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Oh my goodness thank you sooo much!!!! This was really well done!!!!! I also really understood it!!! THANK YOU!!! 

p. s. in my time zone, when you published the answer, it was already like 3 am in the morning. Sorry i couldn't see it earlier. I was sleeping!!! \(:)\)

A stick has a length of \(5\) units. The stick is then broken at two points, chosen at random.

What is the probability that all three resulting pieces are longer than \(1\) unit?

The probability that all three resulting pieces are longer than \(1\) unit = \(\dfrac{ A_{\color{darkorange}orange} } {A} \)

\(\begin{array}{|rcll|} \hline H^2 + \left(\dfrac{S}{2}\right)^2 &=& S^2 \\ H^2 &=& S^2 -\dfrac{S^2}{4} \\ H^2 &=& \dfrac{3S^2}{4} \qquad \mathbf{H=\dfrac{S\sqrt{3}}{2}} \qquad (1) \\ S^2 &=& \dfrac{4H^2}{3} \\ S &=& \dfrac{2H}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\ \mathbf{S} &=& \mathbf{\dfrac{2H\sqrt{3}}{3}} \qquad (2) \\ \hline S &=& \dfrac{2H\sqrt{3}}{3} \quad | \quad H = 5 \\ \mathbf{S} &=& \mathbf{\dfrac{10\sqrt{3}}{3}} \\ \hline 2A &=& H*S \quad | \quad H=5,\ \mathbf{S=\dfrac{10\sqrt{3}}{3}} \\ \mathbf{2A} &=& \mathbf{\dfrac{50\sqrt{3}}{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \tan(30^\circ) &=& \dfrac{1}{x} \quad | \quad \tan(30^\circ) = \dfrac{1}{\sqrt{3}}\\ \dfrac{1}{\sqrt{3}} &=& \dfrac{1}{x} \\ \mathbf{ x } &=& \mathbf{\sqrt{3}} \\ \hline \mathbf{s} &=& \mathbf{S-2x} \quad | \quad \mathbf{S=\dfrac{10\sqrt{3}}{3}},\ \mathbf{ x =\sqrt{3}} \\ s &=& \dfrac{10\sqrt{3}}{3}-2\sqrt{3} \\ s &=& \dfrac{10\sqrt{3}}{3}-\dfrac{6\sqrt{3}}{3} \\ \mathbf{ s } &=& \mathbf{\dfrac{4\sqrt{3}}{3}} \\ \hline \mathbf{h^2 + \left(\dfrac{s}{2}\right)^2} &=& \mathbf{s^2} \\ h^2 &=& s^2 -\dfrac{s^2}{4} \\ h^2 &=& \dfrac{3s^2}{4} \\ \mathbf{h} &=& \mathbf{\dfrac{s\sqrt{3}}{2}}\quad | \quad \mathbf{ s =\dfrac{4\sqrt{3}}{3}} \\ h &=& \mathbf{\dfrac{\dfrac{4\sqrt{3}}{3}\sqrt{3}}{2}} \\ h &=& \dfrac{4}{2} \\ \mathbf{h} &=& \mathbf{2} \\ \hline 2A_{\color{darkorange}orange}&=& h*s \quad | \quad h=2,\ \mathbf{ s=\dfrac{4\sqrt{3}}{3}} \\ 2A_{\color{darkorange}orange} &=& 2*\dfrac{4\sqrt{3}}{3} \\ \mathbf{2A_{\color{darkorange}orange}} &=& \mathbf{\dfrac{8\sqrt{3}}{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{ A_{\color{darkorange}orange} } {A} &=& \dfrac{ 2A_{\color{darkorange}orange} } {2A} \\\\ &=& \dfrac{ \dfrac{8\sqrt{3}}{3} } {\dfrac{50\sqrt{3}}{3}} \\\\ &=& \dfrac{8} {50} \\\\ \mathbf{ \dfrac{ A_{\color{darkorange}orange} } {A} } &=& \mathbf{ \dfrac{4}{25} } \\ \hline \end{array}\)

Sorry Melody: You were RIGHT after all. There was atupid "bug" in my code. It is 1/8

Let  G  = number of Gello pens and  K  = number of Inko pens.

Number equation:  G + K  =  17

Value equation:     5G + 7K  =  109

G + K  =  17     --->     K  =  17 - G

Substituting:  5G + 7K  =  109     --->     5G + 7(17 - G)  =  109

                                                                5G + 119 - 7G  =  109

                                                                                 -2G  =  -10

                                                                                    G  =  5

Since  G + K  =  17  and  G  =  5,  K  =  12.

There is a formula to find the n^th term of a triangular number:  t_n  =  n(n + 1)/2

So, as Guest as shown, the 50^th term is:  50(50 + 1)/2  =  1275.

Triangular numbers begin like this:

(1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275)  - And 1275 is the 50th triangular number.

\(A regular dodecagon $P_1 P_2 P_3 \dotsb P_{12}$ is inscribed in a circle with radius $1.$ Compute \[(P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2.\]\)

Hi Melody: 1/8 cannot be right because 1/8 would be multiple of all natural numbers from 0 to 9. But, you  are only dealing with 6 numbers from 1 to 6!

The permutations begin like this:
11111112 , 11111136 , 11111144 , 11111152 , 11111176 , 11111184 , 11111192 , 11111216.........and end like this: 66616592 , 66616616 , 66616624 , 66616632 , 66616656 , 66616664 , 66616672 , 66616696, which are ALL multiples of 8. My computer did list all 54,432 permutations which are multiples of 8.

As you can see, you cannot have a number beginning with  5 sixes, or 66666, because you cannot make a multiple of 8 out of 5 sixes in a row! If you want me to, I can print all 54,432 multiples of 8 out of 8 dice.

You have linear functions \(p(x)\) and \(q(x)\). You know \(p(2)=3\), and \(p\Big(q(x)\Big)=4x+7\) for all x.
Find \(q(-1)\).

\(\text{Let $p(x)=ax+b$}\)

\(\begin{array}{|rcll|} \hline p(x) &=& ax+b \quad | \quad x=2 \\ p(2) &=& 2a+b \quad | \quad p(2) = 3 \\ 3 &=& 2a+b \\ \mathbf{2a} &=& \mathbf{3-b} \\ \hline \end{array} \begin{array}{|rcll|} \hline p\Big(q(x)\Big) &=& aq(x)+b \quad | \quad p\Big(q(x)\Big) = 4x+7 \\ 4x+7 &=& aq(x)+b \quad | \quad x=-1 \\ 4(-1)+7 &=& aq(-1)+b \\ 3 &=& aq(-1)+b \\ 3-b &=& aq(-1) \quad | \quad * 2 \\ 2(3-b) &=& 2aq(-1) \quad | \quad \mathbf{2a=3-b} \\ 2(3-b) &=& (3-b)q(-1) \quad | :( 3-b) \\ 2 &=& q(-1) \\ \mathbf{q(-1)} &=& \mathbf{2} \\ \hline \end{array}\)

In triangle ABC, AB = BC = 25 and AC = 40. What is sin

Answer with an exact number.

I get 27 too

P(divisable by 8) = \(\frac{27}{6^3}= \frac{3^3}{3^3*2^3}=\frac{1}{8}\)

So I agree with Fiora

NOTE

27 is not the number that are multiples of 8 and 6^3 is not the total number of numbers.

The fact is that only the last 3 rolls have any relevance to this problem.

So Fiora and I just chose to completely ignore the first 5 rolls.

If 7n+1 is a multiple of 4, then what is n mod 4 

7n+1 = 0 (mod 4)

4n+3n+1 = 0 (mod 4)

3n+1 = 0 (mod 4)

3n = -1 (mod 4)

3n = 3 (mod 4)

n = 1 (mod 4)

My computer says there are: 54,432 numbers that are mutiples of 8.

Therefore, the probability is: 54,432 / 6^8 =7 / 216

Ah

I feel so dum

And I think that they are the only 2 missing, right?

THHHHHT

HHHHHTT

HHHHHTH 

TTHHHHH

HTHHHHH 

THHHHHH

HHHHHHT

HHHHHHH

Yes, i think that's it.

So... \(\frac{8}{2^7}=\frac{8}{128}=\frac{1}{16} \)

THANQ

How do you work out what y and z are if x=y (mod z), where 7x+1=0 mod 3, x=0 mod 2,
without guessing and checking, and finding the LCM?

\(\begin{array}{|lrcll|} \hline & 7x+1 &\equiv& 0 \pmod{ 3 } & \text{or} \\ & 7x+1 &=& 0 + 3n,\ n\in \mathbb{Z} \\ & 7x+1 &=& 3n \\ (1) & \mathbf{7x} &=& \mathbf{3n-1} \\ \hline & x &\equiv& 0 \pmod{ 2 } & \text{or}\\ & x &=& 0 + 2m,\ m\in \mathbb{Z} \\ & x &=& 2m \quad | \quad *7 \\ (2) & \mathbf{7x} &=& \mathbf{14m} \\ \hline & \mathbf{7x} = 3n-1 &=& 14m \\ & 3n-1 &=& 14m \\ (3) & \mathbf{3n-14m} &=& \mathbf{1} \\ \hline \end{array} \)

There is a difference between your question and the one answered by heureka at : https://web2.0calc.com/questions/a-fair-coin-is-flipped-7-times-what-is-the-probability-that-at-least-5-of-the-flips-come-up-heads; namely you are asking for 5 consecutive heads, whereas the other questions just asked for 5 heads (not necessarily consecutive)..

For your case, have you ignored HHHHHTH and HTHHHHH ?

Each triangle in this figure is an isosceles right triangle. The length of BD is 2 units. What is the number of units in the perimeter of the quadrilateral ABCD? Express your answer in simplest radical form.

Perimeter  P = 2√2 + √8 + √4  

a)

\(\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\ \text{by counting the number of ordered triples (a, b) of positive integers}, \\\text {where }   1 \le a,b,c \le   \text{in two different ways.}\\ Where \;\;\;n\ge3\;\;\;and \;\;\;n\in Z\)

I am going to prove this using Induction.

STEP 1

Prove true for n=3

\(LHS=n^3 \\ LHS=3^3 \\ LHS=27\\ RHS= n + 3n(n - 1) + 6 \binom{n}{3}\\ RHS= 3 + 3*3(3 - 1) + 6 \binom{3}{3}\\ RHS= 3 + 18 + 6 \\ RHS=27\\ RHS=LHS\\ \text{So the statement is true for n=3}\)

STEP 2

If true for n=k prove true for n=k+1

so we can assume

\(k^3=k+3k(k-1)+6\binom{k}{3}\\ k^3=k+3k^2-3k+6\binom{k}{3}\\ k^3=3k^2-2k+6\binom{k}{3}\\\)

Now I need to prove it will be true for n=k+1

i.e.  Prove

\((k+1)^3=(k+1)+3(k+1)(k+1-1)+6\binom{k+1}{3}\\ (k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\ \qquad \qquad \qquad \text{An aside:}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)!}{3!(k+1-3)!}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{k!(k+1)}{3!(k-3)!(k-2)}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k-2)+3}{(k-2)}*\binom{k}{3}\\ \qquad \qquad \qquad \binom{k+1}{3}=\binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\\\)

\(LHS=(k+1)^3\\ LHS=k^3+3k^2+3k+1\)

\(RHS=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\ RHS=(k+1)+3(k+1)(k)+6\left[ \binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\right]\\ RHS=k+1+3k^2+3k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\ RHS=1+3k^2+4k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\ RHS=3k^2-2k+1+6 \binom{k}{3}+6k+\frac{6*3}{k-2}*\binom{k}{3}\\ substituting\\ RHS=k^3+6k+1+\frac{6*3}{k-2}*\binom{k}{3}\\ RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{k!}{3!(k-3)!}\\ RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{(k-3)!(k-2)(k-1)k}{3!(k-3)!}\\ RHS=k^3+6k+1+\frac{6*3}{1}*\frac{(k-1)k}{3!}\\ RHS=k^3+6k+1+3k^2-3k\\ RHS=k^3+3k^2+3k+1\\ RHS=LHS\\~\\ \text{So if it is true for n=k then it will be true also for n=k+1} \)

Step 3

Since it is true for n=3 it must be true for n=4, n=5, n=6,  ....

Hence it must be true for all integer n where n greater or equal to 3.

LaTex:

\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\
 \text{by counting the number of ordered triples (a, b) of positive integers}, 
\\\text {where }   1 \le a,b,c \le   \text{in two different ways.}\\
Where \;\;\;n\ge3\;\;\;and \;\;\;n\in Z

LHS=n^3        \\                              
LHS=3^3 \\
LHS=27\\
RHS= n + 3n(n - 1) + 6 \binom{n}{3}\\
RHS= 3 + 3*3(3 - 1) + 6 \binom{3}{3}\\
RHS= 3 + 18 + 6 \\
RHS=27\\
RHS=LHS\\
\text{So the statement is true for n=3}

k^3=k+3k(k-1)+6\binom{k}{3}\\
k^3=k+3k^2-3k+6\binom{k}{3}\\
k^3=3k^2-2k+6\binom{k}{3}\\

(k+1)^3=(k+1)+3(k+1)(k+1-1)+6\binom{k+1}{3}\\
(k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\
\qquad \qquad \qquad \text{An aside:}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)!}{3!(k+1-3)!}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{k!(k+1)}{3!(k-3)!(k-2)}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k-2)+3}{(k-2)}*\binom{k}{3}\\
\qquad \qquad \qquad \binom{k+1}{3}=\binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\\

LHS=(k+1)^3\\
LHS=k^3+3k^2+3k+1

RHS=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\
RHS=(k+1)+3(k+1)(k)+6\left[ \binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\right]\\
RHS=k+1+3k^2+3k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\
RHS=1+3k^2+4k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\
RHS=3k^2-2k+1+6 \binom{k}{3}+6k+\frac{6*3}{k-2}*\binom{k}{3}\\
substituting\\
RHS=k^3+6k+1+\frac{6*3}{k-2}*\binom{k}{3}\\
RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{k!}{3!(k-3)!}\\
RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{(k-3)!(k-2)(k-1)k}{3!(k-3)!}\\
RHS=k^3+6k+1+\frac{6*3}{1}*\frac{(k-1)k}{3!}\\
RHS=k^3+6k+1+3k^2-3k\\
RHS=k^3+3k^2+3k+1\\
RHS=LHS\\~\\
\text{So if it is true for n=k then it will be true also for n=k+1}

No one got an answer? 

@P = $3/4($ 2rpi ) +& 2r 

If the questions can be "camouflaged", so can the answers.

Is Alan's answer good enough?

or do you just want the answer? 

Yep you are correct

d is definitely not it

\(\frac{d}{dx}\left(e^{3x}\sin \left(3x\right)\right)\\ \frac{d}{dx}\left(e^{3x}\right)\sin \left(3x\right)+\frac{d}{dx}\left(\sin \left(3x\right)\right)e^{3x}\\ \frac{d}{dx}\left(e^{3x}\right)=e^{3x}\cdot \:3\\ \frac{d}{dx}\left(\sin \left(3x\right)\right)=\cos \left(3x\right)\cdot \:3\\ e^{3x}\cdot \:3\sin \left(3x\right)+\cos \left(3x\right)\cdot \:3e^{3x}\\\)

Hmm...

That's confusing

Report an error if possible

You are right, I think.

(a+bi)^2=2i

then you expand to \(\left(a^2-b^2\right)+2iab=2i\)

\(\left(a^2-b^2\right)+2iab=0+2i\)

then, let's write is as system of equations: (I assume you know what it is)

\(\begin{bmatrix}a^2-b^2=0\\ 2ab=2\end{bmatrix}\)

So, \(x=1+i,\:x=-1-i\)

Hope this made sense! :D

-tigernathan