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There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.
There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.
There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.
There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.
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How can you show that the modular equation 22x41(mod13) has a unique solution and find that solution?
Let 22*41 = 13*a + b where a and b are integers and 0
What is (b c) divided by 8 for b 67 and c 13?
67 plus 13 = 80 80 divided by 8 = 10
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
Simplify this expression 13 -12--5 a -6 b -30 c 6 d 30?
30.. B
Prove If a b and c d then a c b d?
a/b = c/d Multiply both sides by b/c Thus (a/b)*(b/c) = (c/d)*(b/c) ie a/c = b/d
