CPhill

24/ 75    reduces to 8/25 ....  multiplying the top and bottom by 4, we have 32/100  = .32 or 32%

This means we either rolll no fives, one five, two fives, three fives or 4 fives

The probability of rolling no fives is C(12,0)(1/6)^0*(5/6)^12 = 0.1121566547846151

The probability of rolling one five is C(12,1)(1/6)^1*(5/6)^11 = 0.0000000044516784

The probability of rolling two fives is C(12,2)(1/6)^2*(5/6)^10 = 0.2960935686313838

The probability of rolling three fives is C(12,3)(1/6)^3*(5/6)^9 = 0.1973957124209225

The probability of rolling four fives is C(12,4)(1/6)^4*(5/6)^8 = 0.0888280705894151

And summing these we have......about 69.4%  chance of rolling no more than 4 fives

Notice the pattern generated by raising a number ending in 7 to the following powers.....

7^1 = 7      7^2 = 49    7^3 = 343    7^4 = 2401 

And this cyclic pattern repeats every four powers

So    ...   256/4 = 64

And this whole number tells us that 7^256 completes a cycle of 4, so this number ends in a "1"

5 x 5 - 2(1/2) = 25 - 1 = 24

(7 - 1)(6)(2/3) = 6 * 6 * (2/3)  = 36 * (2/3)  = 24

There are 1760 yds. in a mile....so

(2 x 1760)/88  = 40 min.

3 in= .25 ft

So we have

12 x 24 x .25 = 72 cu ft

Each bag covers 1.5 cu ft

So

72 / 1.5 = 48 bags required  x $4 per bag = $192

2x^2 + 9x=-5     add 5 to both sides

2x^2 + 9x + 5 = 0      this will not factor, using the onsite solver, we have

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{41}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{41}}}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{3.850\: \!781\: \!059\: \!358\: \!212\: \!2}}\\
{\mathtt{x}} = -{\mathtt{0.649\: \!218\: \!940\: \!641\: \!787\: \!8}}\\
\end{array} \right\}$$

cot^2θ-csc^2θ

cot^2θ - (1 + cot^2θ) =

-1

-(1/5)(x-5)>-9    multiply both sides by -1 and reverse the inequality sign

(1/5)(x-5) < 9    multiply both sides by 5

x- 5 < 45           add 5 to both sides

x <  50

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