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1 kilogram =

2.20462262 pounds .. i can convert then back later.

Let ABCD be a convex quadrilateral, and let M and N be the midpoints of sides AD and BC, respectively.

Prove that MN<=(AB+CD)/2. 

Plot the 4 points for THIS question and tell me what shape you have.

Or explain anything special about the shape.   

I think this is for your OTHER question Nicholas. 

Here is the starting point.

Questions for you Nicholas,

1)  How long is the interval AB

2) What is the forumla for the area of a rectangle?

Hi Nicholas,

Have you plotted those points on a number plane?

You need to do that first.

Do you know the name of the shape that it is ?  Can you describe it or upload a photo of the pic you have drawn?

The tree is 34.7 feet tall.

Penny is 1 feet tall.

Hi CoolStuffYT

Rom is right but it is good that you think about 'puzzles' like this.

You need to question the maths that is put in front of you.

Here is another example like your one

x=-3,   y=9

I can see that 9=(-3)^2 so I could say these fits the equation   y=x^2

but there are two solutions to y=x^2   when y=9    x could be  -3 or 3  

They are just 2 different solutions to the equation that i synthesized.       

.
So, the measure of the largest angle measure of a triangle is:

Angle C = Arc Cosine [(a^2+b^2-c^2) / 2ab]
.

You have found two of the possibilities EP.

2015 SS 28

1.

cos-theorem:

\(\begin{array}{|rcll|} \hline BM^2&=&x^2+\left(\dfrac{x}{2}\right)^2-2\cdot x \cdot \dfrac{x}{2}\cos(108^\circ) \\ BM^2 &=& x^2\left(1+\dfrac{1}{4}-\cos(108^\circ)\right) \quad |\quad \cos(108^\circ)=\cos(180^\circ-72^\circ)=-\cos(72^\circ) \\ BM^2 &=& x^2\left( \dfrac{5}{4}+\cos(72^\circ)\right) \\ \mathbf{BM} &\mathbf{=}& \mathbf{x \sqrt{\dfrac{5}{4}+\cos(72^\circ)}} \\ \hline \end{array} \)

2.
sin-theorem:

\(\begin{array}{|rcll|} \hline \dfrac{\sin(z)}{x} &=& \dfrac{\sin(36^\circ)}{ZB} \\\\ \mathbf{ZB} &\mathbf{=}& \mathbf{\dfrac{\sin(36^\circ)}{\sin(z)}x }\\ \hline \end{array} \)

3.

sin-theorem:

\(\begin{array}{|rcll|} \hline \dfrac{\sin(180^\circ-z)}{\frac{x}{2}} &=& \dfrac{\sin(72^\circ)}{ZM} \\\\ \dfrac{2\sin(z)}{x} &=& \dfrac{\sin(72^\circ)}{ZM} \\\\ \mathbf{ZM} &\mathbf{=}& \mathbf{\dfrac{\sin(72^\circ)}{2\sin(z)}x }\\ \hline \end{array} \)

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

\(\begin{array}{|rcll|} \hline BM &=& ZB+ZM \\ BM&=& \dfrac{\sin(36^\circ)}{\sin(z)}x+\dfrac{\sin(72^\circ)}{2\sin(z)}x \\\\ BM&=& \dfrac{x}{\sin(z)} \left( \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2}\right) \quad | \quad BM=x \sqrt{\dfrac{5}{4}+\cos(72^\circ)} \\\\ x \sqrt{\dfrac{5}{4}+\cos(72^\circ)}&=& \dfrac{x}{\sin(z)} \left( \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2}\right) \\\\ \sqrt{\dfrac{5}{4}+\cos(72^\circ)}&=& \dfrac{1}{\sin(z)} \left( \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2}\right) \\\\ \mathbf{\sin(z)} & \mathbf{=} & \mathbf{ \dfrac{ \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2} }{\sqrt{\dfrac{5}{4}+\cos(72^\circ)}} } \\\\ z &=& 180^\circ- \arcsin\left( \dfrac{ \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2} }{\sqrt{\dfrac{5}{4}+\cos(72^\circ)}}\right) \\ &=& 180^\circ- \arcsin\left( \dfrac{ 1.06331351044 }{1.24860602048} \right) \\ &=& 180^\circ- \arcsin\left( 0.85160049928 \right) \\ &=& 180^\circ- 58.3861775592^\circ \\ \mathbf{z} &\mathbf{=}& \mathbf{121.613822441^\circ} \\ \hline \end{array}\)

\(\text{Let $\angle ABM =180^\circ-(36^\circ+z) $} \\ \text{Let $ 36^\circ+z = 36^\circ+ 121.613822441^\circ=157.613822441^\circ$} \)

4.
sin-theorem:

\(\begin{array}{|rcll|} \hline \dfrac{\sin(z)}{x} &=& \dfrac{\sin(ABM)}{3} \quad | \quad \angle ABM =180^\circ-(36^\circ+z) \\\\ \dfrac{\sin(z)}{x} &=& \dfrac{\sin\Big(180^\circ-(36^\circ+z)\Big)}{3} \\\\ \dfrac{\sin(z)}{x} &=& \dfrac{\sin( 36^\circ+z )}{3} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{3\cdot \dfrac{\sin(z)}{\sin( 36^\circ+z )} } \\\\ x & = & 3\cdot \dfrac{0.85160049928}{\sin( 157.613822441^\circ )} \\\\ x & = & 3\cdot \dfrac{0.85160049928}{0.38084732121} \\\\ x & = & 3\cdot 2.23606797751 \quad | \quad 2.23606797751 = \sqrt{5} \\ \mathbf{x} &\mathbf{=}& \mathbf{3\cdot \sqrt{5}} \\ \hline \end{array}\)

Now correct me if I'm wrong but this site people come to to get answers. (sic)

Well ... yes, it is. I’m sure it wasn’t the creator’s original intent, but over time, it’s morphed into a homework answer and a defacto “cheating” forum. Diligent students can certainly learn from this, but for most, it presents only an illusion of knowledge, with no real substance of learning.     

... Anyone can come to this site and post questions as well as answers..  Nowhere does it state that you are obligated to even click on my question. Muchless read it or even answer it!! (sic)

Yep; you’ve summed it up in three generalized, poorly punctuated, sentences. 

Instead of wasting our time you can simply move on along, leave me alone, quit wasting MY time!!  (sic)

Yes, I could simply have moved on, and you could have simply pursued further explanation on your original post rather than repeatedly posting the same question with the expectation that someone will present a solution that magically gives you an understanding of the prerequisite concepts. 

Your comments on your pervious post indicated a sincere desire to learn this material.   So I spent the time to present detailed information on where you can learn the techniques to solve that particular type of statistical problem and where to learn the prerequisite concepts.  That took about 60 minutes to research, verify, and post –well longer than the five minutes needed to just answer this simple question.   (No comments about this; just bitching about harsh bullying.)

I then see this post, with your repeated question and a lament about three different answers.  I then knew with certainty that the time I invested in responding to your question was wasted.   So I trolled your lame and lazy ass. That pissed you off, so I at least recovered something from my investment of time. 

BTW, I am not a guy.   Most children –even very young children know this. But maybe you lack the prerequisites for this subject, too.  Do you need links that will help in gender recognition?  ...Meh, that’s just a waste of time. 

GA

Really?! How did that happen? Was it magic, or did you actually visit and review the links I posted for you? 

Doing that might give magical-like results, but you should know it’s not really magic.

GA

\(\dbinom{4}{2} \dbinom{10}{2}+2\dbinom{4}{1}\dbinom{10}{2} = 630\)

The first term picks 2 slots from 4 in the code for the first digit and then picks 2 digits from 10 to use.

The second term picks 1 slot from the 4 to use for the first digit and then picks 2 digits from 10.

The factor of 2 is there because either digit can be the singleton, the other will be the triplet.

10 x 9 x 9 x 9

|9x-4|=x

so  9x -4 = +x       x = 1/2

or   9x-4 = -x        x = 2/5