Tribe 1
Liabala tribe
Tribe 2
Yakima tribe
2 answers
(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 answer
If a = 2 and b = 2, then a + b = 2 + 2 = 4. Further, if a = 2 and b = 2, then a x b = 2 x 2 = 4. The answer is a and b are equal to 2.
1 answer
let a=b
=>
a^2=a*b
subtract by b^2 on both sides
=>
(a^2)-(b^2)=(a*b)-(b^2)
(a+b)(a-b)=b(a-b)
=>
a+b=b
since a=b
2a=a
=>
2=1......
1 answer
To factor a^4 - b^4 completely, you can use the formula for the difference of squares, which states that a^2 - b^2 = (a + b)(a - b). In this case, a^4 - b^4 is a difference of squares twice: (a^2)^2 - (b^2)^2. So, you can factor it as (a^2 + b^2)(a^2 - b^2). Then, factor a^2 - b^2 further using the difference of squares formula to get (a^2 + b^2)(a + b)(a - b), which is the complete factorization of a^4 - b^4.
2 answers
Given any number b > 2, let l = 2b/(b-2)
then the rectangle with sides of l and b will have an area
= l*b = 2b/(b-2) * b = 2b2/(b-2)
and perimeter = 2*(l+b) = 2*[2b/(b-2) + b]
=2*[2b/(b-2) + b(b-2)/(b-2)] = 2/(b-2)* [2b+b2-2b] = 2b2/(b-2)
So that the area and perimeter have the same numeric value.
1 answer
consider:
(A+B) x (A + B)
= A(A+B) + B(A+B)
= A^2 + AB + BA + B^2
= A^2 + 2AB + B^2
(A-B) x (A-B)
= A(A-B) - B(A-B)
=A^2-AB-BA+B^2
= A^2 - 2AB + B^2
The answers are similar however square of sum has positive component and square of difference has negative component.
1 answer
The cast of Tokyo Tribe 2 - 2014 includes: Young Dais
1 answer
let a and b be non-zero quantities a=b, multiply through by a a^2=ab, subtract b^2 from both sides a^2-b^2=ab-b^2, factor (a+b)(a-b)=b(a-b), divide both sides by (a-b) a+b=b, a=b so write... 2b=b, divide both sides by b 2=1
1 answer
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
1 answer
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)
1 answer
Jesus is in the tribe of Judah
this is found in Matthew 1:2
1 answer
there is blood and blackfoot tribes from the plains people
1 answer
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
1 answer
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.
1 answer
a^(3) - b^(-3) =
a^(3) - 1/b^(3)
This factors to
(a - 1/b)(a^2 + a/b + (1/b)^2))
2 answers
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
1 answer
a^2 - b^2 = (a - b)(a + b)
a^2 + b^2 doesn't factor rationally.
1 answer
Ripcord - 1961 The Lost Tribe 2-22 was released on:
USA: 1963
1 answer
a squared - b squared = (a+b)(a-b)
2 answers
A^2 + b = 7 A + B^2 =11 -------------- A=2 and B=34(2^2) + 3 = 7 2 + 9(3^2) = 11
1 answer
12 = -4/2 * -2 + b
12 = -2 * -2 + b
12 = 4 + b
8 = b
1 answer
Conmutative:
a + b = b + a
5 + 2 = 2 + 5
(a)(b) = (b)(a)
(2)(3) = (3)(2)
Associative:
(a + b) + c = a + (b + c)
(1 + 2) + 3 = 1 + (2 + 3)
1 answer
a2 + b2 = (a + b)2 - 2ab = (a - b)2 + 2ab.a2 - b2 = (a + b)*(a - b)
2 answers
Mordecai, Esther's uncle, came from the tribe of Benjamin (Esther 2:5).
While not directly mentioned, it can be inferred from this that Esther was also from the tribe of Benjamin.
2 answers
u can play it on facebook, but it wont b the same
1 answer
Here are the steps:
ax^2 + bx + c = 0 Subtract c and divide by a
x^2 + (b/a)x = -(c/a) Take the square of (b/a)/2 and add it to both sides
(x + ((b/a)/2))^2 = -(c/a) + ((b/a)/2)^2 Take the square root of both sides
Subtract ((b/a)/2) and you have your solutions:
x = -(c/a) + ((b/a)/2)^2 - ((b/a)/2)
x = (c/a) - ((b/a)/2)^2 - ((b/a)/2)
1 answer
[a+b]2 = [a+b]*[a+b]
= a*a + a*b + b*a + b*b
= a2 + 2ab + b2
1 answer
Switching System
1.Manual Switching
2.Automatic Switching
2.a Electromechanical
2.a.1 Strowger
2.a.2 Crossber
2.b Electronics
2.b.1 Space Division
2.b.2 Time Division
2.b.2.1 Analog
2.b.2.2 Digital
2.b.2.2.1 Space
2.b.2.2.2 Time
2.b.2.2.3 Combination
1 answer
Blackfoot is the name of a Native American tribe. It begins with the letter B.
5 answers
The sum of two squares cannot be factored.
If that's -a^2 + b^2, that's b^2 - a^2, which factors to (b - a)(b + a)
1 answer