Melody

Four letters are selected at random with replacement from the word MATHCOUNTS. What is the probability that the letters in the order they are selected will form the word NUTS?

This is what I think.  Sounds correct to me.

1/10 * 1/10 * 2/10 *1/10 = 2/(10^4)  = 2/10000  = 1/5000  = 0.0002

This is a concave up parabola.

The axis of symmetry is    -b/2a  =  -8/2  = -4

when x=-4

f(-4)= 16-32+c = -16+c

-16+c <= -2

c <= 14

\( \dfrac{8x-4}{x+3} - \dfrac{12x-20}{2x+6}\\ =\dfrac{8x-4}{x+3} - \dfrac{2(6x-10)}{2(x+3)}\\ =\dfrac{8x-4-(6x-10)}{x+3} \\ =\dfrac{8x-4+[-1(6x-10)]}{x+3} \\ \)

You can take it from there, questions are welcome.

Thanks proyaop for finding that answer.

This was a great question and the answer by Oofrence was really good.

\(1/(1 + 1/(sqrt(3) + 2))\\ =\frac{1}{1 + \frac{1}{(\sqrt3 + 2)}}\\ =1\div( 1 + \frac{1}{(\sqrt3 + 2)})\\ =1\div( \frac{\sqrt3+2+1}{(\sqrt3 + 2)})\\ =1\div \frac{3+\sqrt3}{2+\sqrt3 }\\\)

Invert and multiply

then

multiply top and bottom by the conjugate of the bottom.

Can you take it from there?

You could just use a calculator.

factor(1000343) = 107*9349

Answered on the original

Hi MGoo

2b = 2a+c       The external angle of a triangle is equal to the sum of the opposite internal angles

b=   a+37        same reason

now its easy to finish.  :)

The general term is:

\(\displaystyle \binom{9}{n}\;(2z)^{(9-n)}(\frac{-1}{\sqrt z})^n\\~\\ =\displaystyle \binom{9}{n}\;(-1)^n(2)^{(9-n)}(\frac{1}{\sqrt z})^n*z^{(9-n)}\\~\\ =\displaystyle \binom{9}{n}\;(-1)^n(2)^{(9-n)}*z^{( \frac{-n+18-2n}{2} )}\\~\\ =\displaystyle \binom{9}{n}\;(-1)^n(2)^{(9-n)}*z^{( \frac{18-3n}{2} )}\\~\\ \qquad \text{You want the constant so}\\ \qquad (18-3n)/2=0 \quad \therefore n=6\\~\\ constant = \displaystyle \binom{9}{6}\;(2z)^{(9-6)}(\frac{-1}{\sqrt z})^6\\~\\ constant = \displaystyle \binom{9}{6}\;(2z)^3 (\frac{-1}{\sqrt z})^6\\~\\ constant = \displaystyle \binom{9}{6}\;*\;8\\~\\ \)

Answers without explanations are useless.