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How do the squares of sums and differences relate?
(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
If a and b are rational numbers and a plus b equals a times b what are a and b?
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.
How do you prove that one is equals to two?
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......
What are the rectangles with an area the same as perimeter?
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.
How will you differentiate the square of a sum from the square of a difference?
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.
What is the ratio of 2 to b?
It is written 2/b or 2:b
How do you proof that 2 is actually 1?
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
What are the different special product formulas?
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
What is proof of Heron's Formula?
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)
What is 2 B in a B?
2 p in a b = 2 pieces in a bikini
How do you prove 2 equals 1?
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
A and b are numbers where a equals b plus 2 the sum of a and v is equal to the product of a and b show that a and b are not integers?
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.
