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A joke thread would be nice too!

Same for me probably. 

Worst is definitely when my close relatives die, e.g. (grandparents, parents, etc.)

Best is probably when I successfully start my own company?

may you please type out the second part, I cannot read it :)

Alright, here we go:

In n/m, both are relatively prime and cannot simplify the fraction. 

To prove that √3 is irrational from n^2 = 3m^2

From n^2 = 3m^2, we can say that n is a multiple of 3, since there is a 3 on the right side of the equation. 

We can set n equal to 3k

So n^2 = 9k^2 = 3m^2

3k^3 = m^2

So this means that m is also a multiple of 3.

In the beginning we stated that they are relatively prime and cannot simplify, but if both are n and m are multiples of 3, how is this possible?

This leads to a contradiction. 

Sure.

The greatest for me when I pass colllege. Worst is when my parents die.

Also help me and the gang out by telling us if you guys would appreciate a joke thread

This may be right if there were no restrictions, but as I stated in the problem, you must be able to circle to letters R-S-M going left to right without anything in between. 

The last few example were not correct because R-S-M are not in that order next to one another. 

Thanks for trying!

Here’s what google says:

A compound translation is a transformation that encapsulates a sequence of transformations so that when the compound transformation is applied to an object the entire sequence of constituent transformations are applied and then the new object is displayed.

Write a rule for g that represents a reflection in the x-axis, followed by a vertical stretch by a factor of 6 and a translation 4 units left of the graph of 

y=(2/3)^x

refection in x axis.

y=-(2/3)^x

Vertical stretch factor 6

y=-6*(2/3)^x

Translation 4 units left

y=-6*(2/3)^(x+4)

g(x)=-6*(2/3)^(x+4)

check

That is good!.

not   9*8=72 AND  9+8=17

\(\frac{e^2+17e+72}{e+9}=\frac{(e+9)(e+8)}{e+9}=e+8 \qquad e\in R \;\;where\;\;e\ne -9\)

A shape is translated when it is slid up or down or left or right. 

It can not be rotated or reflected or enlarged. It can only be slid left right up or down or a combination of these, like up and left.

And that's the graph.

Line    \(\ell_1\) is horizontal and passes through (1,1).

Line  \(\ell_2\) is vertical and passes through (2,2). Find the point of intersection of  \(\ell_1\)  and \(\ell_2\)

Have you tried sketching this. It is super easy!  You can do it yourself!

A sequence of positive integers with a_1 = 1 and a_9 + a_{10} = 646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all \(n\geq1\) , the terms, a_{2n-1}, a_{2n}, a_{2n + 1} are in geometric progression, and the terms, a_{2n}, a_{2n + 1}, and a_{2n + 2} , and  are in arithmetic progression. Let a_n be the greatest term in this sequence that is less than 1000. Find a_n.

NOTE there is an error in this question, The y value is given as two different values. I used the first one.

Suppose you graphed every single point of the form (2t+3, 3-t). For example, when t=2, we have 2t+3=7 and 3-3t=-3, so (7,3) is on the graph. Explain why the graph is a line, and find an equation whose graph is this line.

In order for this graph to be a line, we need to verify two things: first, that all the points on the graph are on the proposed line and second, that all points on the proposed line are on the graph. Be sure to deal with both.

Hint: If the point (2t+3, 3-t) is on the graph, then x=2t+3 and y=3-3t.   

But the equations we're used to graphing don't have any variables besides x and y.

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I really do not know what is wanted in this question but I shall discuss it a little.

I have used the value of y given in the original co-ordinates 

(2t+3, 3-t)

This a a set of points where x and y are defined by the variable t such that

x=2t+3   and  y=3-t

if I make x the subjects of these then I get

\(t=\frac{x-3}{2}\qquad and \qquad t=3-y\\ hence\\ \frac{x-3}{2}=3-y\\ y=3-\frac{x-3}{2}\\ y=\frac{6}{2}-(\frac{x}{2}-\frac{3}{2})\\ y=\frac{6}{2}-\frac{x}{2}+\frac{3}{2}\\ y=-\frac{x}{2}+\frac{9}{2}\\\)

This is clearly a linear graph with a gradient of -0.5 and a y intercept of 4.5

All the points on this line can be represented as  (2t+3,3-t) 

Any points not on this line cannot be represented by  (2t+3,3-t)      Where t is in the set of real numbers.

I do not know if something else is needed or wanted 

The sum of two longest altitudes in a right-angled triangle of 8, 15, 17 should be:

8 + 15 = 23 units

The Fibonacci sequence, with \(F_0 = 0\), \(F_1 = 1\) and \(F_n = F_{n - 2} + F_{n - 1}\), had a closed form \(F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right),\) where \(\phi = \frac{1 + \sqrt{5}}{2} \; \text{and} \; \widehat{\phi} = \frac{1 - \sqrt{5}}{2}.\) The Lucas numbers are defined in a similar way. Let \(L_0\) be the zeroth Lucas number and \(L_1\) be the first. If \(\begin{align*} L_0 &= 2 \\ L_1 &= 1 \\ L_n &= L_{n - 1} + L_{n - 2} \; \text{for}\; n \geq 2 \end{align*}\) Find \((a,b)\) such that \(L_n = a\phi^n + b\widehat{\phi}^n.\)

\(\text{Fibonacci sequence $0,1,1,2,3,5,8,13,21,34,55,\ldots$} \\ \begin{array}{|lcr|} \hline a_0 = a_0 &=& 0\cdot a_1 + 1 \cdot a_0 \\ a_1 = a_1 &=& 1\cdot a_1 + 0 \cdot a_0 \\ a_2 = a_1+a_0 &=& 1\cdot a_1 + 1 \cdot a_0 \\ a_3 = a_2+a_1 &=& 2\cdot a_1 + 1 \cdot a_0 \\ a_4 = a_3+a_2 &=& 3\cdot a_1 + 2 \cdot a_0 \\ a_5 = a_4+a_3 &=& 5\cdot a_1 + 3 \cdot a_0 \\ a_6 = a_5+a_4 &=& 8\cdot a_1 + 5 \cdot a_0 \\ a_7 = a_6+a_5 &=& 13\cdot a_1 + 8 \cdot a_0 \\ a_8 = a_7+a_6 &=& 21\cdot a_1 + 13 \cdot a_0 \\ \ldots \\ a_n = a_{n-1}+a_{n-2} &=& F_n\cdot a_1 + F_{n-1} \cdot a_0 \\ && \boxed{a_n = F_n\cdot a_1 + F_{n-1} \cdot a_0 } \\\\ \text{Lucas numbers:} \\ a_0 = L_0 = 2 \\ a_1 = L_1 = 1 \\ a_n = L_n =F_n\cdot 1 + F_{n-1} \cdot 2 \\ && \boxed{L_n =F_n + 2F_{n-1}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline F_n &=& \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right) \\ F_{n-1} &=& \frac{1}{\sqrt{5}} \left( \phi^{n-1} - \widehat{\phi}^{n-1} \right) \\ \boxed{L_n =F_n + 2F_{n-1}} \\ L_n &=& \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right) + 2\frac{1}{\sqrt{5}} \left( \phi^{n-1} - \widehat{\phi}^{n-1} \right) \\ &=& \dfrac{ \left( \phi^n - \widehat{\phi}^n \right)+2\left( \phi^{n-1} - \widehat{\phi}^{n-1} \right) } { \sqrt{5} } \\ &=& \dfrac{ \phi^n - \widehat{\phi}^n+2 \phi^{n-1} - 2\widehat{\phi}^{n-1} } { \sqrt{5} } \\ &=& \dfrac{ \phi^n+2 \phi^{n-1} - \widehat{\phi}^n - 2\widehat{\phi}^{n-1} } { \sqrt{5} } \\ &=& \dfrac{ \phi^n\left( 1+\dfrac{2}{\phi} \right) - \widehat{\phi}^n\left( 1+\dfrac{2}{\widehat{\phi}} \right) }{ \sqrt{5} } \quad | \quad \frac{1}{\phi} = \phi -1 \quad \frac{1}{\widehat{\phi}} = \widehat{\phi} -1 \\ &=& \dfrac{ \phi^n\Big( 1+2(\phi -1 ) \Big) - \widehat{\phi}^n\left( 1+2(\widehat{\phi} -1) \right) }{ \sqrt{5} } \\ &=& \dfrac{ \phi^n( 2\phi -1 ) - \widehat{\phi}^n\left( 2\widehat{\phi} -1 \right) }{ \sqrt{5} } \quad | \quad 2\phi -1 = \sqrt{5} \quad 2\widehat{\phi} -1= -\sqrt{5} \\ &=& \dfrac{ \phi^n\sqrt{5} - \widehat{\phi}^n\left( -\sqrt{5} \right) }{ \sqrt{5} } \\ &=& \dfrac{ \phi^n\sqrt{5} + \widehat{\phi}^n \sqrt{5} } { \sqrt{5} } \\ &=& \phi^n + \widehat{\phi}^n \\ \boxed{L_n =\phi^n + \widehat{\phi}^n } \\ \hline \end{array}\)

(a,b) = (1,1)
 

I was not convinced there was an answer to this question so I decided to experiment with Geogebra to get an answer.

Bummer!!

 I only just saw the word perpendicular. What a goose I am.    

I guess the pic I drew is not relevant after all!! 

Suppose that $f(x)=\frac{1}{2x+b}$. For what value of $b$ does $f^{-1}(x)=\frac{1-2x}{2x}$?

\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{1}{2x+b} \\ \hline y &=& \dfrac{1}{2x+b} \\\\ \dfrac{1}{y} &=& 2x+b \\\\ 2x &=& \dfrac{1}{y}-b \\\\ x &=& \dfrac{\dfrac{1}{y}-b}{2} \quad & | \quad x \leftrightarrow y \\\\ y &=& \dfrac{\dfrac{1}{x}-b}{2} \\\\ f^{-1}(x) &=& \dfrac{\dfrac{1}{x}-b}{2} \\ \hline f^{-1}(x) = \dfrac{\dfrac{1}{x}-b}{2} &=& \dfrac{1-2x}{2x} \\\\ \dfrac{\dfrac{1}{x}-b}{2} &=& \dfrac{1-2x}{2x} \\\\ \dfrac{1}{x}-b &=& \dfrac{1-2x}{x} \\\\ \dfrac{1}{x}-b &=& \dfrac{1}{x}-2 \\\\ -b &=& -2 \\\\ \mathbf{b} & \mathbf{=}& \mathbf{2} \\ \hline \end{array}\)