2n -1)^2 ]  =   8n 

The same result as geno's   !!!!!

Any digit "a" could be used 4 times..aaaa..so that's 4

And any two of the digits - "a" and "b" - can be selected out of 4 = 6 ways

And of the second, we have two ways to arrange the digits.... abba, baab

So...4 + 6(2)  = 16 palindromes

There are 4 choices for the first digit, 4 choices for the second digit, 4 choices for the third digit, and 4 choices for the last digit; thus, there will by 4 x 4 x 4 x 4 choices.

To determine the number of palindromes: there are 4 choices for the first digit and 4 choices for the second digit, however, there is only 1 choice for the third digit (it must match the second digit), and there is only 1 choice for the last digit (it must match the first digit). Thus, there will be 4 x 4 x 1 x 1 choices.

If the hypotenuse is 16, then the shorter side leg will be 8 and the longer leg will be 8√3.

Since it's a right triangle, the area will be ½·8·8√3  =  32√3  = 55.43 (approx)

I agree with you!

Notice something....that in combinatorics....

C(n, k)  is always the same thing as  C(n, n-k)

So if n = 16 and k = 4

C(16, 4) = C(16, 16-4) = C(16, 12)

And....if two things are equal, then, when we subtract one from another, the result is 0 !!

Notice that if you had 11 friends and you choose any four of them, there would be

C (11,4)  = 330 different combinations.....not enough!!!

And 12 friends would produce

C(12, 4) = 495 different combinations

So you would need at least 12

Thank you Geno!

The number of ways that you can have 8 heads out of 10 flips, without caring about order is:  ₁₀C₈.

The number of ways that you can have 9 heads out of 10 flips, without caring about order is:  ₁₀C₉.

The number of ways tht you can have 10 heads out of 10 flips, without caring about order is:  ₁₀C₁₀.

To find the number of ways to have at least 8 out of 10, add those three answers together.

Since there are 5 letters, the numerator will be 5!

However, since there are two M's, this must be divided by 2! and since there re two A's, it must also be divided by another 2!.

5! / (2!·2!)

We could choose 2 of 10 boys and the same wth the girls....so we have

[C(10,2)]^2 = 2025 ways

Another symbolism is:  ₁₆C₄ - ₁₆C₁₂

₁₆C₄ =  16!/(4!·12!)

₁₆C₁₂ =  16!/(12!·4!)

So, how much is  16!/(4!·12!)  -  16!/(12!·4!) ?

i meant h= 8^(sqrt(3)) oops

C(52,5) = 2,598,960  possible hands

The first ring has 8 squares           = 1 x 8

The second ring has 16 squares   = 2 x 8

The third ring has 24 squares       = 3 x 8

Continuing with this pattern, can you find the number of squares in the 100^th ring?

The common denominator for all the denominators is b(b - 1).

Multiply both the numerator and the denominator by this expression.

Numerator: [ (1/b - 2/(b - 1) ] x (b)(b - 1)  =  1(b - 1) - 2(b)  =  b - 1 - 2b  =  -1 - b

Denominator:  [ 3/b + 1/(b - 1) ] x (b)(b - 1)  =  3(b - 1) + 1(b)  =  3b - 3 + b  =  4b - 3

Answer:  (-1 - b) / (4b - 3) 

I think you might be talking about something like this

√(y¹⁰) / √(y⁶)

Notice that we could simplify this as

l y⁵ l /  l y³ l   this would account for the fact that y might be negative

But notice .....if we just simplified the original thing as this

√(y¹⁰) / √(y⁶) = √(y¹⁰ / y⁶)   √(y⁴) = y²

Then we don't have to worry about the absolute value signs at all....!!! 

I think you misunderstood my previous answer

2^n-1  for n = 1, 2, 3, 4.....

Note ....if we put 1 into this for "n" we get

2^1-1 = 2⁰ = 1

And if we put 2 into it, we have

2^2-1 = 2¹= 2

etc.

2x+9y=4

2x-6y=-1     multiply this equation by - 1   = -2x + 6y = 1

So we have

 2x +9y=4

-2x +6y=1    now.....add these together

      15y = 5        divide both sides by 15

y = 5/15 = 1/3

And using the frist equation, we can find x  by substituting 1/3 for y

2x + 9(1/3)  = 4

2x + 3  = 4    subtract 3 from both sides

2x  = 1          divide by 2  on both sides

x = 1/2

So the solution is (x, y) = (1/2, 1/3)

$${\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{12}}$$ =339.2920065876976698cm cubed

Hit the"2nd" button......you will see "asin" which is the tangent inverse

3.8 / 500 = .0076 cm

The "formula"  is

2^n-1  for n = 1, 2, 3, 4........

life is good! whats up? :)

y = (x + 2)(x - 2)^2       y = 3x + 5      setting the y's equal, we have

(x + 2) (x - 2)^2   =  3x + 5    simplify

(x + 2) (x - 2) ( x - 2)  = 3x + 5

(x^2 - 4)(x - 2) = 3x + 5

x^3 - 4x - 2x^2 + 8 = 3x + 5

x^3 - 2x^2 - 7x + 3 = 0    this isn't capable of being factored and the Rational Roots Theorem doesn't produce any results. either.......thus....a graphical solution seems to be a good option.....here you are....

16= 2k1 + 3k2

25= 5k1 + 4k2      

Multiply the first equation by 4 and the second equation by -3.....this will eliminate k2.....so we have

64 =    8k1     + 12k2

-75 = -15k1    -  12k2       add these together

-11 = -7k1      divide by -7 on both sides

11/7  = k1      substitute  this into either equation to find k2  ...I'll use the first one

16 = 2(11/7) + 3k2

16= 22/7 + 3k2      subtract 22/7 from borh sides

16 - 22/7  = 3k2

[112 - 22]/7  = 3k2

90/7  = 3k2     divide by 3 on both sides

90/21 = 30/7 = k2

So...our solution is (k1, k2)  = (11/7, 30/7 )