a2
The reciprocal of a + bi is a - bi:1/(a + bi) since the conjugate is a - bi:= 1(a - bi)/[(a + bi)(a - bi)]= (a - bi)/[a2 - (b2)(i2)] since i2 equals to -1:= (a - bi)/(a2 + b2) since a2 + b2 = 1:= a - bi/1= a - bi
Pythagorean Theorem
All you need to do is substitute the given values of a and c into the equation, then solve for c: a2 + c2 = b2 102 + 302 = b2 100 + 900 = b2 b2 = 1000 b = √1000 b = 10√10
a2 + b2 + c2 - ab - bc - ca = 0 => 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0 Rearranging, a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ca + a2 = 0 => (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0 or (a - b)2 + (b - c)2 + (c - a)2 = 0 so a - b = 0, b - c = 0 and c - a = 0 (since each square is >=0) that is, a = b = c
A2 + B2 = C2 If C=8, then A2 + B2 = 64
a2
Pythagoras' theorem for a right angle triangle.
l a2 b2 is c2!!Its completely norma
Pythagoras' theorem. This is telling us that, for a right angled triangle, the square of the hypotenuse (c2) is equal to the sum of the square of the other two sides (a2 + b2).
The Theorem of Pythagoras.
c2 or what i like to call the pathagoream thereom to the triangle
The reciprocal of a + bi is a - bi:1/(a + bi) since the conjugate is a - bi:= 1(a - bi)/[(a + bi)(a - bi)]= (a - bi)/[a2 - (b2)(i2)] since i2 equals to -1:= (a - bi)/(a2 + b2) since a2 + b2 = 1:= a - bi/1= a - bi
Pythagorean Theorem
No. The pythagorean theorem is a2+b2=c2 so if you want to solve for b2 you'd have to subract a2 from both side. Doing so leaves b2 = c2 - a2. In addition, a2 cannot equal 0 since a, b, and c represent sides of a triangle and you cannot have a triangle with a side length of 0.
That it is a right triangle with the longest side c facing the right angle.
Yes If it is a right angle triangle then c2 = a2 + b2: c2 = 262 = 676 a2 + b2 = 242 + 102 = 576 + 100 = 676 = c2