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P varies directly as q and inversely as r?
P=q/r* * * * *The correct answer is P = k*q/r where k is the constant of proportionality.
Prove that if a and b are rational numbers then a multiplied by b is a rational number?
If a is rational then there exist integers p and q such that a = p/q where q>0. Similarly, b = r/s for some integers r and s (s>0) Then a*b = p/q * r/s = (p*r)/(q*s) Now, since p, q r and s are integers, p*r and q*s are integers. Also, q and s > 0 means that q*s > 0 Thus a*b can be expressed as x/y where p and r are integers implies that x = p*r is an integer q and s are positive integers implies that y = q*s is a positive integer. That is, a*b is rational.
If p is 50 percent of q and r is 40 percent of q what percent of r is p?
p = 50q/100 = 1/2 q r = 40q/100 = 2/5 q p = (1/2)/(2/5) = (1/2)(5/2) = 5/4 r or 1 1/4 r Thus, p is 125% of r.
P q plus r pq pr?
p(q + r) = pq + pr is an example of the distributive property.
What is the proof of transitivity in logic NOT with truth table if p is q and if r is s and either p or r is true therefore either q or s is true?
Prove: [ P -> Q AND R -> S AND (P OR R) ] -> (Q OR S) -> NOT, --- 1. P -> Q ___ hypothesis 2. R -> S ___ hypothesis 3. P OR R ___ hypothesis 4. ~P OR Q ___ implication from hyp 1. 5. ~R OR S ___ implication from hyp 2 6. ~P OR Q OR S ___ addition to 4. 7. ~R OR Q OR S ___ addition to 5. 8. Let T == (Q OR S) ___ substitution 9. (~P OR T) AND (~R OR T) ___ Conjunction 6,7 10. T OR (~P AND ~R) ___ Distribution from 9 11. T OR ~(P OR R) ___ De Morgan's theorem 12. Let M == (P OR R) ___ substitution 13. (T OR ~M) AND M ___ conjunction 11, hyp 3 From there, you can use distribution to get (T AND M) OR (~M AND M). The contradiction goes away leaving you with T AND M, which can simplify to T.
