9 Provinces in South Africa Warm Regards, Samar Haider
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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.
P=s r t , so, s= P/(st)
To solve for p, set up the equation 5p + 9 = p - 3 5p = p - 12 4p = -12 p = -3 to prove the answer, substitute -3 for p and prove 5(-3) + 9 = (-3) -3 -15 + 9 = -6 -6 = -6
If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.
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