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The slope - except at (0,0) -  is always positive so there are no  "extremes"

The inflection point (0,0) is not an "extreme"......the slope would have to change on either side of this point, but it doesn't.....the inflection point just represents the point where - in this  case - the curve changes from concave down to concave up

Thanks Heureka, a great one for me to study :)

Much appreciated :)

A factor of what?  You should read your question after you post it then you would know it is not complete!

 Rotate 90 degrees counterclockwise about the orgin.

Translate 4 units left.

Thanks X^2 

What you say makes sense but there is still room for confusion ://

Thanks, X².....that notation often confuses me, too...!!!!

Your  pic and presentation of the answer on this geometry question made it pretty easy to see......most of these proofs rely one one key thing....that exterior angle-opposite interior angle thing was the key...

I saw that one that rosala posted......and your answer...I think that one is probably above my puny level....>>>LOL!!!!!

Yes....I'm never sure are about "equal"  vs. "congruent, either.....

Your answer, Melody, was crafted with plenty of time and care. Good job! I do agree that answers requiring additional thought can go unnoticed, too.

I see that you are confused on the notation. Maybe this will help. 

For angles

1) If you are specifically referencing the measure of the angle, then you use the equal sign. For example, $m\angle ABC=m\angle CBD=125^{\circ}$. I generally place the m in front to denote that it is the measure of the angle.

2) If you are referencing the figure, in general, then you use the congruency symbol. $\angle ABC\cong \angle CBD$. Here, one has stated that the angles are congruent, but you may not necessarily know what the measures are.

Generally, I saw it out loud. If I say that angles are of equal measure, then you are certainly using an equal sign. 

For segments

1) If you are specifically referencing the length of the segment, then you would use the equal sign. $BD=QA=AX=25cm$

2) If you are referencing the figures themselves (maybe you know the figures are congruent because of a theorem, for example), then you will use the congruency symbol.

$\overline{BD}\cong\overline{QA}\cong\overline{AX}$

find the sum of n terms of the series:

A) 1+ 2x + 3x^2 + 4x^3 +.....

like Euler:

$\begin{array}{|rcll|} \hline \mathbf{s_n} &\mathbf{=}& \mathbf{1+ 2x + 3x^2 + 4x^3 +\ldots +n x^{n-1}} \quad & | \quad \cdot dx \\ s_n dx &=& dx+ 2xdx + 3x^2dx + 4x^3dx +\ldots +nx^{n-1}dx \quad & | \quad \int{} \\ \int{s_n dx} &=& x+ 2\frac{x^2}{2} + 3\frac{x^3}{3} + 4\frac{x^4}{4} +\ldots +n\frac{x^n}{n} \\ \int{s_n dx} &=& x+ x^2 + x^3 + x^4 +\ldots + x^n = \frac{x-x^{n+1}} {1-x} \\ \int{s_n dx} &=& \frac{x-x^{n+1}} {1-x} \\ \int{s_n dx}&=& \frac{x-x^{n+1}} {1-x} = [x-x^{n+1}][(1-x)^{-1}] \quad & | \quad \text{derivate} \\ s_n &=& [1-(n+1)x^n](1-x)^{-1}+ (x-x^{n+1})(-1)(1-x)^{-2}(-1) \\ s_n &=& \frac{1-(n+1)x^n} {1-x} + \frac{x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1-(n+1)x^n} {1-x} \left(\frac{1-x}{1-x}\right) + \frac{x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{[1-(n+1)x^n](1-x)+x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1-x-(n+1)x^n+(n+1)x^{n+1}+x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1 -(n+1)x^n+(n+1)x^{n+1} -x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1 -(n+1)x^n+[-1+(n+1)]x^{n+1} } {(1-x)^2} \\ \mathbf{s_n} & \mathbf{=}& \mathbf{\dfrac{1 -(n+1)x^n+n x^{n+1} } {(1-x)^2} } \\ \hline \end{array}$

B ) 1.2^2 + 2.3^2 + 3.4^2 + ......

$\begin{array}{|rcll|} \hline \mathbf{s_n} &\mathbf{=}& \mathbf{1\cdot 2^2 + 2\cdot3^2 + 3\cdot4^2 +\ldots +n \cdot (n+1)^2 } \\ s_n &=& \sum \limits_{k=1}^{n} k \cdot (k+1)^2 \\ &=& \sum \limits_{k=1}^{n} k \cdot (k^2+2k+1) \\ &=& \sum \limits_{k=1}^{n} (k^3+2k^2+k) \\ &=& \sum \limits_{k=1}^{n} (k^3) + \sum \limits_{k=1}^{n} (2k^2) + \sum \limits_{k=1}^{n} (k) \\ &=& \sum \limits_{k=1}^{n} (k^3) + 2\sum \limits_{k=1}^{n} (k^2) + \sum \limits_{k=1}^{n} (k) \\ && \begin{array}{|rcll|} \hline \sum \limits_{k=1}^{n} (k^3) &=& 1^3+2^3+3^3+ \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \\ \sum \limits_{k=1}^{n} (k^2) &=& 1^2+2^2+3^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \\ \sum \limits_{k=1}^{n} (k) &=& 1+2+3+ \ldots + n = \frac{n(n+1)}{2} \\ \hline \end{array} \\ &=& \left(\frac{n(n+1)}{2}\right)^2 + 2\cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \\ &=& \left(\frac{n(n+1)}{2}\right)\left( \frac{n(n+1)}{2} + \frac23\cdot(2n+1) +1\right) \\ &=& \left(\frac{n(n+1)}{2}\right)\left(\frac{3n(n+1)+4(2n+1)+6}{6}\right) \\ &=& \left(\frac{n(n+1)}{12}\right) \Big( 3n(n+1)+4(2n+1)+6 \Big) \\ &=& \frac{n(n+1)[3n(n+1)+4(2n+1)+6]}{12} \\ &=& \frac{n(n+1)(3n^2+3n+8n+4+6)}{12} \\ &=& \frac{n(n+1)(3n^2+11n+10)}{12} \\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{n(n+1)(n+2)(3n+5)}{12}} \\ \hline \end{array}$

Thanks for complementing me on my answer Chris. 

I'm sure you know better than most that when we put a lot of effort into an answer like this we do want someone to notice.

I mean we get our own satisfaction but still it is nice if we have a small appreciative audience as well.   

You provide so many excellent geometry answers, I did not think you would notice this one.  

".I did not know the thing about the exterior angle of a cyclic quad = the opposite interior angle"

I kind of extrapolated that ...

One of the most important features of a cyclic quad is that opposite angles are supplementary.

I was going to use that.

But then I realised that since this is true it means, by extension, that the exterior angle of a cyclic quad is equal to the opposite internal angle. 

It just meant that I could skip one step in the proof, that is all. :)

------------------------------------

With these proofs I am sometimes a bit confused about when I should use the 'congruent to' sign and when I should just use the equal sign. Do you have any confusion over this?

-----------------------------------

Did you see Rosala's question about series that are combinations of Arithmetic Progressions and Geometric progressions?

She asked for the formula derivation to be explained and then she had 2 questions using it.

I answered the first, but I couldn't do the second.

It was 'new' maths for me and quite interesting. 

I'll see if I can find the link.

Arr, I can see Heureka is answering it now :))

WDQA

AAS Congruence Theorem

CT

Angle S

Angle L

LT

D. With the compass point on D, construct an arc that intersectsEF¯¯¯¯¯ twice.

Scale factor  =    4.8 / 24    =   1/5   = .2

E is located at  (−4, 5) and point F is located at (7, 2) 

[ -4 +  (6/10)(7 - -4), 5 + (6/10)(2 -5) ]

[ -4 + (6/10) (11) , 5 + (6/10)(-3) ] 

[ -4 + (3/5)(11) , 5 + (3/5)(-3) ] 

[ -4 + 33/5, 5 - 9/5 ]

[ -20/5 + 33/5 , 25/5 - 9/5 ]

[ 13/5 , 16/5 ]

[ 2.6 , 3.2]

if 11/25 = 4/5, what is 5/5? I tried to do it and can't. please help.

If in some "IFTHEN LAND", 11/25 was equal to 4/5, then what would 5/5 be?????

Well, (4/5)  / (11/25) =1 9/11. So 1 9/11 x 5/5 =1 9/11 = 20/11.

So, 5/5 =20/11 in the "LAND of IFTHEN" !!!!!!!!

Well, you can see that WK is congruent (is that the right word?) to RN.

They even gave the measurements.

KT would be congruent to PR.

Make a ratio:

WK : RN = KT : PR

Now put in the measurements, with x as PR

14 : 24.5 = 18 : x

This will equal to 14x = 24.5 * 18.

14x = 441

x = 441/14

x = 31.5441/14 = 31.5

x = 31.5

So PR is 31.5 in.

That is excellent, Melody!!!!.....I did not know the thing about the exterior angle of a cyclic quad = the opposite interior angle.....I'm going to have to prove that to myself.....LOL!!!!!

I'm adding this one to my "Watchlist".......it's a very nice one  !!!!!

One dimension   =  sqrt [ ( -4 - - 1)^2 + ( 1 - -2)^2 ]  = sqrt [ (-3)^2 + (3)^2 ] =

sqrt [ 9 + 9 ]  =  sqrt [ 18]   =   sqrt [ 9 * 2  ] =  3sqrt[2]

The other dimension  = sqrt [ ( -4 -  -6)^2 + ( 1 -  - 1)^2 ]  =  sqrt [ (2)^2 + (2)^2 ] =

sqrt [ 4 + 4] =  sqrt [8]  =  sqrt [ 4 * 2 ]  =  2sqrt[2]

So....the area  is     [  3sqrt (2)  * 2 sqrt (2) ]  =  6 * 2   = 12 units^2

(4, 9)   (12, -5)

[ 4 + (1/4) ( 12 - 4),  9 + (1/4)(-5 - 9) ]  = [ 4 +  8 /4 , 9 + -14/4]  = 

[ 4 + 2 , (36 - 14) /4 ]  =  [ 6, 22/4 ]  = [ 6, 11/2 ] = ( 6, 5.5 )

I have labeled the points T, X and R as shown in my diagram.

I have also let  $\angle DPX =\alpha$   $\;\;and\;\;\angle AQX=\beta\;\;and \;\;\angle ADC=\theta$ 

Now:

$\angle DPX\cong \angle CPX= \alpha \qquad \qquad \overline{PX}\;\;bisects \angle DPC\\ \angle AQX\cong \angle DQX= \beta \qquad \qquad \overline{QX}\;\;bisects \angle AQD\\~\\ \angle ADC\cong \angle QBR=\theta \\ \qquad \text{Exterior angle of cyclic quad= opposite internal angle}\\~\\ \angle BRX=\angle QBR+\angle BQR=\theta + \beta\\ \qquad \text{Exterior angle of triangle = sum of opp internal angles in } \triangle BRQ\\~\\ \angle ATX=\angle TDQ+\angle TQD = \theta+\beta\\ \qquad \text{Exterior angle of triangle = sum of opp internal angles in } \triangle TQD\\~\\ \therefore \angle BRX=\angle ATX\\$

$Consider\;\; \triangle PXT \;\;and\;\; \triangle PXR\\ \angle PTX=\angle PRX=\theta + \beta\\ \angle TPX=\angle RTX=\alpha PX = PX \;\; common side\\ \therefore \triangle PXT=\triangle PXR \\ \qquad \text{Two angles and corresponding side test.}\\~\\ \therefore PXT=\angle PXR\\ \qquad \text{Corresponding angles in congruent triangles}\\ But \;\;\angle PXT+\angle PXR=180^\circ \\\qquad \text{Adjacent supplementary angles}\\ \therefore \angle PXT=\angle PXR=90^\circ\\ \therefore QT\;\;and \;\;PX \text{ are perpendicular.}$

$\text{Therefore the angle bisectors of  DPC and  AQD are perpendicular.         QED}$

[ Well that is assuming I have not put letters in stupid places anyway ]

Since AB  is parallel to BC, then  chords AD and BC are equal....thus

x^2 + 7x  = 60 - 4x        rearrange as

x^2 + 11x  - 60  =   0       factor

(x - 4) ( x + 15)   = 0     setting each factor to 0 and solving for x we have that

x = 4    or  x  = -15

Taking the positive solution, then BC = AB  = 60 - 4(4)  = 60 - 16  = 44

Thus  arcs  AD + DC + CB   =   [ 44  + 30  + 44 ] =  118°

Thus arc AEB  =  360 - 118  = 242°

Is something missing from the question?

11/25  does not equal  4/5  .    4/5  is bigger.

So...   11/25  =  4/5   this isn't true

Thx, man.