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I've edited my post. :) 

Thanks Melody and Alan!

In that case, in your first answer you got a sign wrong when you skipped it!

28+ 7x = 0 isn't the same as 28 = 7x, it's the same as 28 = -7x

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the real calculator is located in the Home section. then something that is white with a bunch of numbers and other factors of math is where the calculator rests today

The guy who inherited a center piece of the land, must be in a good relationship with, at least, one of his brothers.

happy bday, Heather feather!

see, I told you a face reveal was a brilliant idea. everyone thinks you're more beautiful than summer.

yea, i agree with you. the bidding at the howl was fun, and our gang had far more reputation than even our feeder empire:

yes, the Empire Of Royality really is a 3 empire alliance

there's GVMS Royality (my empire)

then in Ch is 1 more empire

CH Royality II

in RHS (high school), there's a empire reserved for us when we go to high school ruled by my good friend Matt called

Royality Mustangs

I know that Alan, but i just skipped it or is that part vital for the equation?

That's the hard way to do it.

The easy way: call d the length of the diagonal, and s the length of the side.

The ratio $${\frac{{\mathtt{d}}}{{\mathtt{c}}}}$$ is equal to φ for any regular Pentagon.

When you add 9x to both sides MG you should get 28 + 7x = 0, not 28 = 7x

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Thanks Melody, but why was it wrong to + 9x

How did you know that it was wrong and you had to add 2x?

Here is a method to find out exact value of cos72 degrees.My idea is come from C.F.Gauss.

let cos(a) =2pi/5=x

sin(a)=-sin(2kPi-a) if k=1

sin(a)=-sin(2pi-a)=-sin(5a-a)=-sin4a=-2sin(2a)cos(2a)=-4sin（a)cos(a)cos(2a)

because -4sin(a)  equal 0,so divide the equation by -4sin(a)

simplify we have -1/4=cos(a)cos(2a)

                        -1/4=1/2*[cos(a+2a)+cos(a-2a)]

                        -1/2=cos3a+cos(-a）

because cos(3a)=cos(-3a)=cos(2kpi-3a),when k=1,we have cos(3a)=cos(5a-3a)=cos(2a),and cos(-a)=cos(a)

                    so -1/2=cos3a+cos(-a) $$\Rightarrow$$ -1/2=cos(2a)+cos(a)

remember,I setted up cos(a)=x at the start 

so cos(2a)=cos^2(a)-sin^2(a)=cos^2(a)-[1-cos^2(a)]=2cos^2(a)-1=2x^2-1

change the equation -1/2=cos(2a)+cos(a) to -1/2=2x^2-1+x$$\Rightarrow$$-1=4x^2+2x-2$$\Rightarrow$$0=4x^2+2x-1

ax^2+bx+c=0 a=4 b=2 c=-1 let x1 and x2 are the two soloutionof the equation

x1=[-b+(b^2-4ac)^1/2]/2a=[-2+(2^2-4*4(-1)^1/2]/(4*2)=[-2+20^(1/2)]/8=[-1+5^(1/2)]/4

x2=[-b-(b^2-4ac)^1/2]/2a=[-2-(2^2-4*4(-1)^1/2]/(4*2)=[-2-20^(1/2)]/8=[-1-5^(1/2)]/4

Beacause x=cos(a)=cos 72degrees>0,so cos 72 degrees=-1/4+5^(1/2)/4

Yes, but there's another way to find the length of the diagonal. An easier way...

Hi MG

Your understanding of equations and your LaTex presentations are coming along brilliantly.

All the adults on the forum a are very pleased that you are learning here with us!

Your first post here was incorrect (you did not add 9x correctly) and you have redone it so you should delete your first post so it will not be confusing.

Better comments for the correct post would have been  

(-5 from both sides)

(+2x to both sides)

(divide both sides by -7)

To change a decimal into a percentage, multiply by 100.  

To change a percentage into a decimal, divide by 100.

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Use 2nd followed by asin

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The relationship between a side length (s) and a diagonal (d) for a regular pentagon is: sin(18°) = s/(2d)

Rearrange and plug in the numbers as required.

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also, as the Ancient Romans said:

The cover up is worse than the crime!

YES YOU CAN MG  

I have changed you second list of arguments to {rlll}

This means

push the first things to the right                              this is right align BEFORE the first ampersand (&)

then allign the next left   that is the equal sign       this is left align AFTER the first ampersand

then align the next left    that is the answer             this is left align AFTER the second ampersand

then align the next left    that is the comment         this is left align AFTER the third ampersand

See how i have added a third ampersand. 

Heureka makes fantastic use of aligning in many of his posts!

$$\begin{array}{rlll}\\
(y/3)+5&=&-9\qquad&(take\;away\;5)\\
(y/3)&=&-14\qquad&(times\;3)\\
y&=&-42\\
\end{array}$$

\begin{array}{rlll}\\

(y/3)+5&=&-9\qquad&(take\;away\;5)\\

(y/3)&=&-14\qquad&(times\;3)\\

y&=&-42\\

\end{array}

Your answer is in the wording of question 1... But that's because I used the word "size" instead of "lenght".

I know pretty much nothing on Pythagoras.

But i remember watching a video, is it that A= triangle numbers 

b= even

c=prime

its probably wrong like i said i know nothing about it.

But i'm sure its for right and triangles?

Yes, 4*20=80

20,40,60,80

4*2=8

4*20=80

I had to google FigNewton to find out what you were talking about Chris.   

It is a pity I was not on the forum today.  You have obviously been in a very fun mood :)   

What Is the length of the side of this pentagon?

she got in a dangerous fight with my rival

and she bit my rival's hand

she could've died that day

she could've gone to jail that day

she got expelled instead, which was more upsetting

Add 4 to both sides    

1. 6e+8f=38

2. 6e+3f=18 *(-1)

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1. 6e+8f=38

2. -6e-3f=-18

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(then you subtract 1. and 2.)

5f=20

f=4

(then replace f in 1. equation)

6e+8*4=38

6e=38-32

6e=6

e=1