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bapulana names for boys?

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Tribe 1

Liabala tribe

Tribe 2

Yakima tribe

2 answers


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(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


A British Christian dance band in the 1990s.

An R&B band from the 1970s.

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


It is written 2/b or 2:b

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


2 p in a b = 2 pieces in a bikini

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^2 + b^2 + 2ab = (a + b)^2

1 answer


a^3 + b^3 = (a + b)(a^2 - ab + b^2)

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

1 answer


It is a^2 + b^2, or a*a + b*b.

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


(A+b)^2

1 answer


u can play it on facebook, but it wont b the same

1 answer


(a^2 - b)(a^2 + b)(a^4 + b^2)

1 answer


(b + 2)(b + 2) or (b + 2)2

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


It is (B/100)*(B/100)*5000 = (B^2)/2

1 answer


b(b^2 + 1)(b^4 - b^2 + 1)

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