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Suppose that ABCD is a trapezoid in which AD is parallel to BC. Given AC is perpendicular to CD, AC bisects ∠BAD, and the area of ABCD is 20, then compute the area of triangle ACD.

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AB = BC = 1/2 (AD)

[ACD] = 2/3 (20) 

asymptope at   1 and -2

(x+4) / [(x+2)(x-1)]

(x+4)/(x^2 +x-2)    tak it from here ....

4 dials with digits  1 2 3 4

   4 x 4 x 4 x 4 = 4⁴  possible combinations    = 256 possibles     

       you have a 1 in 256 chance of guessing it on one try

I got it 

it is 45

This is the correct question 

\($In trapezoid $ABCD$, sides $\overline{AB}$ and $\overline{CD}$ are parallel, $\angle A = 2\angle D$, and $\angle C = 3\angle B$. Find $\angle B$.\)

\($In trapezoid $ABCD$, sides $\overline{AB}$ and $\overline{CD}$ are parallel, $\angle A = 2\angle D$, and $\angle C = 3\angle B$. Find $\angle B$.\)

Area of a kite      A = (12 * 20) / 2 = 120 u²

The diagonals of a rhombus divide its area into 4 congruent right-angled triangles. Since the area of the whole rhombus is 120 u², the area of a single triangle is 30 u². In order to find sides of this triangle, we can double its area.

1st attempt:               6 * 10 = 60               sqrt(6² + 10²) = 11.662        no good

2nd   attempt:            5 * 12 = 60               sqrt(5² + 12²) = 13         good 

So, the length of a side of quadrilateral P is 13 units.

Merely Unlucky?  No! Not merely. This unlucky series of failures is astronomically unlikely.

Ignoring the roulette wheel for the moment, this question has the same probability as flipping a fair coin 59 times and incorrectly guessing the outcome every time –or correctly guessing the outcome every time. It’s also the same probability as having 59 heads appear or 59 tails, or...

For this question, a failure is a statistical success. 

With a 50% probability of success/fail, the probability of 59 sequential failures is (0.5)^59 is: 

\(\dfrac{1}{576460752303423488} = 1.73472347597680709441192448139190673828125E-18\)

 Here are some perspectives comparing other probabilities to the probability of 59 sequential failures in guessing coin flips.

Lotteries: PowerBall, MegaMillions, and Keno 80/20

The probability of simultaneously winning both jackpots in the PowerBall and MegaMillions lotteries with a single ticket for each lottery is 6.5 times greater than (59) sequential failures in guessing coin flips.

The Math:

\(\dfrac {1}{292201338} * \dfrac {1}{302575350}= \dfrac {1}{88412922115818300}\)

\(\dfrac{576460752303423488}{88412922115818300} \approx 6.52\) times more likely than (59) sequential failures in guessing coin flips.

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Keno 80/20

Matching all 20 numbers on Keno 80/20 is 163 times more likely than (59) sequential failures in guessing coin flips.

The Math:

 Probability of matching all 20 numbers on Keno: 

\(\dfrac {1} {\dbinom{80}{20}} = 3535316142212174320 \)

\( \underbrace {576460752303423488}_{59 \;sequential \;coin \;fails} \div \underbrace {3535316142212174320 } _{20 \; Keno \; matches} \approx 163.13 \text{ times more likely}\)

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Blackjack:

The probability of 13.4064 sequential (natural) BlackJack wins is slightly less than the probability for (59) sequential failures in guessing coin flips.

Blackjack Math:

The probability of a single deck (natural) Blackjack (Blackjack on the initial deal of two (2) cards) is 

\(2* \dfrac {4}{13}* \dfrac {4}{51} = \dfrac {8}{169} \approx 0.047619\\\)

Inverse of  \(\dfrac {8}{169} = 21.125\)  (This is used as the base of a Logarithm)

\(\log_{21.125} (576460752303423488) \approx 13.4064\) sequential Blackjack wins for every (59) sequential failures in guessing coin flips.

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Royal Flush:

The probability of 3.0555  sequential (natural) Royal Flush wins in 5-card draw poker will occur slightly more than 3 times before (59) sequential failures in guessing coin flips.

Royal Flush Math:

The probability of a (natural) Royal Flush for 5-card draw poker is \(\dfrac {1}{649740}\)

This gives

 \(\large log_{649740}(576460752303423488) =3.0555\) sequential Blackjack wins for every (59) sequential failures in guessing coin flips.

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Testing the sample space:

Conducting this experiment 576,460,752,303,423,488 times gives an expectation of one (1) statistical success.   However, expectation is not probability. The probability of realizing this expectation is:

\(1- \left(\frac {576460752303423487}{576460752303423488}\right )^{576460752303423488} \approx  63.21 \% \)

So the probability of NOT realizing this expectation is:

\(\approx 36.79 \%\) (Not coincidentally, this is equal the inverse of Euler's Number (e) 2.71828...)

Yep, doing this experiment 576,460,752,303,423,488 times will result in failure well more than \(\dfrac {1}{3}\) of the time. 

How long to conduct an experiment?

Figure it will take about 40 seconds, with an experienced croupier spinning the (special, no green pocket zero(s)), red-black, even-odd, Roulette wheel and ball; and the gambler flipping the coin and placing the wager while the ball orbits the wheel.  That’s just one spin in the experimental attempt. The flipping and spinning continues until the gambler matches (wins) (or fails to match (loses) 59 times in sequence). If the gambler wins, the experiment begins again.   Most of the experiments will fail before the fifth flip and spin.  Remember, 576,460,752,303,423,488 is the number of experiments, NOT the number of coin flips and wheel spins, which will be many multiples.     

So how long would it take to do 576,460,752,303,423,488 coin flips and wheel spins, figuring 40 seconds for each one?

That’s 23,058,430,092,136,939,520 seconds. That’s a long time. It’s (53) times the age of the universe. Calculated using the Big Bang event as a starting point –the formation/creation of the universe –more formally, the starting point when the universe inflated and space-time began. (This should not be confused with the big bang of your mum and pop, the starting point that procreated you, even though that was also long ago.)

So Ron, do you still think the gambler is Merely Unlucky? 

If you do, I’m willing to bet a Rouble that some Russian fucked with the Roulette wheel? 

Known in Russia as Гребаная русская рулетка    

GA

Follow the steps and you will get your and answer

Such trapezoid is NOT possible!!!           52 - (5 + 10) = 37        CD = 36

10,000 combinations because you can go up to 9999

10,000

7+11+13=31

Haha don't worry you will get better

Future Value =  1000 (1+.05)⁴     Calculate this value

                                                         ...then subtract the original $ 1000 to find her interest earned

2/7 ÷ 3/5   =     2/7 x 5/3  =  10/21

N=4; R=0.05;PV=1000; FV=PV*(1 + R)^N; print"FV =$",FV

FV =$ 1215.51 - Value of the investment after 4 years.

$1215.51  -  $1,000 =$215.51 - Interest earned after 4 years. [Money should be rounded to the nearest hundred].

It is 2.1

I did

       2/7

      -3/5

    .............

       1.2

then i reversed the numbers

       2.1

 (this does not always work)

you could just use the caculator on here too

Any number divisible by 6 is also divisible by 3.

If you add 2 to any multiple of 3 the result is not divisible by 3.

Hence none of the numbers in the sequence are divisible by 6.

A sequence is formed by adding 2 to the triple of the previous term. If the first term is 1 , how many multiples of 6 less than 10000  would be terms of the sequence?

Do you undestand the question?

First term = 1

2nd term  =1 x 3 + 2 = 5

3rd term  =5 x 3 + 2 = 17

4th term   =17 x 3 + 2 = 53

5th term =53 x 3 + 2 = 161

6th term=161 x 3 + 2 = 485

7th term=485 x 3 + 2 = 1457

8th term=1457 x 3 + 2 =4373

9th term=4373 x 3 + 2 =Goes over 10,000

Of the 8 terms, how many of them are multiple of 6? Or are divisible by 6? Use your calculator and find out!.

In triangle ABC, AB=c, BC=a, CA=b,
\(\angle A = \alpha\),
\(\angle B = \beta\), and \(\angle C = \gamma\)
where all angles are measured in degrees.
Suppose that \(b^2-a^2=ac\) and \(\beta^2 = \alpha \gamma\).
Find \(\lfloor 10\beta \rfloor\).

1.
\(\begin{array}{|rcll|} \hline b^2-a^2 &=&ac \\ (b-a)(b+a) &=& ac \\ \mathbf{(a-b)(a+b)} &=& \mathbf{-ac} \\ \hline \end{array}\)

Mollweide's formula

\(\begin{array}{|rcll|} \hline (a-b)\cos\left(\dfrac{\gamma}{2}\right) &=& c\cdot \sin\left(\dfrac{\alpha-\beta}{2} \right) \qquad (1) \\ (a+b)\sin\left(\dfrac{\gamma}{2}\right) &=& c\cdot \cos\left(\dfrac{\alpha-\beta}{2} \right) \qquad (2) \\ \hline \end{array}\)

(1) x (2)

\(\begin{array}{|rcll|} \hline (a-b)(a+b)\sin\left(\dfrac{\gamma}{2}\right)\cos\left(\dfrac{\gamma}{2}\right) &=& c^2\sin\left(\dfrac{\alpha-\beta}{2} \right)\cos\left(\dfrac{\alpha-\beta}{2} \right) \\ -ac\dfrac{\sin(\gamma)}{2} &=& c^2\dfrac{\sin(\alpha-\beta)}{2} \\ -a\sin(\gamma) &=& c\sin(\alpha-\beta) \\ a\sin(\gamma) &=& c\left(-\sin(\alpha-\beta)\right) \\ a\sin(\gamma) &=& c\left(\sin(-(\alpha-\beta))\right) \\ a\sin(\gamma) &=& c\sin(\beta-\alpha) \\ \dfrac{a}{c} &=& \dfrac{\sin(\beta-\alpha)}{\sin(\gamma)} \\ && \boxed{\dfrac{a}{c}=\dfrac{\sin(\alpha)} {\sin(\gamma)} } \\ \dfrac{\sin(\beta-\alpha)}{\sin(\gamma)} &=& \dfrac{\sin(\alpha)} {\sin(\gamma)} \\ \sin(\beta-\alpha) &=& \sin(\alpha) \\ \beta-\alpha &=& \alpha \\ \beta &=& 2\alpha \quad \text{or} \quad \mathbf{\alpha = \dfrac{\beta}{2}} \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline \beta^2 &=& \alpha \gamma \quad | \quad \mathbf{\alpha = \dfrac{\beta}{2}} \\ \beta^2 &=& \dfrac{\beta}{2} \gamma\\ \beta&=& \dfrac{\gamma}{2} \\ \mathbf{\gamma} &=& \mathbf{2\beta} \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline \mathbf{ \alpha + \beta + \gamma } &=& \mathbf{ 180^\circ } \quad | \quad \alpha = \dfrac{\beta}{2} \\ \dfrac{\beta}{2} + \beta + \gamma &=& 180^\circ \\ \dfrac{3}{2}\cdot \beta + \gamma &=& 180^\circ \quad | \quad \gamma = 2\beta \\ \dfrac{3}{2}\cdot\beta + 2\beta &=& 180^\circ \quad | \quad :2 \\ \dfrac{3}{4}\cdot\beta + \beta &=& 90^\circ \\ \dfrac{7}{4}\cdot\beta &=& 90^\circ \\ \beta &=& \dfrac{4}{7}\cdot 90^\circ \\ \beta &=& 51.4285714286^\circ \\ 10\cdot \beta &=& 514.285714286^\circ \\ \mathbf{ \lfloor 10\beta \rfloor } &=& \mathbf{ 514 } \\ \hline \end{array}\)