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In this case, you probably typed something like, "If a>b, then a-c > b-c". This would be called the subtraction property of inequality, and is one of the rules used to solve inequalities.
B c bc bc bc bcbcbcbc
CA CAG A CAG A GAGG F* (THATS THE TOP BLACK F OR WATEVER IF IT HAS A * IT MEANS ON TOP) D B* DB*AB* DB*AB* CGG B*A BC BCDC DACB* B*B*AC B* B* B* A CB*AG EFGG F urs faithfully ...... ......
'Meet the Flintstones' was performed by the B-52s (calling themselves the BC-52s).
"B. C. Rich Warlock" is the name of electric guitars and bass guitars made by B. C. Rich. They have a distinct shape and are used by some popular musicians.
From the letters A, B, C, D, and E, you can form several segments by selecting combinations of these letters. For example, you can create segments like AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE. Additionally, single letters like A, B, C, D, and E themselves can also be considered segments. In total, there are 10 unique two-letter segments and 5 single-letter segments, leading to 15 segments altogether if counting all combinations.
If a < b, and c is positive, then ac < bcIf a < b, and c is negative, then ac > bc(inequality swaps over!)
A+BC+AC+B=A+BC+AC+B unless any of these variables has an assigned value.
There is no distributive property of addition over multiplication. The equation works if a + (b * c) = (a + b)*(a + c) = a2 + ab +ac +bc => a + bc = a2 + ab +ac +bc ie a = a2 + ab + ac = a*(a+b+c) and that, in turn requires that a = 0 or a+b+c = 1 If a, b and c are fractions than the second condition requires the fractions to sum to 1 - not be equal to 1.
The probability of ac and bc is 1/5.
associative? single replacement
a+bc --> ac+b
transitive
Yes, for example (a + bi)(c + di) = ac + adi + bic + bidi, and commutative property works as well --> ac + adi + bci + bdi² --> ac + (ad + bc)i + bd(-1) = (ac - bd) + (ad + bc)i
(a + b)/(a - b) = (c + d)/(c - d) cross multiply(a + b)(c - d) = (a - b)(c + d)ac - ad + bc - bd = ac + ad - bc - bd-ad + bc = -bc + ad-ad - ad = - bc - bc-2ad = -2bcad = bc that is the product of the means equals the product of the extremesa/b = b/c
A+bc---> b+ac
Yes, the expression AC + AD + BC + BD can be factored as (A + C)(B + D). This is evident by applying the distributive property, where expanding (A + C)(B + D) yields AB + AD + BC + CD. The terms AC and BD are not present, so the expression can be expressed in a different form, but the original expression itself represents a different factoring structure.
AC=5 AB=8 A=1 B=8 C=5 BC=40