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- Mathematics Stack Exchange In boolean algebra, why is a+a'b = a+b? [duplicate] Asked 7 years, 3 months ago Modified 3 years, 11 months ago Viewed 87k times 9 This question already has answers here : 'Algebraic' way to prove the boolean identity a +a¯¯¯ ∗ b = a + b a + a ¯ ∗ b = a + b (2 answers) Closed 7 years ago.
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Viewed 220 times. 3. Prove: |a| −|b| | a | − | b | is less than or equal to a − b| a − b |. I sloved this problem by assuming it was right, than checking if a − b > 0 a − b > 0, a a positive\negative b$$ positive\negative than doing the same thing with a − b < 0 a − b < 0. I have a feeling it has a more simple way. absolute-value.
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Formula ( a + b) ( a − b) = a 2 − b 2 It is read as the a plus b times a minus b is equal to the a squared minus b squared. Introduction Let the two literals a and b be two variables. The addition of them form a binomial a + b. The subtraction of them form another binomial a − b. The two binomials are sum and difference basis binomials.
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The basic algebraic properties of real numbers a,b and c are: 1. Closure: a + b and ab are real numbers 2. Commutative: a + b = b + a, ab = ba 3. Associative: (a+b) + c = a + (b+c), (ab)c = a(bc) 4. Distributive: (a+b)c = ac+bc 5. Identity: a+0 = 0+a = a 6. Inverse: a + (-a) = 0, a(1/a) = 1 7. Cancelation: If a+x=a+y, then x=y 8. Zero-factor.
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Algebra Reduce (a-b)/ (b-a) a − b b − a a - b b - a Factor −1 - 1 out of a a. −1(−a)− b b−a - 1 ( - a) - b b - a Factor −1 - 1 out of −b - b. −1(−a)− (b) b−a - 1 ( - a) - ( b) b - a Factor −1 - 1 out of −1(−a)− (b) - 1 ( - a) - ( b). −1(−a+ b) b−a - 1 ( - a + b) b - a Reorder terms. −1(−a+ b) −a+b - 1 ( - a + b) - a + b Cancel the common factor.
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View solution steps Quiz Algebra a(a+b)−b(a−b)= Similar Problems from Web Search How do you solve the system 3a − 3b + 4c = −23 , a + 2b − 3c = 25 , and 4a − b + c = 25 ? https://socratic.org/questions/how-do-you-solve-the-system-3a-3b-4c-23-a-2b-3c-25-and-4a-b-c-25
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Algebra Simplify (a-b) (a+b) (a − b) (a + b) ( a - b) ( a + b) Expand (a−b)(a+b) ( a - b) ( a + b) using the FOIL Method. Tap for more steps. a⋅a+ab− ba−b⋅b a ⋅ a + a b - b a - b ⋅ b Simplify terms. Tap for more steps. a2 − b2 a 2 - b 2
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Basic Math Simplify a/ (a-b)+b/ (b-a) a a − b + b b − a a a - b + b b - a Factor −1 - 1 out of b b. a a−b + b −1(−b)−a a a - b + b - 1 ( - b) - a Factor −1 - 1 out of −a - a. a a−b + b −1(−b)−(a) a a - b + b - 1 ( - b) - ( a) Factor −1 - 1 out of −1(−b)− (a) - 1 ( - b) - ( a). a a−b + b −1(−b+a) a a - b + b - 1 ( - b + a) Simplify the expression.
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Otherwise, the result of the binary operation is converted to the type of the left-hand variable, subjected to value set conversion (§5.1.13) to the appropriate standard value set (not an extended-exponent value set), and the result of the conversion is stored into the variable. So your expression does: a^=b^=a^=b; evaluate a.
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In normal regular expression grammar, (a+b)* means zero or more of any sequence that start with a, then have zero or more a, then a b. This discounts things like baa (it doesn't start with a ), abba, and a (there must be one exactly b after each a group), so is not correct. (a*b*)* means zero or more of any sequence that contain zero or more a.
