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Which of the following could be the sum of the reciprocals of two different prime numbers?

(a) 7/13 (b) 10/21 (c)11/30 (d)23/50 (e)19/77

B..sum of reciprocals of 7 and 3.. _________________

--------------------------------------------------------------------------------------- If you think you can,you can If you think you can't,you are right.

Which of the following could be the sum of the reciprocals of two different prime numbers?

(a) 7/13 (b) 10/21 (c)11/30 (d)23/50 (e)19/77

Two primes - a and b. [m]1/a + 1/b = (b+a)/(ab)[m]

So: (a) - not possible, because 13 is a prime - ab cannot equal 13. (b) - 7,3 work 7*3=31, 7+3=10, the answer is B...

(c) 30 -- 15,2 (not prime), 10,3 (not prime), 5,6 (not prime), 30,1 (not prime)... 30 does not factor into only two numbers (i.e. primes), so not possible. (d) Like 30, 50 will not factor into solely two primes. (e) 77 = 7 * 11, so two primes. 7+11 = 18, NOT 19, so not E. _________________

________________________________________________________________________ Andrew http://www.RenoRaters.com

(a) \(-x<-12\) --> \(x>12\) not true; (b) \(-x-2<14\) --> \(x>-16\) not always true; (c) \(-x+2<-10\) --> \(x>12\) not true; (d) \(x+2<10\) --> \(x<8\), not always true; (e) \(x-2<11\) --> \(x<13\) now if \(x<12\) (given), \(x\) will definitely be less than \(13\). Always true.

E.Add the 2 fractions and replace the value of xy. _________________

--------------------------------------------------------------------------------------- If you think you can,you can If you think you can't,you are right.

a*b = a + b - ab given a*b = 0, then 0 = a + b -ab -b = a(1-b) b = a(b-1) a = b / (b-1) if b=1, then a becomes undefined. So b=1 cannot be valid. IMO B