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 Numeric_Fractions/189229: = I dont know how to find x for this. Please help. Thanks so much ^_^1 solutions Answer 141964 by jim_thompson5910(28598)   on 2009-04-01 18:38:13 (Show Source): You can put this solution on YOUR website! Start with the given equation. Cross multiply Multiply Divide both sides by 7 to isolate "x". So the solution is which is approximately
 Complex_Numbers/189232: Write the complex number 2-i (over) 5+i in the usual a + bi form1 solutions Answer 141963 by jim_thompson5910(28598)   on 2009-04-01 18:36:14 (Show Source): You can put this solution on YOUR website! Start with the given expression. Multiply the fraction by . Combine the fractions. FOIL the numerator. FOIL the denominator. Multiply. Combine like terms. Break up the fraction. Reduce. So . So the expression is now in standard form where and
 Rectangles/189234: Hello I have tried to figure out this problem for like two or three days and it just doesn't make since to me so if you could be so kind as to help me figure it out, I would be greatly appriciated. The question or the problem is: The length of a rectangle is 1 less than three times it's width. The perimeter is 30 ft. Find the length of this rectangle. You might not be able to figure it out either but if you can't I appriciate the offer anyways. Thank you so much for your help.1 solutions Answer 141961 by jim_thompson5910(28598)   on 2009-04-01 18:32:24 (Show Source): You can put this solution on YOUR website!Let L=length and W=width Since "The length of a rectangle is 1 less than three times it's width", this means that and because "The perimeter is 30 ft", this tells us that Start with the perimeter formula Plug in and Distribute Add 2 to both sides. Combine like terms. Divide both sides by 8 to isolate W. Reduce. Rearrange the equation. So the width is 4 feet. Go back to the first equation Plug in Multiply. Subtract. So the length is 11 feet.
 Proofs/189186: ~(A or B) or ~(C or D) (E or F) -> D therefore A-> ~E1 solutions Answer 141960 by jim_thompson5910(28598)   on 2009-04-01 18:25:40 (Show Source): You can put this solution on YOUR website!Note: I'm going to use the symbol "v" for "or" So ~(A or B) = ~(A v B) 1. ~(A v B) v ~(C v D) 2. (E v F) -> D / A -> ~E ------------------------------- 3. (~A & ~B) v (~C & ~D) 1 De Morgan's Law 4. [(~A & ~B) v ~C] & [(~A & ~B) v ~D] 3 Distribution 5. [(~A & ~B) v ~D] & [(~A & ~B) v ~C] 4 Commutation 6. (~A & ~B) v ~D 5 Simplification 7. ~D v (~A & ~B) 6 Commutation 8. (~D v ~A) & (~D v ~B) 7 Distribution 9. ~D v ~A 8 Simplification 10. ~A v ~D 9 Commutation 11. A -> ~D 10 Material Implication 12. ~(E v F) v D 2 Material Implication 13. (~E & ~F) v D 12 De Morgan's Law 14. D v (~E & ~F) 13 Commutation 15. (D v ~E) & (D v ~F) 14 Distribution 16. D v ~E 15 Simplification 17. ~~D v ~E 16 Double Negation 18. ~D -> ~E 17 Material Implication 19. A -> ~E 11,18 Hypothetical Syllogism 
 Proofs/188928: I cannot solve this: I can use: Transportation, Material Implication, Material Equivalence, Exportation, Tautology, Double Negation, Commutation, Association, Distribution, Demorgan's Theorem, Modus Ponems, Modus Tollens, Hypothetical Syllogism, Conjunstion, Simplification, Addition, Constructive Dilemma, Absorption and Disjuntive Syllogism. 1. ~(p <--> s) 2. (p v q) -> r 3. (t v ~s) -> (~s & ~r) Therefore: ~t & ~p 1 solutions Answer 141959 by jim_thompson5910(28598)   on 2009-04-01 18:16:20 (Show Source): You can put this solution on YOUR website!This is a very long proof. I hope that I'm not too late. There are a number of ways to derive this, but the main goal is isolate ~t and ~p (see lines 41 and 44) and then form a conjunction of the two atomic components. Note: when I start deriving from new premise, I separate the lines (to make things a little bit cleaner). Normally those dividing lines wouldn't be in the proof. 1. ~(p <--> s) 2. (p v q) -> r 3. (t v ~s) -> (~s & ~r) Therefore: ~t & ~p ------------------------------------------------- 4. ~[(p & s) v (~p & ~s)] 1 Material Equivalence 5. ~(p & s) & ~(~p & ~s) 4 De Morgan's Law 6. ~(~p & ~s) & ~(p & s) 5 Commutation 7. ~(p & s) 5 Simplification 8. ~p v ~s 7 De Morgan's Law 9. ~s v ~p 8 Commutation 10. s -> ~p 9 Material Implication 11. ~(~p & ~s) 6 Simplification 12. ~~p v ~~s 11 De Morgan's Law 13. ~~p v s 12 Double Negation 14. ~p -> s 13 Material Implication ------ 15. ~(p v q) v r 2 Material Implication 16. (~p & ~q) v r 15 De Morgan's Law 17. r v (~p & ~q) 16 Commutation 18. (r v ~p) & (r v ~q) 17 Distribution 19. r v ~p 18 Simplification 20. ~p v r 19 Commutation 21. p -> r 20 Material Implication ------ 22. ~(t v ~s) v (~s & ~r) 3 Material Implication 23. (~t & ~~s) v (~s & ~r) 22 De Morgan's Law 24. (~t & s) v (~s & ~r) 23 Double Negation 25. [(~t & s) v ~s] & [(~t & s) v ~r] 24 Distribution 26. [(~t & s) v ~r] & [(~t & s) v ~s] 25 Commutation 27. (~t & s) v ~s 25 Simplification 28. ~s v (~t & s) 27 Commutation 29. s -> (~t & s) 28 Material Implication 30. (~t & s) v ~r 26 Simplification 31. ~r v (~t & s) 30 Commutation 32. r -> (~t & s) 31 Material Implication 33. p -> (~t & s) 21,32 Hypothetical Syllogism 34. ~p v (~t & s) 33 Material Implication 35. (~p v ~t) & (~p v s) 34 Distribution 36. (~p v s) & (~p v ~t) 35 Commutation 37. ~p v s 35 Simplification 38. p -> s 37 Material Implication 39. p -> ~p 38,10 Hypothetical Syllogism 40. ~p v ~p 39 Material Implication 41. ~p 40 Tautology 42. s 14,41 Modus Ponens 43. ~t & s 29,42 Modus Ponens 44. ~t 43 Simplification 45 ~t & ~p 44,41 Conjunction 
 Proofs/189101: I need to construct a proof using all of the rules: (ex. commutation, association, material implication, exportation, hypothetical syllogism, DeMorgans, Double Negation, Distribution, etc.) 1. (p <--> q) -> s 2. ~(~r -> t) 3. ~q v ~s Therefore: (t v p) -> (~t & ~ q)1 solutions Answer 141917 by jim_thompson5910(28598)   on 2009-04-01 13:19:46 (Show Source): You can put this solution on YOUR website!This is one tricky derivation, so you need to be a little creative with this one. Note: each premise is used and when a new premise is derived from, I separated it to keep things looking clean. 1. (p <--> q) -> s 2. ~(~r -> t) 3. ~q v ~s Therefore: (t v p) -> (~t & ~ q) -------------------------------------------- 4. ~(~~r v t) 2 Material Implication 5. ~(r v t) 4 Double Negation 6. ~r & ~t 5 De Morgan's Law 7. ~t & ~r 6 Commutation 8. ~t 7 Simplification --- 9. ~s v ~q 3 Commutation 10. s -> ~q 9 Material Implication --- 11. (p <--> q) -> ~q 1,10 Hypothetical Syllogism 12. [(p & q) v (~p & ~q) ] -> ~q 11 Material Equivalence 13. ~[(p & q) v (~p & ~q) ] v ~q 12 Material Implication 14. [~(p & q) & ~(~p & ~q) ] v ~q 13 De Morgan's Law 15. ~q v [~(p & q) & ~(~p & ~q) ] 14 Commutation 16. [~q v (~p v ~q)] & [~q v (p v q) ] 15 Distribution 17. ~q v (~q v ~p) 16 Simplification 18. (~q v ~q) v ~p 17 Association 19. ~q v ~p 18 Tautology 20. ~p v ~q 19 Commutation --- 21. ~t & (~p v ~q) 8,20 Conjunction 22. (~t & ~p) v (~t & ~q) 21 Distribution 23. ~(t v p) v (~t & ~q) 22 De Morgan's Law 24. (t v p) -> (~t & ~q) 23 Material Implication 
 Rational-functions/188930: This question is from textbook algebra2 Find (f+g)(x), (f-g)(x), (f*g)(x) snd (f/g)(x) for each f(x) and g(x) 3. f(x)= x^2+7x+12 g(x)= x^2-91 solutions Answer 141875 by jim_thompson5910(28598)   on 2009-03-31 22:10:54 (Show Source): You can put this solution on YOUR website! Start with the given expression. Expand Plug in and Combine like terms. So ----------------------------------------- Start with the given expression. Expand Plug in and Distribute Combine like terms. So ----------------------------------------- Start with the given expression. Expand Plug in and Rearrange the terms. Expand. Note: Distribute Multiply Combine like terms. So ----------------------------------------- Start with the given expression. Expand Plug in and Factor to get . Factor to get . Highlight the common terms. Cancel out the common terms. Simplify. So where or -----------------------------------------
 Polynomials-and-rational-expressions/189121: can you help me please? factor the equation 1 solutions Answer 141857 by jim_thompson5910(28598)   on 2009-03-31 20:41:17 (Show Source): You can put this solution on YOUR website! Start with the given equation. Factor out the GCF "r" From here, you cannot factor further.
 Graphs/189113: Calculate the value of the discriminant of x^(2 )+2x+1=0 By examining the sign of the discriminant in part a, how many x-intercepts would the graph of have? Why? 1 solutions Answer 141854 by jim_thompson5910(28598)   on 2009-03-31 20:28:17 (Show Source): You can put this solution on YOUR website! From we can see that , , and Start with the discriminant formula. Plug in , , and Square to get Multiply to get Subtract from to get Since the discriminant is equal to zero, this means that there is one real root. This also means that there is only one x-intercept (since a root is an x-intercept)
 Exponential-and-logarithmic-functions/189102: How do I simplify: 4^{log [2^(log 5)]} *both logs have a base of 2* Thanks1 solutions Answer 141853 by jim_thompson5910(28598)   on 2009-03-31 20:26:04 (Show Source): You can put this solution on YOUR website!Remember, is one identity that is useful for simplifying logarithms. So (in this case, and ) This means that simplifies to In other words, ----------------------------------------------------------------- Now let's simplify ... Rewrite as ... Multiply the exponents. ... Rewrite the inner log using the identity ... Square 5 to get 25 ... Use the first identity given to simplify So =========================================================== Answer: This means that
 Proofs/188933: I cannot solve this: I can use: Transportation, Material Implication, Material Equivalence, Exportation, Tautology, Double Negation, Commutation, Association, Distribution, Demorgan's Theorem, Modus Ponems, Modus Tollens, Hypothetical Syllogism, Conjunstion, Simplification, Addition, Constructive Dilemma, Absorption and Disjuntive Syllogism. 1. p -> (q & r) 2. ~t -> (~r v ~s) Therefore: (p & s) -> t I know that you must somehow get: ~p ~p v ~s Add ~(p & s) Demorgans ~(p & s) v t Addition (p & s) -> t Implication OR ~t -> t t v t Implication t Tautology t v ~(p & s) Addition ~(p & s) v t Commutation (p & s) -> t Implication But I cannot get the steps before these. Please help!1 solutions Answer 141836 by jim_thompson5910(28598)   on 2009-03-31 18:13:19 (Show Source): You can put this solution on YOUR website!Note: try to involve every premise (the given lines) to find the conclusion. 1. p -> (q & r) 2. ~t -> (~r v ~s) Therefore: (p & s) -> t ---------------------------------------------- 3. ~~t v (~r v ~s) 2 Material Implication 4. t v (~r v ~s) 3 Double Negation 5. (~r v ~s) v t 4 Commutation 6. ~r v (~s v t) 5 Association 7. ~r v (s -> t) 6 Material Implication 8. r -> (s -> t) 7 Material Implication 9. ~p v (q & r) 1 Material Implication 10. (~p v q) & (~p v r) 9 Distribution 11. (~p v r) & (~p v q) 10 Commutation 12. ~p v r 11 Simplification 13. p -> r 12 Material Implication 14. p -> (s -> t) 13,8 Hypothetical Syllogism 15. (p & s) -> t 14 Exportation 
 Polynomials-and-rational-expressions/189043: Divide x^2+7x+6/x+61 solutions Answer 141833 by jim_thompson5910(28598)   on 2009-03-31 17:56:37 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor to get . Highlight the common terms. Cancel out the common terms. Simplify. So simplifies to . In other words, where
 Radicals/189060: square root 36a^9b^12 / 3ab^71 solutions Answer 141832 by jim_thompson5910(28598)   on 2009-03-31 17:54:49 (Show Source): You can put this solution on YOUR website! Start with the given expression. Divide 36 into 3 to get 12 Divide the variable terms by subtracting the corresponding exponents. Subtract Factor into Factor into Factor into Break up the square root using the identity . Take the square root of to get . Take the square root of to get . Take the square root of to get . Rearrange and multiply the terms. ================================================== Answer: So simplifies to In other words, where and
 Polynomials-and-rational-expressions/189074: This question is from textbook Find each product x^-25 x+5 ------ * ------- 9 x-51 solutions Answer 141831 by jim_thompson5910(28598)   on 2009-03-31 17:50:58 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor to get (use the difference of squares). Combine the fractions. Highlight the common terms. Cancel out the common terms. Simplify. FOIL So simplifies to . In other words, where
 Distributive-associative-commutative-properties/189080: I am having a problem solving this math equation. Could you help me and show me step by step how you get the answer? Here is the problem: 8x +3 { 2 -6 (2x -3)} Thank you Cynthia Bryant1 solutions Answer 141829 by jim_thompson5910(28598)   on 2009-03-31 17:45:00 (Show Source): You can put this solution on YOUR website!8x +3 { 2 -6 (2x -3)} ... Start with the given expression. 8x +3 { 2 -6(2x) -6(-3)} ... Distribute 8x +3 { 2 -12x + 18} ... Multiply 8x +3 {-12x + 20} ... Combine like terms. 8x +3 {-12x} + 3{20} ... Distribute 8x -36x + 60 ... Multiply -28x + 60 ... Combine like terms. So 8x +3 { 2 -6 (2x -3)} = -28x + 60
 Complex_Numbers/189079: Please help me solve this equation: Directions: tell whether the statement is always true, sometimes true, or never true. Explain your reasoning. Problem: The LCD of two rational expressions is the product of the denominators. Thanks! =)1 solutions Answer 141827 by jim_thompson5910(28598)   on 2009-03-31 17:37:11 (Show Source): You can put this solution on YOUR website!Let's look at some examples: Ex 1: LCD of and is . Also, the product of the denominators is . So in this case, statement is NOT true. So it CANNOT be always true (all it takes is one counter example). So the statement is either sometimes true or never true. Ex 2: LCD of and is . Since is the LCD, this shows us that in this case, the statement is true. So this means that we've eliminated the "never true" possibility (as at least one case is true) So the statement "The LCD of two rational expressions is the product of the denominators. " is sometimes true. Note: the statement is only true if the GCF (or GCD) of the denominators is equal to 1, but we don't have to worry about that technicality.
 Polynomials-and-rational-expressions/189078: Hi there,I'm new so im sorry if i do not do something correctly. I am too embarrassed to ask questions in class and... Can you please explain to me how to work out the following problem? ______ Minus ______ I tried this and here is my answer: _____ 1 solutions Answer 141826 by jim_thompson5910(28598)   on 2009-03-31 17:28:07 (Show Source): You can put this solution on YOUR website!You should never feel embarrassed to ask questions. You're asking them now aren't you? Start with the given expression. Factor to get (use the difference of squares) Factor a negative 1 from to get . Also, rearrange the terms. Cancel out the common terms. Simplify Reduce Multiply both the numerator and denominator of the first fraction by FOIL Multiply both the numerator and denominator of the second fraction by Distribute Add the fractions. Combine like terms. So where
 Functions/189012: This question is from textbook f(2)=-3, f(-2)=5 Find the slope of the graph of the linear function f1 solutions Answer 141764 by jim_thompson5910(28598)   on 2009-03-30 23:27:21 (Show Source): You can put this solution on YOUR website!Since f(2)=-3, this means that we have the ordered pair (2,-3). This also means that if , then Also, since f(-2)=5 this means that we have the ordered pair (-2,5). This also means that if , then So this makes the first point and the second point . Start with the slope formula. Plug in , , , and Subtract from to get Subtract from to get Reduce So the slope of the line that goes through the points and is
 Complex_Numbers/189013: True of false? The number (1/2)i is the reciprocal of 2i. Could you please provide a detailed explanation on how you achieved you answer, and show me how the product of 2i and its reciprocal equal 1. 1 solutions Answer 141761 by jim_thompson5910(28598)   on 2009-03-30 23:20:44 (Show Source): You can put this solution on YOUR website!Let The reciprocal of "x" is simply Plug in So the reciprocal of is (not ) So the statement is false. Now multiply by its reciprocal Combine the fractions. Cancel out the common terms. Simplify Multiply EVERY term by "i" to make the denominator real (this doesn't change the expression) Multiply Replace with Note: . So Reduce So
 Money_Word_Problems/189004: This question is from textbook saxon algebra 2 A savings account earns interest at an annual rate of 5%, compounded continuously. If the account begins with a value of $2000, what will its value be after 4 years? A=Pe^rt A=(2000)e^(0.05)(4) A=? When I type this into my calculator, I know I'm getting the wrong answer. Can someone please help!1 solutions Answer 141747 by jim_thompson5910(28598) on 2009-03-30 21:33:55 (Show Source): You can put this solution on YOUR website! Start with the given equation. Plug in , and Multiply 0.05 and 4 to get 0.2 Raise "e" (which is roughly 2.718...) to the 0.2 power to get approximately 1.2214 Multiply (this is also an approximate value) So the account will have about$2,442.80 in 4 years.
 Proofs/188823: I can use: Transportation, Material Implication, Material Equivalence, Exportation, Tautology, Double Negation, Commutation, Association, Distribution, Demorgan's Theorem, Modus Ponems, Modus Tollens, Hypothetical Syllogism, Conjunstion, Simplification, Addition, Constructive Dilemma, Absorption and Disjuntive Syllogism. 10. 1. p -> (q <--> r) 2. ~q -> r 3. ~(q <--> s) 4. p -> s Therefore: ~p1 solutions Answer 141620 by jim_thompson5910(28598)   on 2009-03-29 19:45:51 (Show Source): You can put this solution on YOUR website!I'm sorry, I didn't realize I made a mistake. I'll get this fixed asap... Sorry for the delay, but here's the corrected solution: 1. p -> (q <--> r) 2. ~q -> r 3. ~(q <--> s) 4. p -> s Therefore: ~p ------------------------ 5. ~[(q & s) v (~q & ~s)] 3 Material Equivalence 6. ~(q & s) & ~(~q & ~s) 5 De Morgan's Law 7. ~(~q & ~s) & ~(q & s) 6 Commutation 8. ~(q & s) 6 Simplification 9. ~q v ~s 8 De Morgan's Law 10. q -> ~s 9 Material Implication 11. ~~s -> ~q 10 Transposition 12. s -> ~q 11 Double Negation 13. s -> r 12,2 Hypothetical Syllogism 14. ~(~q & ~s) 7 Simplification 15. ~~q v ~~s 14 De Morgan's Law 16. ~q -> ~~s 15 Material Implication 17. ~q -> s 16 Double Negation 18. ~q -> r 17,13 Hypothetical Syllogism -------------- 19. p -> [(q -> r) & (r -> q)] 1 Material Equivalence 20. ~p v [(~q v r) & (~r v q)] 19 Material Implication 21. [~p v (~q v r)] & [~p v (~r v q)] 20 Distribution 22. [~p v (~r v q)] & [~p v (~q v r)] 21 Commutation 23. ~p v (~r v q) 22 Simplification 24. p -> (~r v q) 23 Material Implication 25. p -> (r -> q) 24 Material Implication 26. (p & r) -> q 25 Exportation 27. (p & r) -> ~s 26,10 Hypothetical Syllogism 28. ~~s -> ~(p & r) 27 Transposition 29. s -> ~(p & r) 28 Double Negation 30. ~q -> ~(p & r) 17,29 Hypothetical Syllogism 31. q v (~p v ~r) 30 Material Implication 32. (q v ~p) v ~r 31 Association 33. ~r v (q v ~p) 32 Commutation 34. ~r v (~p v q) 33 Commutation 35. r -> (p -> q) 34 Material Implication 36. ~q -> (p -> q) 18,35 Hypothetical Syllogism 37. (~q & p) -> q 36 Exportation 38. ~(~q & p) v q 37 Material Implication 39. (~~q v ~p) v q 38 De Morgan's Law 40. (q v ~p) v q 39 Double Negation 41. q v (q v ~p) 40 Commutation 42. (q v q) v ~p 41 Association 43. q v ~p 42 Tautology 44. ~p v q 43 Commutation 45. p -> q 44 Material Implication 46. ~s -> ~p 4 Transposition 47. q -> ~p 10,46 Hypothetical Syllogism 48. p -> ~p 45,47 Hypothetical Syllogism 49. ~p v ~p 48 Material Implication 50. ~p 49 Tautology 
 Proofs/188825: 1. p <--> q 2. p -> r 3. p -> (r -> s) Therefore q -> s ...so far I've tried this but I got stuck... 4. (p -> q) & (q ->p) 1 equivalence 5. p -> q 4 simplification 6. (q -> p) & (p -> q)4 Commutation 7. q -> p 6 simplification 8. q -> (r ->s) 7, 3 Hypothetical Syllogism 9. (q & r) -> s 8 exportation 10. (r & q) -> s 9 commutation 11. p -> (p & r) 2 Absorption please help :(1 solutions Answer 141611 by jim_thompson5910(28598)   on 2009-03-29 17:05:42 (Show Source): You can put this solution on YOUR website!Here's one way to do it... 1. p <--> q 2. p -> r 3. p -> (r -> s) Therefore: q -> s ------------------------------------ 4. (p -> q) & (q -> p) 3 Material Equivalence 5. (q -> p) & (p -> q) 4 Commutation 6. q -> p 5 Simplification 7. q -> r 6,2 Hypothetical Syllogism 8. q -> (r -> s) 6,3 Hypothetical Syllogism 9. (q & r) -> s 8 Exportation 10. ~(q & r) v s 9 Material Implication 11. (~q v ~r) v s 10 DeMorgan's Theorem 12. ~q v (~r v s) 11 Association 13. ~q v (s v ~r) 12 Commutation 14. (~q v s) v ~r 13 Association 15. (q -> s) v ~r 14 Material Implication 16. ~r v (q -> s) 15 Commutation 17. r -> (q -> s) 16 Material Implication 18. q -> (q -> s) 7,17 Hypothetical Syllogism 19. (q & q) -> s 18 Exportation 20. q -> s 19 Tautology 
 Proofs/188805: I cannot solve this: I can use: Transportation, Material Implication, Material Equivalence, Exportation, Tautology, Double Negation, Commutation, Association, Distribution, Demorgan's Theorem, Modus Ponems, Modus Tollens, Hypothetical Syllogism, Conjunstion, Simplification, Addition, Constructive Dilemma, Absorption and Disjuntive Syllogism. 4. 1. p <--> q 2. p -> r 3. p -> (r -> s) Therefore: q -> s 1 solutions Answer 141610 by jim_thompson5910(28598)   on 2009-03-29 16:50:21 (Show Source): You can put this solution on YOUR website!1. p <--> q 2. p -> r 3. p -> (r -> s) Therefore: q -> s ------------------------------------ 4. (p -> q) & (q -> p) 3 Material Equivalence 5. (q -> p) & (p -> q) 4 Commutation 6. q -> p 5 Simplification 7. q -> r 6,2 Hypothetical Syllogism 8. q -> (r -> s) 6,3 Hypothetical Syllogism 9. (q & r) -> s 8 Exportation 10. ~(q & r) v s 9 Material Implication 11. (~q v ~r) v s 10 DeMorgan's Theorem 12. ~q v (~r v s) 11 Association 13. ~q v (s v ~r) 12 Commutation 14. (~q v s) v ~r 13 Association 15. (q -> s) v ~r 14 Material Implication 16. ~r v (q -> s) 15 Commutation 17. r -> (q -> s) 16 Material Implication 18. q -> (q -> s) 7,17 Hypothetical Syllogism 19. (q & q) -> s 18 Exportation 20. q -> s 19 Tautology 
 Proofs/188807: I cannot solve this: I can use: Transportation, Material Implication, Material Equivalence, Exportation, Tautology, Double Negation, Commutation, Association, Distribution, Demorgan's Theorem, Modus Ponems, Modus Tollens, Hypothetical Syllogism, Conjunstion, Simplification, Addition, Constructive Dilemma, Absorption and Disjuntive Syllogism. 6. 1. (p v q) -> r 2. r -> (s <--> t) 3. s -> ~t Therefore: ~p v ~t1 solutions Answer 141609 by jim_thompson5910(28598)   on 2009-03-29 16:17:00 (Show Source): You can put this solution on YOUR website!1. (p v q) -> r 2. r -> (s <--> t) 3. s -> ~t Therefore: ~p v ~t ------------------------------- 4. r -> [ (s -> t) & (t -> s) ] 2 Material Equivalence 5. ~r v [ (s -> t) & (t -> s) ] 4 Material Implication 6. [~r v (s -> t)] & [~r v (t -> s)] 5 Distribution 7. [~r v (t -> s)] & [~r v (s -> t)] 6 Commutation 8. ~r v (t -> s) 7 Simplification 9. r -> (t -> s) 8 Material Implication 10. (r & t) -> s 9 Exportation 11. (r & t) -> ~t 3,10 Hypothetical Syllogism 12. r -> (t -> ~t) 11 Exportation 13. r -> (~t v ~t) 12 Material Implication 14. r -> ~t 13 Tautology 15. (p v q) -> ~t 1,14 Hypothetical Syllogism 16. ~(p v q) v ~t 15 Material Implication 17. (~p & ~q) v ~t 16 DeMorgan's Theorem 18. ~t v (~p & ~q) 17 Commutation 19. (~t v ~p) & (~t v ~q) 18 Distribution 20. ~t v ~p 19 Simplification 21. ~p v ~t 20 Commutation