Well first off, you need to wear pants that sag to your knees, then you need to have a hat that has some sort of money sign on it. Oh yeah, not to mention it has to be a flatbill. Also, your shirt must be as tall as you are otherwise you just wont make the cut. Your car must be crappy and your stereo must be bumpin, and you must shake the neighborhood to let people know you hit harder than your system,when really you don't and your hiding behind a pair of brass knucks or a "gat".
(a + b) = a^2 + 2ab + b^2 (a - b)^2 = a^2 - 2ab + b^2 or you can work like this: [a + (-b)]^2 = a^2 + 2a(-b) + (-b)^2 (a - b)^2 = a^2 - 2ab + b^2
1. Square of a binomial (a+b)^2 = a^2 + 2ab + b^2 carry the signs as you solve 2. Square of a Trinomial (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc carry the sings as you solve 3. Cube of a Binomial (a+b)^3 = a^3 + 3(a^2)b + 3a(b^2) + b^3 4. Product of sum and difference (a+b)(a-b) = a^2 - b^2 5. Product of a binomial and a special multinomial (a+b)(a^2 - ab + b^2) = a^3-b^3 (a-b)(a^2 + ab + b^2) = a^3-b^3
This is a proof that uses the cosine rule and Pythagoras' theorem. As on any triangle with c being the opposite side of θ and a and b are the other sides: c^2=a^2+b^2-2abcosθ We can rearrange this for θ: θ=arccos[(a^2+b^2-c^2)/(2ab)] On a right-angle triangle cosθ=a/h. We can therefore construct a right-angle triangle with θ being one of the angles, the adjacent side being a^2+b^2-c^2 and the hypotenuse being 2ab. As the formula for the area of a triangle is also absinθ/2, when a and b being two sides and θ the angle between them, the opposite side of θ on the right-angle triangle we have constructed is 4A, with A being the area of the original triangle, as it is 2absinθ. Therefore, according to Pythagoras' theorem: (2ab)^2=(a^2+b^2-c^2)^2+(4A)^2 4a^2*b^2=(a^2+b^2-c^2)^2+16A^2 16A^2=4a^2*b^2-(a^2+b^2-c^2)^2 This is where it will start to get messy: 16A^2=4a^2*b^2-(a^2+b^2-c^2)(a^2+b^2-c^2) =4a^2*b^2-(a^4+a^2*b^2-a^2*c^2+a^2*b^2+b^4-b^2*c^2- a^2*c^2-b^2*c^2+c^4) =4a^2*b^2-(a^4+2a^2*b^2-2a^2*c^2+b^4-2b^2*c^2+c^4) =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.1) We will now see: (a+b+c)(-a+b+c)(a-b+c)(a+b-c) =(-a^2+ab+ac-ab+b^2+bc-ac+bc+c^2)(a^2+ab-ac-ab-b^2+bc+ac+bc-c^2) =(-a^2+b^2+2bc+c^2)(a^2-b^2+2bc-c^2) =-a^4+a^2*b^2-2a^2*bc+a^2*c^2+a^2*b^2-b^4+2b^3*c-b^2*c^2+2a^2*bc-2b^3*c+(2bc)^2-2bc^3+a^2*c^2-b^2*c^2+2bc^3-c^4 =-a^4+2a^2*b^2+2a^2*c^2-b^4+(2bc)^2-c^4-2b^2*c^2 =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.2) And now that we know that Eq.1=Eq.2, we can make Eq.1=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) Therefore: 16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16 =[(a+b+c)/2][(-a+b+c)/2][(a-b+c)/2][(a+b-c)/2] And so if we let s=(a+b+c)/2 A^2=s(s-a)(s-b)(s-c)
It can if you divide by zero. 1. Let a and b be equal non-zero quantities a = b 2. Multiply both sides by a a^2 = ab 3. Subtract b^2 a^2 - b^2 = ab - b^2 4. Factor both sides (a - b)(a + b) = b(a - b) 5. Divide out (a - b) a + b = b 6. Since a = b ... b + b = b 7. Combine like terms on the left 2b = b 8. Divide by the non-zero b 2 = 1
Basically, you have the following linear equation system: a = b + 2 a + b = a * b (I assume "v" is a typo for "b"). Substituting (b+2) for a in the second equation, we get: 2b+2 = b(b+2) = b^2+2b Subtracting (2b) from both sides, we get: 2 = b^2 and we can conclude that: b = sqrt(2) which is not an integer (in fact, it is not even rational). And it is pretty easy to prove that sqrt(2) is not an integer, if needed.
gangster clothse
The Gangster Two Six Nation, is a gang under the Folk Nation, which means they wear their flag to the right, because ALL Folk Nation gangs wear their flag to the right.
Tiny Raskals Gang wear grey and Gangster Disciples Nation wear black.
you are technically supposed 2 b 17 2 hav it but any1 can download it w/o restrictions
Gangster ones and leather
You Cant Be A Gangster But You Could Purchase And Wear All Black Clothes And Act Like A Gangster That's As Gangster As You Can Be I Think And Also People Always 'Steal' (No Actual Money Is Made Or Stolen) You Could Also Do That
thinking your a gangster when you wear a sideways hat and saying yo
no
no it is not weird, because everybody was timberland boots not just gangster's or where I am from we call them wannabe gangster's. "Normal" people were wearing Timberland boots long before "gangsters" were anyway.
A gangster fedora. Yep, it's true.
069 BrickSquad (B$) Gangster Disiple
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