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It is the second time I have posted links to these questions.  And the questions where answered by that guest that always gets it wrong (not to be rude but thats the truth) I even tested the guests answers and my website showed them as wrong.

anyways i'll answer them

I'm pretty sure all those questions are already answered and i've seen you posting links to these questions a lot

we have 180-94-41=4 so the angle x is 180-(111+45)=24 and we have x=24.

What's the picture? and could you please type in the question to let us know? because we can't see the picture

What's the picture? and what's the question asking?

We could prime factorize the numbers below 30 and get the list
2,3,5,7,11,13,17,19,23,29 

and  that is \(\boxed{10\text{ numbers}}\)

11)   x and y  add with the 90 angle to form a straight line, so x+ y = 90

          AND they are in a ratio of 3:2      so x is   3/(3+2) * 90    and y is   2/(3+2) *90

12)   ALL of the angles   4z  and 3y   and 6x   are the same value : 108

               4z = 108       3y = 108     6x = 108   

Wow! After some research, apparently googol factorial is larger than the amount of particals in the universe! But, if I were to geuss, it should be undefinded? I'm probably wrong, as numbers are infinite.

It seems that my LaTeX has not loaded properly. Wherever it says \(2n, I mean 2n < n-4

Solving directly for n is pretty hard. Which is why we can divide this up into two inequalities:

\(2n

and

\(n-4\leq3n+8\)

Let's solve the first one:

\(2n

\(\text{Subtract n from both sides}\)

\(n<-4\)

Now, let's solve the second one:

\(n-4\leq3n+8\)

\(\text{Subtract n from both sides}\)

\(-4\leq2n+8\)

\(\text{Subtract 8 from both sides}\)

\(-12\leq2n\)

\(\text{Divide by 2}\)

\(-6\leq n\)

So, our inequality is now:

\(-6 \leq n < -4\)

Therefore, the integer solutions are:

\(\boxed{-6, -5}\)

Here is a challenge for you to occupy your time!

Can you find the first 20 digits of (10^100)!, or Googol factorial? Good luck.

Wow! Wonderfully presented answer, as usual. Just wondering, how did you get the boxes? With the \boxed{} command? For me, that only worked for one line, so how did you put all those lines of math into the box? 

Thanks Alan ...  but I do not know how to calculate that sum to get 3 either.  

Alan, can you explain more how you got from second to last step to the answer? I don't seem to quite understand. :( I do understand the beginnning and middle though. Good reasoning!!! 

This requires some angle chasing, which I think is pretty fun.

So, we see that since m and n are parallel, the angle next to B is also 65 degrees. So, to solve for B, we add 55+65+B and subtract that from 180. Try that.

For A, we know that all the angles in a triangle add up to 180. So,  A+B+65 and subtract that from 180.

Once you have done what is put in green, you will get the answers you need.

To find the inverse of a function, you first need ot swap in "y" for the "f(x)"

Then, we need to switch the x for the y and vice versa.

Then, solve for the y.

Thank you Alan!!! :)

Thanks very much Alan,

I am never quite confident about expected values.  :)

Alan's post is being blocked here but available on the answer page  for some weird reason

so I have taken a pit of it and will try to display it here.

Six squares are inscribed in an 11 by 13 rectangle.  Find the shaded area.

\(\text{Let $P_2=\dbinom{x_2}{y_2} $ } \\ \text{Let $P_4=\dbinom{x_4}{y_4} $ }\)

\(\begin{array}{|rcll|} \hline P_1 &=& a\dbinom{\sin(\varphi)}{\cos(\varphi)} \\ \hline \end{array} \begin{array}{|rcll|} \hline P_2 &=& P_1 + 3a\dbinom{\cos(\varphi)}{-\sin(\varphi)} \\\\ P_2 &=& a\dbinom{\sin(\varphi)}{\cos(\varphi)} + 3a\dbinom{\cos(\varphi)}{-\sin(\varphi)} \\\\ P_2 &=& a\dbinom{\sin(\varphi)+3\cos(\varphi)} {\cos(\varphi)-3\sin(\varphi)} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{x_2} &=& \mathbf{a\Big( \sin(\varphi)+3\cos(\varphi) \Big)} \quad | \quad x_2 = 11 \\ a\Big( \sin(\varphi)+3\cos(\varphi) \Big) &=& 11 \\ a &=& \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline P_3 &=& 2a\dbinom{\cos(\varphi)}{-\sin(\varphi)} \\ \hline \end{array} \begin{array}{|rcll|} \hline P_4 &=& P_3 - 3a\dbinom{\sin(\varphi)}{\cos(\varphi)} \\\\ P_4 &=& 2a\dbinom{\cos(\varphi)}{-\sin(\varphi)} - 3a\dbinom{\sin(\varphi)}{\cos(\varphi)} \\\\ P_4 &=& a\dbinom{2\cos(\varphi)-3\sin(\varphi)} {-2\sin(\varphi)-3\cos(\varphi)} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{|y_4|} &=& \mathbf{a\Big( 2\sin(\varphi)+3\cos(\varphi)\Big)} \quad | \quad y_4 = 13 \\ a\Big( 2\sin(\varphi)+3\cos(\varphi) \Big) &=& 13 \\ a &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1): & a &=& \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } \\ (2): & a &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \\ \hline & a = \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \\\\ & \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \\\\ & 11 \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) &=& 13 \Big( \sin(\varphi)+3\cos(\varphi) \Big) \\\\ & 22\sin(\varphi)+33\cos(\varphi) &=& 13 \sin(\varphi)+39\cos(\varphi) \\\\ & 22\sin(\varphi)- 13 \sin(\varphi) &=& 39\cos(\varphi)-33\cos(\varphi) \\\\ & 9\sin(\varphi) &=& 6\cos(\varphi) \\\\ & \mathbf{\tan(\varphi)} &=& \mathbf{\dfrac{2}{3}} \\ \hline \end{array}\)

\(\mathbf{a=\ ?}\)

\(\begin{array}{|rclrcl|} \hline a &=& \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } \\ &&& \sin(\varphi) &=& \dfrac{\tan(\varphi) }{\sqrt{1+\tan^2(\varphi)}} \quad | \quad \mathbf{\tan(\varphi)=\dfrac{2}{3}} \\ &&& \sin(\varphi) &=& \dfrac{\dfrac{2}{3}}{\sqrt{1+\dfrac{4}{9}}} \\ &&& \mathbf{\sin(\varphi)} &=& \mathbf{\dfrac{2\sqrt{13}}{13}} \\\\ &&& \cos(\varphi) &=& \dfrac{1}{\sqrt{1+\tan^2(\varphi)}} \quad | \quad \mathbf{\tan(\varphi)=\dfrac{2}{3}} \\ &&& \cos(\varphi) &=& \dfrac{1}{\sqrt{1+\dfrac{4}{9}}} \\ &&& \mathbf{\cos(\varphi)} &=& \mathbf{\dfrac{3\sqrt{13}}{13}} \\\\ a &=& \dfrac{11}{ \dfrac{2\sqrt{13}}{13}+3*\dfrac{3\sqrt{13}}{13} } \\\\ a &=& \dfrac{11}{ \dfrac{2\sqrt{13}}{13}+\dfrac{9\sqrt{13}}{13} } \\\\ a &=& \dfrac{11}{ \dfrac{11\sqrt{13}}{13} } \\\\ a &=& \dfrac{13}{\sqrt{13}} \\\\ \mathbf{a} &=& \mathbf{\sqrt{13}} \\ \hline \end{array}\)

The shaded area:

\(\begin{array}{|rcll|} \hline \text{The shaded area} &=& 11*13 - 6a^2 \quad | \quad \mathbf{a=\sqrt{13}} \\ \text{The shaded area} &=& 11*13 - 6*13 \\ \text{The shaded area} &=& 5*13 \\ \mathbf{\text{The shaded area}} &=& \mathbf{65} \\ \hline \end{array}\)

\(\displaystyle a^{4}+(ab)^{2}+b^{4}=\{(a^{2}+b^{2})-ab\} \{(a^{2}+b^{2})+ab\}=900, \\ \text{substituting the second equation,}\\ (a^{2}+b^{2}-ab).45=900,\\ \text{so,}\\ a^{2}+b^{2}-ab=20,\\ \text{and now subtract this from the second equation.}\)

Let the sides be a, b and sqrt (a^2+b^2)

So the radii are       \(\frac{a}{2},\quad \frac{b}{2},\quad \frac{\sqrt{a^2+b^2}}{2}\)

Area of triangle =  ab/2

Area of big semicircle = \(0.5 \pi*\frac{(a^2+b^2)}{4}=\frac{(a^2+b^2)\pi}{8}\)

Sum of little white segments = \(\frac{(a^2+b^2)\pi}{8}-\frac{ab}{2}=\frac{(a^2+b^2)\pi-4ab}{8}\)

Sum of the 2 smaller semicircle =       \(0.5*\pi((\frac{a}{2})^2+(\frac{b}{2})^2)= \pi(\frac{a^2+b^2}{8})\\ \)

You can finish it

First rearrange both equations in slope-intercept form:

1st eqn:  y = -Vx/(2*(V-1))  + 1/(2*(V-1))   slope =  -V/(2*(V-1))

2nd eqn: y =  -x  + 2/(V - 5)   slope = -1

The two lines will never intersect if they are parallel.  They are parallel if their slopes are the same, so set the two slopes equal to each other and solve for V.  

Try the following reasoning: