Imcool

You can't

2 variables one equation, no way to solve the 2 variables unless ur looking for something else

Let's see what i can do..

So you need lengths AE and CE

Apparently Angle ADC is not a right angle?

wow! Nice job thank you! I checked all the work, and it was perfect!

Method 1: Brute Force

Usually not a good idea, but in this case it is simple so it will work.

Lets say one color is x and other is y.

xxy

xyx

yyx

yxy

4 ways.

Method 2: Logical thinking

So you can think. If i put xxy as a way, then any way with 2 x's touching is not going to be a solution, since you can rotate it. so the only other way with 2 x's is xyx. Then the same thing with y's giving you 4 solutions.

Alright, this is gonna be a long one.

first expand this whole thing

\(27t^3+36t^2+27t^2\sqrt{t}+19t\sqrt{t}+12t+3\sqrt{t}+1+\left(\sqrt{t}-3t+1\right)^3\)

\(=27t^3+36t^2+27t^2\sqrt{t}+19t\sqrt{t}+12t+3\sqrt{t}+1-27t^3+18t^2+27t^2\sqrt{t}-17t\sqrt{t}-6t+3\sqrt{t}+1\)

\(=27t^3-27t^3+36t^2+18t^2+27t^2\sqrt{t}+27t^2\sqrt{t}+19t\sqrt{t}-17t\sqrt{t}+12t-6t+3\sqrt{t}+3\sqrt{t}+1+1\) grouping like terms

simplify to get: \(54t^2+54t^2\sqrt{t}+2t\sqrt{t}+6t+6\sqrt{t}+2\)

you technically could group them together in parenthesis and stuff but im not doing that

I already answered this question, and the answer is 1.

Why? Check out my other post on this question by going into my profile and going to answers

Guest, im pretty sure you missed one, which is when m = 2, which would result in 20, and it is valid because the sign is less than or equal to.

Im confused, which one is correct? i feel like the second one but im not sure...

Alright, the first step is very simple

So basically subtract 6ab^2 from the first equation and you will get the second equation so 6ab^2 = 6

\(6ab^2 = 6\)

\(ab^2 = 1\)

now, you are asking for a - b

\(b^2 = 1/a\)

\(\cdot 1/b\) on both sides

\(b = 1/ab\)

im just doing every possible thing i can do bc this is my first time seeing this so im doing it as well as you sorry

\(ab = 1/b\) ok that might be useful

oh......

\(a^3 + 3 = 679\)

\(a^3 = 676\)

\(a \approx 8.77639\)

so b is 1/ whatever that is. Square rooted.....

its approximately 8.43883.... probably not correct, looking at the numbers. Maybe its like the square root of something?

 its like \(\sqrt{71.212}\) not exacly but close

\(\frac{\left(2z-1\right)^4}{\left(z\sqrt{z}+1\right)^4}\)this 

is what you get okay now for the expansion.......

\(\frac{16z^4-32z^3+24z^2-8z+1}{z^6+4z^{\frac{9}{2}}+6z^3+4z\sqrt{z}+1}\)

i agree, the numbers are huge but u have your answer. multiply it out and you will get 1 as your constant term because of the +1 in your original equation