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in the sampler I got an example 22 ^ 10 / (2 ^ 7 * 11 ^ 9) = 88 I want to understand the solution, how did it happen?

\(\frac{22^{10}}{2^7*11^9}\\~\\ =\frac{(2*11)^{10}}{2^7*11^7*11^{2}}\\~\\ =\frac{(2*11)^{10}}{(2*11)^7*11^{2}}\\~\\ =\frac{(2*11)^{10-7}}{11^{2}}\\~\\ =\frac{(2*11)^{3}}{11^{2}}\\~\\ =\frac{2^3*11^3}{11^{2}}\\~\\ =\frac{2^3*11^{(3-2)}}{1}\\~\\ =8*11\\~\\ =88 \)

Thanks Ginger, that is very interesting :)

to polar form

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{12\sqrt{2}\mathop{cis}\left(\dfrac{7\pi}{6}\right)} {3\sqrt{3}\mathop{cis}\left(\dfrac{\pi}{4}\right)} } \\\\ &=& \dfrac{4\sqrt{2}\mathop{cis}\left(\dfrac{7\pi}{6}\right)} {\sqrt{3}\mathop{cis}\left(\dfrac{\pi}{4}\right)} \\\\ &=& \dfrac{4\sqrt{2}\sqrt{3}\mathop{cis}\left(\dfrac{7\pi}{6}\right)} {\sqrt{3}\sqrt{3}\mathop{cis}\left(\dfrac{\pi}{4}\right)} \\\\ &=& \dfrac{4\sqrt{6}\mathop{cis}\left(\dfrac{7\pi}{6}\right)} {3\mathop{cis}\left(\dfrac{\pi}{4}\right)} \\\\ &=& \dfrac{4\sqrt{6} e^{i\frac{7\pi}{6}} } {3 e^{i\frac{\pi}{4}} } \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\frac{7\pi}{6}}e^{-i\frac{\pi}{4}} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\frac{7\pi}{6}-i\frac{\pi}{4}} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\pi\left(\frac{7}{6}- \frac{1}{4}\right)} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\pi \left(\frac{28-6}{24} \right)} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\pi \frac{22}{24}} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\pi \frac{11}{12}} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\frac{11\pi}{12}} \\\\ &=& \mathbf{\dfrac{4\sqrt{6}}{3} \mathop{cis}\left( \dfrac{11\pi}{12}\right)} \\ \hline \end{array}\)

AELN, your answer is wrong! Apparently, you don’t understand what an inertial moment is.

There are 24 possible arrangements for the weights, and none of these arrangements will balance the see saw.

There is no solution to this problem.

law of sines

\(\begin{array}{|rcll|} \hline \dfrac{AB}{\sin(49^\circ)} &=& \dfrac{18}{\sin\Big(180^\circ-(49^\circ+78^\circ)\Big)} \\\\ && \boxed{\text{Formula: }\ \sin(180^\circ-\theta) = \sin(\theta)} \\\\ \dfrac{AB}{\sin(49^\circ)} &=& \dfrac{18}{\sin(49^\circ+78^\circ)} \\\\ \dfrac{AB}{\sin(49^\circ)} &=& \dfrac{18}{\sin(127^\circ)} \\\\ AB &=& \dfrac{18\sin(49^\circ)}{\sin(127^\circ)} \\\\ AB &=& \dfrac{18*0.75470958022}{0.79863551005} \\\\ AB &=& \dfrac{13.5847724440}{0.79863551005} \\\\ AB &=& 17.0099779851 \\ \mathbf{ AB } &\approx& \mathbf{17.0\ \text{in}} \\ \hline \end{array}\)

Evaluate the infinite geometric series: \(1-\dfrac{2}{7}+\dfrac{4}{49}-\dfrac{8}{343}+\dots\)

\(\begin{array}{|rcll|} \hline && \mathbf{1-\dfrac{2}{7}+\dfrac{4}{49}-\dfrac{8}{343}+\ldots} \\\\ &=& 1+\left(-\dfrac{2}{7}\right) +\left(-\dfrac{2}{7}\right)^2 +\left(-\dfrac{2}{7}\right)^3 +\ldots \quad | \quad \text{ratio}\ = -\dfrac{2}{7} \\ \hline \end{array}\)

Formula:

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\dfrac{1}{1-r}} \quad | \quad r=\text{ratio},\ s=\text{infinite sum} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\dfrac{1}{1-r}} \quad | \quad r = -\dfrac{2}{7} \\\\ s &=& \dfrac{1}{1-\left(-\dfrac{2}{7}\right)} \\\\ s &=& \dfrac{1}{1+ \dfrac{2}{7} } \\\\ s &=& \dfrac{1}{\dfrac{7+2}{7} } \\\\ s &=& \dfrac{7} {7+2} \\\\ \mathbf{s} &=& \mathbf{\dfrac{7}{9}} \\ \hline \end{array}\)

WOAH! If you look at the computer screen, there are beautiful mathematical curves there. If anyone with the knowledge on why the curves look the way they do and their color, it would be very appreciated. I am very curious right now .

Guest has unfortunately counted every segment twice O_O

I think I like Melody's method the best because it is a good way to visualize the solution. The official solution is the best if you are familiar with these types of problems.

Wow. I bookmarked it. And yes! I don't think Loki is dead. 

In the small triangle....h is to 4        In the larger triangle   3 is to  (2+4)  

Thank you :)

\(f(x) = \frac{3x + 4}{x^2 + 3x + 5} \)

If I look at top and bottom separately I can see that the bottom would be a concave up parabola with no roots.

Therefore,for all real values of x the denominator will always be positive.

Find the turning points. This is where  \(f'(x)=0\)

\(f(x) = \frac{3(x^2 + 3x + 5)-(2x+3)(3x + 4)}{(x^2 + 3x + 5)^2}\\ f(x) = \frac{-3x^2-8x+3}{(x^2 + 3x + 5)^2}\\ f'(x)=0\quad when\quad -3x^2-8x+3=0\\ \text{That happens when } x=-3,\;and \;\;when\;x=\frac{1}{3}\)

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I only just realized that Geno had done that much and you showed no sign of understanding that his question was incomplete and a little wrong.

Do you understand the graph Desmos drew?    Can you tell me the answer from that graph?

You will need to use a little algebra as well but you do not have to do ALL the algebra to get the correct answer from the graph.

HINTS:

You need to find the exact max and min.

If you do the whole thing algebraically you would also need to get the y value as x approaches + and - infinity.  You can use L'hopital's rule to do that.  BUT you can see the answer from the graph anyway.

Come back with what you discover.

LaTex:

f(x) = \frac{3(x^2 + 3x + 5)-(2x+3)(3x + 4)}{(x^2 + 3x + 5)^2}\\
f(x) = \frac{-3x^2-8x+3}{(x^2 + 3x + 5)^2}\\
f'(x)=0\quad when\quad -3x^2-8x+3=0\\
\text{That happens when } x=-3,\;and \;\;when\;x=\frac{1}{3}

I think it is ten choose four, but I'm not so sure.

BTW when you see a right triangle a thing I always do is imagine it is inscribed in a circle with the hypotenuse as the diameter. That's just what I do, no need to do what I do. But is always nice to do what I do, because doing what I do makes me feel like people exist. I like when people exist, especially people who do what I do.

Anyways, draw a segment from \(A\) to to a point on \(BC\) in such it way that it is perpendicular to the hypotenuse. We call the point of intersection on the hypotenuse \(L\). 

This means \(AL\perp{BC}\). 

 - BC = 5 because it is a 3-4-5 triangle.

By AA similarity, we know that triangle \(ALC\) is similar to triangle \(ABC\). To find AL, we write the following proportion:

\(\frac{AL}{3}=\frac{4}{5}\)

Solving for AL: 

1. \(5AL=12\)

2. \(AL=\frac{12}{5}\)

Now here is how you solve the rest:

1. Consider point \(M\), that is the midpoint of \(BC\). Because it is the midpoint, we know that \(CM=\frac{5}{2}\)

2. Use the pythagorean theorem to find \(CL\)     hint:\(\sqrt{AC^2-AL^2}\)

3. Now that you found \(CL\), we can find \(LM\) by evaluating \(CM-CL\)

4. Since you know \(AL\) and \(LM\), you can easily find \(AM\) by pythagorean theorem.     hint: \(\sqrt{AL^2+LM^2}\)

Yay

I don't know how to use desmos to graph over interval ?? 

Another way:

Working off of lokiisnotdead, we know that triangle \(BCE\) is isosceles. That means angle EBC is 180 - 35 * 2 = 110

 - By supplements we know that FBA is 180-110=70

 - By supplements we know that AFB is 180-102 = 78

Now that you know two angles of triangle AFB, you can find the third angle, which is angle \(A\).

     *All angles in a triangle add up to 180

Thanks! Not gonna edit it because people can learn from my mistake.