Q: What is the reciprocal of a plus bi if a2 plus b2 equals 1?

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a2

l a2 b2 is c2!!Its completely norma

Pythagorean Theorem

a right triangle

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

a2+2a2b+2ab2+b2

You just typed it.

a2+b2+c2=x2+y2+z2 divide each side by 2 (a2+b2+c2)/2=(x2+y2+z2)/2 a+b+c=x+y+z

Pythagoras' theorem for a right angle triangle.

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

This is known as the Cosine Rule.

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 - 4a + 4

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.

a2+b2=c2 a squared plus b squared equals c squared

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

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).

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)

The answer is x = 3i and x = -3i. {Where i= âˆš(-1)}An expression in the form a2 - b2 can be factored into (a - b)(a + b), but you have a2 + b2 so this factors into (a - bi)(a + bi). Check by multiplying the binomials: a2 + abi - abi - (bi)2 the [abi]'s cancel, and i2 = -1, so you have a2 + abi - abi - -b2 which is a2 + b2, so it checks out. In this case, a is x and b is 3.

Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )Assuming the sale is in B2 and the cost in A2, you could use the following formula to do it:=IF( B2>=A2*1.25, B2*7%, 0 )

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

v2 = b2 (a2 - x2)Divide each side by b2 :v2/b2 = a2 - x2Subtract a2 from each side:v2/b2 - a2 = -x2Multiply each side by -1 :x2 = a2 - v2/b2Take the square root of each side:x = Ã‚Â± sqrt ( a2 - v2/b2 )

The Theorem of Pythagoras.