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What is the argument that A equals B B equals C so A equals C?
A=B , A-B=B-B , A-B =0 B=C , B-B=C-B, 0=C-B So A-B=0 but also C-B=0 A-B=C-B ...add +b ...A-B+B=C-B+B , A=C
What is the answer to A plus B plus C equals A times B equals C?
A=0 b=0 c=0
If a equals positive and b equals negative and c equals zero how many solutions does a Quadratic Equation have?
Two: one is 0, the other is -b/a ax2 + bx + c = 0, but c = 0 ⇒ ax2 + bx + 0 = 0 ⇒ ax2 + bx = 0 ⇒ x(ax + b) = 0 ⇒ x = 0 or (ax + b) = 0 ⇒ x = -b/a
Why when you multiply two negative numbers it turns positive?
Here is one explanation. Suppose A and B are two positive numbers.-A*0 = 0(B-B) = 0, so substitutong this value in place of 0 gives: -A*(B-B) = 0expanding the bracket, -A*B -A*(-B) = 0Adding A*B to both sides, A*B - A*B = -A*(-B) = A*BBut A*B - A*B = 0, which leave -A*(-B) = A*B
If 2 rational number a and b are ordered such that a is less than b then what must be true about the order for their opposites -a and -b?
a < b → a - a < b - a → 0 < b - a → 0 - b < b - a - b → -b < -a → if a < b then -b < -a which can also be expressed as -a > -b
