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you don't need to post a link to a question, if its very recent.

"DRAFTY" =6 letters =6! =720 permutations. Each permutation will begin with one of the 6 letters in: 720 / 6 = 120 ways. And so the letters "AY", in that order, will occur 120 times in the 720 permutations. Similarly, the letters "YA", in that order, will also occur 120 times in 720 permutation of the word "DRAFTY".

So, the probability of "A followed by Y, or "AY'' is: 120 /720 = 1/6

You're welcome..... 

605,8=(Pi/1530)²*21000*i      divide both sides by  (pi/1530)^2

143685315.3  = 21000 i        now divide both sides by 21000

i = 6842.16

I edited my question, it's not 1/6 it is 1/8... I'm so sorry I don't know why it didn't show it, you don't have to solve it again though, I think I understood how I was supposed to do it. Thank you!

This is a sphere with   r = 2 "       volume of a sphere    4/3 pi r^3

  PLUS a cylinder with r = 2"        volume of cylinder     pi r^2 h               h = 8 -2 -2 = 4 inches

Plug and Play.......

Multiply conjugates.

\(\quad \dfrac1{1 + \sqrt 3 + \sqrt 5}\\ = \dfrac1{1 + \sqrt 3 + \sqrt 5} \cdot \dfrac{1 + \sqrt 3 - \sqrt 5}{1 + \sqrt 3 - \sqrt 5}\\ = \dfrac{1 + \sqrt 3 - \sqrt 5}{(1 + \sqrt 3)^2 - \sqrt5^2}\\ = \dfrac{1 + \sqrt 3 - \sqrt 5}{4 + 2\sqrt 3 - 5}\\ = \dfrac{1 + \sqrt 3 - \sqrt 5}{2\sqrt 3 - 1}\\ = \dfrac{1 + \sqrt 3 - \sqrt 5}{2\sqrt 3 - 1}\cdot \dfrac{2\sqrt 3 + 1}{2\sqrt 3 + 1}\\ = \dfrac{(1 + \sqrt 3 - \sqrt 5)(2 \sqrt 3 + 1)}{(2\sqrt 3)^2 - 1^2}\\ = \dfrac{1 + 2\sqrt 3 + 6 + \sqrt 3 - 2\sqrt{15} - \sqrt 5}{11}\\ = \dfrac{7 + 3 \sqrt 3 - \sqrt 5 - 2\sqrt {15}}{11}\)

I think I recall that the height of the head is 1/6 of the total.....(it isn't in your Q)

1/6  *   6' 10"  =

1/6   * 82" =  14"   head height

Enumerate possible combinations:

{8}

{4, 4}

{4, 2, 2}

{4, 2, 1, 1}

{4, 1, 1, 1, 1}

{2, 2, 2, 2}

{2, 2, 2, 1, 1}

{2, 2, 1, 1, 1, 1}

{2, 1, 1, 1, 1, 1, 1}

{1, 1, 1, 1, 1, 1, 1, 1}

Total 10 ways.

We calculate side lengths by Pythagorean theorem.

\(DE = \sqrt{17}\\ EF = \sqrt{17}\\ DF = \sqrt{34}\)

Obviously, \(DE^2 + EF^2 = DF^2\)

So by converse of Pythagorean theorem, m∠DEF = 90^o.

Let 

\(a = x^2 - 10x - 45\\ b = x^2 + 10x - 29\\ c = x^2 - 10x - 69\)

\(\dfrac1a + \dfrac1b = \dfrac2c\\ 2ab = c(a + b)\\ 2(x^2 - 10x - 45)(x^2 + 10x - 29) = 2(x^2 - 10x - 69)(x^2 - 37)\)

\((x^2 - 10x - 45)(x^2 + 10x - 29) = (x^2 - 10x - 69)(x^2 - 37)\)

Then expand and simplify.

\(5x^3 - 34x^2 - 265x - 624 = 0\)

Solving by numerical methods gives \(x\approx 12.055\).

\(ab^4 = 384\\ a^2 b^5 = 4608 \)

Considering the second equation: 

\(ab^4 \cdot ab = 4608\\ ab = \dfrac{4608}{384} = 12\)

Considering the first equation:

\(ab^4 = 384\\ ab\cdot b^3 = 384\\ b = \sqrt[3]{\dfrac{384}{12}} = 2\sqrt[3]4\)

Substituting directly gives \(a =3 \sqrt[3]{2}\)

oooh, I see now

\(\sqrt{8} = \sqrt{2^2 \cdot 2} = 2 \sqrt2\\ \sqrt{32} = \sqrt{4^2 \cdot 2} = 4\sqrt 2\)

So

\(\quad\sqrt8 + 3\sqrt 2 + \sqrt{32}\\ = 2\sqrt 2 + 3\sqrt 2 + 4\sqrt 2\\ = 9\sqrt 2\)

By considering power of points at X,

\(2(2 + 7) = 3(3 + ED)\\ ED = 3\)

\(\quad\left(\sqrt{3x} + \sqrt 5\right)\left(\sqrt{15x} + 2\sqrt{30}\right)\\ = \sqrt{45x^2} + \sqrt{75x} + 2\sqrt{90x} + 2\sqrt{150}\\ =3x\sqrt5 + 5\sqrt{3x} + 6\sqrt{10x} + 10\sqrt 6\)

\(f(f^{-1}(x)) = f(4)\\ x = 3\cdot 4^3 + 2 = 194\)

can anyone help me 

605,8=(Pi/1530)²*21000*i

i = 6842 cm^4 

ist that right?if yes can you show me the way

Thank you as well!

Amanda's portion is 1/10 of the total, which is $20. Then the total amount is $200. 

Thank You!

Amanda's is    1 out of   1+2 + 7 = 10      1/10th

10 x 20 = $ 200 total

See the answer here:  https://web2.0calc.com/questions/help-please-been-stuck-on-this-for-weeks#r1

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