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Q: If a plus b plus c equals x plus y plus z then prove a2 plus b2 plus c2 equals x2 plus y2 plus z2?

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a2

l a2 b2 is c2!!Its completely norma

a right triangle

Pythagorean Theorem

a2+2a2b+2ab2+b2

A2 + B2 = C2 If C=8, then A2 + B2 = 64

You just typed it.

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

Pythagoras' theorem for a right angle triangle.

a2 - 4a + 4

This is known as the Cosine Rule.

Yes, as long as either number is 0. a2 + b2 = (a+b)2 a2 + b2 = a2 + 2ab + b2 0 = 2ab

No. If you expand (a + b)2 you get a2 + 2ab + b2. This is not equal to a2 + b2

Pythagoras. That's why it's called the Pythagorean Theorem.

sqrt(a2 + b2) can't be simplified. Neither can (a2 + b2) .

Lupang hinirang intro:g f# a2 g d/a2 b2 c2 b2 a2 b2 g/g f# a2 g d/ a2 b2 c2 b2 a2 g/g f# a2 g d/a2 b2 c2 b2 a2 b2 g/ g f# a2 g d/a2 b2 c2 b2 a2 g/ refrain:g f# g a2 a2 d d a2 a2 d d/b2 c2 b2 e2 d2/ g f# g a2 a2 d d a2 a2 d d/b2 c2 b2 a2 b2 a2 g/(2x) g f# g c2 c2 d2 d2 e2 d2 c2 d2 e2/f2 e2 d2 e2 c2 d2 b2 b2 c2r/(2x)

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).

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 a squared plus b squared equals c squared

(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)

If you mean a2+b2 = c2 then it's Pythagoras' theorem

The Theorem of Pythagoras.

It is a*a + b*b + c*c

Pythagoras, the Greek Mathematician. The Pythagorean theorem (P Theorem) ~ a2 + b2 = c2 Also used to find the hypotenuse of certain triangles.

Cannot be simplified