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Numeric_Fractions/189229: 7%2F15=4%2Fx
I dont know how to find x for this. Please help.
Thanks so much ^_^
1 solutions

Answer 141964 by jim_thompson5910(28598) About Me  on 2009-04-01 18:38:13 (Show Source):
You can put this solution on YOUR website!
7%2F15=4%2Fx Start with the given equation.


7x=4%2815%29 Cross multiply


7x=60 Multiply


x=60%2F7 Divide both sides by 7 to isolate "x".


So the solution is x=60%2F7 which is approximately


Complex_Numbers/189232: Write the complex number
2-i
(over)
5+i
in the usual a + bi form
1 solutions

Answer 141963 by jim_thompson5910(28598) About Me  on 2009-04-01 18:36:14 (Show Source):
You can put this solution on YOUR website!
%282-i%29%2F%285%2Bi%29 Start with the given expression.


%28%282-i%29%2F%285%2Bi%29%29%28%285-i%29%2F%285-i%29%29 Multiply the fraction by %285-i%29%2F%285-i%29.


%28%282-i%29%285-i%29%29%2F%28%285%2Bi%29%285-i%29%29 Combine the fractions.


FOIL the numerator.


FOIL the denominator.


%2810-2i-5i%2Bi%5E2%29%2F%2825-5i%2B5i-i%5E2%29 Multiply.


%289-7i%29%2F%2826%29 Combine like terms.


%289%29%2F%2826%29%2B%28%28-7%29%2F%2826%29%29i Break up the fraction.


9%2F26-%287%2F26%29i Reduce.


So %282-i%29%2F%285%2Bi%29=9%2F26-%287%2F26%29i.


So the expression is now in standard form a%2Bbi where a=9%2F26 and b=-7%2F26


Inequalities/189233: Please help.....
-4x - 23 >= 25
solve the inequality for w and simplify answer as much as possible.
1 solutions

Answer 141962 by jim_thompson5910(28598) About Me  on 2009-04-01 18:34:10 (Show Source):
You can put this solution on YOUR website!
I'm assuming you meant to solve for "x"


-4x-23%3E=25 Start with the given inequality.


-4x%3E=25%2B23 Add 23 to both sides.


-4x%3E=48 Combine like terms on the right side.


x%3C=%2848%29%2F%28-4%29 Divide both sides by -4 to isolate x. note: Remember, the inequality sign flips when we divide both sides by a negative number.


x%3C=-12 Reduce.


----------------------------------------------------------------------

Answer:

So the answer is x%3C=-12


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) About Me  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 L=3W-1 and because "The perimeter is 30 ft", this tells us that P=30


P=2L%2B2W Start with the perimeter formula


30=2%283W-1%29%2B2W Plug in P=30 and L=3W-1


30=6W-2%2B2W Distribute


30%2B2=6W%2B2W Add 2 to both sides.


32=8W Combine like terms.


32%2F8=W Divide both sides by 8 to isolate W.


4=W Reduce.


W=4 Rearrange the equation.

So the width is 4 feet.


L=3W-1 Go back to the first equation


L=3%284%29-1 Plug in W=4


L=12-1 Multiply.


L=11 Subtract.


So the length is 11 feet.


Proofs/189186: ~(A or B) or ~(C or D)
(E or F) -> D therefore A-> ~E
1 solutions

Answer 141960 by jim_thompson5910(28598) About Me  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) About Me  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) About Me  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^2-9
1 solutions

Answer 141875 by jim_thompson5910(28598) About Me  on 2009-03-31 22:10:54 (Show Source):
You can put this solution on YOUR website!
%28f%2Bg%29%28x%29 Start with the given expression.


f%28x%29%2Bg%28x%29 Expand


%28x%5E2%2B7x%2B12%29%2B%28x%5E2-9%29 Plug in f%28x%29=+x%5E2%2B7x%2B12 and g%28x%29=+x%5E2-9


2x%5E2%2B7x%2B3 Combine like terms.


So %28f%2Bg%29%28x%29=2x%5E2%2B7x%2B3


-----------------------------------------


%28f-g%29%28x%29 Start with the given expression.


f%28x%29-g%28x%29 Expand


%28x%5E2%2B7x%2B12%29-%28x%5E2-9%29 Plug in f%28x%29=+x%5E2%2B7x%2B12 and g%28x%29=+x%5E2-9


x%5E2%2B7x%2B12-x%5E2%2B9 Distribute


7x%2B21 Combine like terms.


So %28f-g%29%28x%29=7x%2B21


-----------------------------------------



%28f%2Ag%29%28x%29 Start with the given expression.


f%28x%29%2Ag%28x%29 Expand


%28x%5E2%2B7x%2B12%29%2A%28x%5E2-9%29 Plug in f%28x%29=+x%5E2%2B7x%2B12 and g%28x%29=+x%5E2-9



%28x%5E2-9%29%28x%5E2%2B7x%2B12%29 Rearrange the terms.


x%5E2%28x%5E2%2B7x%2B12%29-9%28x%5E2%2B7x%2B12%29 Expand. Note: %28a%2Bb%29%28c%2Bd%2Be%29=a%28c%2Bd%2Be%29%2Bb%28c%2Bd%2Be%29


x%5E2%28x%5E2%29%2Bx%5E2%287x%29%2Bx%5E2%2812%29-9%28x%5E2%29-9%287x%29-9%2812%29 Distribute


x%5E4%2B7x%5E3%2B12x%5E2-9x%5E2-63x-108 Multiply


x%5E4%2B7x%5E3%2B3x%5E2-63x-108 Combine like terms.


So %28f%2Ag%29%28x%29=x%5E4%2B7x%5E3%2B3x%5E2-63x-108


-----------------------------------------



%28f%2Fg%29%28x%29 Start with the given expression.


f%28x%29%2Fg%28x%29 Expand


%28x%5E2%2B7x%2B12%29%2F%28x%5E2-9%29 Plug in f%28x%29=+x%5E2%2B7x%2B12 and g%28x%29=+x%5E2-9


%28%28x%2B4%29%28x%2B3%29%29%2F%28x%5E2-9%29 Factor x%5E2%2B7x%2B12 to get %28x%2B4%29%28x%2B3%29.


%28%28x%2B4%29%28x%2B3%29%29%2F%28%28x-3%29%28x%2B3%29%29 Factor x%5E2-9 to get %28x-3%29%28x%2B3%29.


%28%28x%2B4%29highlight%28%28x%2B3%29%29%29%2F%28%28x-3%29highlight%28%28x%2B3%29%29%29 Highlight the common terms.


%28%28x%2B4%29cross%28%28x%2B3%29%29%29%2F%28%28x-3%29cross%28%28x%2B3%29%29%29 Cancel out the common terms.


%28x%2B4%29%2F%28x-3%29 Simplify.



So %28f%2Fg%29%28x%29=%28x%2B4%29%2F%28x-3%29 where x%3C%3E-3 or x%3C%3E3


-----------------------------------------


Polynomials-and-rational-expressions/189121: can you help me please?

factor the equation +d=.05r%5E2%2Br+
1 solutions

Answer 141857 by jim_thompson5910(28598) About Me  on 2009-03-31 20:41:17 (Show Source):
You can put this solution on YOUR website!
+d=.05r%5E2%2Br+ Start with the given equation.


+d=r%28.05r%2B1%29+ 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) About Me  on 2009-03-31 20:28:17 (Show Source):
You can put this solution on YOUR website!

From x%5E2%2B2x%2B1 we can see that a=1, b=2, and c=1


D=b%5E2-4ac Start with the discriminant formula.


D=%282%29%5E2-4%281%29%281%29 Plug in a=1, b=2, and c=1


D=4-4%281%29%281%29 Square 2 to get 4


D=4-4 Multiply 4%281%29%281%29 to get %284%29%281%29=4


D=0 Subtract 4 from 4 to get 0


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*
Thanks
1 solutions

Answer 141853 by jim_thompson5910(28598) About Me  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, b=2 and x=5)


This means that





simplifies to





In other words,


-----------------------------------------------------------------


Now let's simplify



... Rewrite 4 as 2%5E2


... Multiply the exponents.


... Rewrite the inner log using the identity y%2Alog%28b%2C%28x%29%29=log%28b%2C%28x%5Ey%29%29


... 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) About Me  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+6
1 solutions

Answer 141833 by jim_thompson5910(28598) About Me  on 2009-03-31 17:56:37 (Show Source):
You can put this solution on YOUR website!

%28x%5E2%2B7x%2B6%29%2F%28x%2B6%29 Start with the given expression.


%28%28x%2B6%29%28x%2B1%29%29%2F%28x%2B6%29 Factor x%5E2%2B7x%2B6 to get %28x%2B6%29%28x%2B1%29.


%28highlight%28%28x%2B6%29%29%28x%2B1%29%29%2F%28highlight%28%28x%2B6%29%29%29 Highlight the common terms.


%28cross%28%28x%2B6%29%29%28x%2B1%29%29%2F%28cross%28%28x%2B6%29%29%29 Cancel out the common terms.


x%2B1 Simplify.


So %28x%5E2%2B7x%2B6%29%2F%28x%2B6%29 simplifies to x%2B1.


In other words, %28x%5E2%2B7x%2B6%29%2F%28x%2B6%29=x%2B1 where x%3C%3E-6


Radicals/189060: square root 36a^9b^12 / 3ab^7
1 solutions

Answer 141832 by jim_thompson5910(28598) About Me  on 2009-03-31 17:54:49 (Show Source):
You can put this solution on YOUR website!
sqrt%28%2836a%5E9b%5E12%29%2F%283ab%5E7%29%29 Start with the given expression.


sqrt%28%2812a%5E9b%5E12%29%2F%28ab%5E7%29%29 Divide 36 into 3 to get 12


sqrt%2812a%5E%289-1%29b%5E%2812-7%29%29 Divide the variable terms by subtracting the corresponding exponents.


sqrt%2812a%5E8b%5E5%29 Subtract


sqrt%284%2A3%2Aa%5E8%2Ab%5E5%29 Factor 12 into 4%2A3


sqrt%284%2A3%2Aa%5E2%2Aa%5E2%2Aa%5E2%2Aa%5E2%2Ab%5E5%29 Factor a%5E8 into a%5E2%2Aa%5E2%2Aa%5E2%2Aa%5E2


sqrt%284%2A3%2Aa%5E2%2Aa%5E2%2Aa%5E2%2Aa%5E2%2Ab%5E2%2Ab%5E2%2Ab%29 Factor b%5E5 into b%5E2%2Ab%5E2%2Ab


Break up the square root using the identity sqrt%28A%2AB%29=sqrt%28A%29%2Asqrt%28B%29.


Take the square root of 4 to get 2.


2%2Asqrt%283%29%2Aa%2Aa%2Aa%2Aa%2Asqrt%28b%5E2%29%2Asqrt%28b%5E2%29%2Asqrt%28b%29 Take the square root of a%5E2 to get a.


2%2Asqrt%283%29%2Aa%2Aa%2Aa%2Aa%2Ab%2Ab%2Asqrt%28b%29 Take the square root of b%5E2 to get b.


2a%5E4b%5E2%2Asqrt%283b%29 Rearrange and multiply the terms.

==================================================

Answer:


So sqrt%28%2836a%5E9b%5E12%29%2F%283ab%5E7%29%29 simplifies to 2a%5E4b%5E2%2Asqrt%283b%29


In other words, sqrt%28%2836a%5E9b%5E12%29%2F%283ab%5E7%29%29=2a%5E4b%5E2%2Asqrt%283b%29 where a%3E0 and b%3E0


Polynomials-and-rational-expressions/189074: This question is from textbook
Find each product
x^-25 x+5
------ * -------
9 x-5
1 solutions

Answer 141831 by jim_thompson5910(28598) About Me  on 2009-03-31 17:50:58 (Show Source):
You can put this solution on YOUR website!

%28%28x%5E2-25%29%2F%289%29%29%28%28x%2B5%29%2F%28x-5%29%29 Start with the given expression.


%28%28%28x-5%29%28x%2B5%29%29%2F%289%29%29%28%28x%2B5%29%2F%28x-5%29%29 Factor x%5E2-25 to get %28x-5%29%28x%2B5%29 (use the difference of squares).


%28%28x-5%29%28x%2B5%29%28x%2B5%29%29%2F%289%28x-5%29%29 Combine the fractions.


%28highlight%28%28x-5%29%29%28x%2B5%29%28x%2B5%29%29%2F%289%2Ahighlight%28%28x-5%29%29%29 Highlight the common terms.


%28cross%28%28x-5%29%29%28x%2B5%29%28x%2B5%29%29%2F%289%2Across%28%28x-5%29%29%29 Cancel out the common terms.


%28%28x%2B5%29%28x%2B5%29%29%2F%289%29 Simplify.


%28x%5E2%2B10x%2B25%29%2F%289%29 FOIL


So %28%28x%5E2-25%29%2F%289%29%29%28%28x%2B5%29%2F%28x-5%29%29 simplifies to %28x%5E2%2B10x%2B25%29%2F%289%29.


In other words, %28%28x%5E2-25%29%2F%289%29%29%28%28x%2B5%29%2F%28x-5%29%29=%28x%5E2%2B10x%2B25%29%2F%289%29 where x%3C%3E5


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 Bryant
1 solutions

Answer 141829 by jim_thompson5910(28598) About Me  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


Polynomials-and-rational-expressions/189076: i need your help! Can you help me solve this equation?

Solve 2x(x+4)(4x-9)=0

1 solutions

Answer 141828 by jim_thompson5910(28598) About Me  on 2009-03-31 17:41:44 (Show Source):
You can put this solution on YOUR website!
2x%28x%2B4%29%284x-9%29=0 Start with the given equation.


2x=0, x%2B4=0, 4x-9=0 Set each factor equal to zero (use the zero product property)

----------------------

Let's solve the first equation:


2x=0 Start with the first equation.


x=0%2F2 Divide both sides by 2.


x=0 Reduce


So the first solution is x=-4

----------------------

Let's solve the second equation:


x%2B4=0 Start with the second equation.


x=-4 Subtract 4 from both sides.


So the second solution is x=-4
----------------------

Let's solve the third equation:


4x-9=0 Start with the third equation.


4x=9 Add 9 to both sides.


x=9%2F4 Divide both sides by 4.


So the third solution is x=9%2F4


========================================


So the three solutions are x=0, x=-4, or x=9%2F4 (note: the order of the solutions does not matter)


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) About Me  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 1%2Fx and 1%2F%282x%29 is 2x. Also, the product of the denominators is x%2A%282x%29=2x%5E2. 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 y%2F%283z%29 and r%5E2%2F%287w%29 is 21wz. Since %283z%29%287w%29=21wz 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?
3%2By
______
y%5E2%2B9
Minus
y-3
______
9-y%5E2
I tried this and here is my answer:
+++1
_____
+%28y-3%29%28y%2B3%29

1 solutions

Answer 141826 by jim_thompson5910(28598) About Me  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?


%283%2By%29%2F%28y%5E2%2B9%29-%28y-3%29%2F%289-y%5E2%29 Start with the given expression.


%283%2By%29%2F%28y%5E2%2B9%29-%28y-3%29%2F%28%283-y%29%283%2By%29%29 Factor 9-y%5E2 to get %283-y%29%283%2By%29 (use the difference of squares)


%28y%2B3%29%2F%28y%5E2%2B9%29-%28y-3%29%2F%28-%28y-3%29%28y%2B3%29%29 Factor a negative 1 from 3-y to get -%28y-3%29. Also, rearrange the terms.


%28y%2B3%29%2F%28y%5E2%2B9%29-cross%28%28y-3%29%29%2F%28-cross%28%28y-3%29%29%28y%2B3%29%29 Cancel out the common terms.


%28y%2B3%29%2F%28y%5E2%2B9%29-1%2F%28-%28y%2B3%29%29 Simplify


%28y%2B3%29%2F%28y%5E2%2B9%29%2B1%2F%28y%2B3%29 Reduce


%28%28y%2B3%29%28y%2B3%29%29%2F%28%28y%5E2%2B9%29%28y%2B3%29%29%2B1%2F%28y%2B3%29 Multiply both the numerator and denominator of the first fraction by y%2B3


%28y%5E2%2B6y%2B9%29%2F%28%28y%5E2%2B9%29%28y%2B3%29%29%2B1%2F%28y%2B3%29 FOIL


Multiply both the numerator and denominator of the second fraction by y%5E2%2B9


Distribute


%28y%5E2%2B6y%2B9%2By%5E2%2B9%29%2F%28%28y%5E2%2B9%29%28y%2B3%29%29 Add the fractions.


%282y%5E2%2B6y%2B18%29%2F%28%28y%5E2%2B9%29%28y%2B3%29%29 Combine like terms.


So where y%3C%3E-3


Functions/189012: This question is from textbook
f(2)=-3, f(-2)=5
Find the slope of the graph of the linear function f
1 solutions

Answer 141764 by jim_thompson5910(28598) About Me  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 x=2, then y=-3


Also, since f(-2)=5 this means that we have the ordered pair (-2,5). This also means that if x=-2, then y=5


So this makes the first point and the second point .


m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29 Start with the slope formula.


m=%285--3%29%2F%28-2-2%29 Plug in y%5B2%5D=5, y%5B1%5D=-3, x%5B2%5D=-2, and x%5B1%5D=2


m=%288%29%2F%28-2-2%29 Subtract -3 from 5 to get 8


m=%288%29%2F%28-4%29 Subtract 2 from -2 to get -4


m=-2 Reduce


So the slope of the line that goes through the points and is m=-2



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) About Me  on 2009-03-30 23:20:44 (Show Source):
You can put this solution on YOUR website!
Let x=2i


The reciprocal of "x" is simply 1%2Fx


1%2F%282i%29 Plug in x=2i


So the reciprocal of 2i is 1%2F%282i%29 (not %281%2F2%29i)


So the statement is false.


Now multiply 2i by its reciprocal 1%2F%282i%29


%282i%29%281%2F%282i%29%29


%282i%29%2F%282i%29 Combine the fractions.


%28cross%282%29i%29%2F%28cross%282%29i%29 Cancel out the common terms.


i%2Fi Simplify


%28i%2Ai%29%2F%28i%2Ai%29 Multiply EVERY term by "i" to make the denominator real (this doesn't change the expression)


%28i%5E2%29%2F%28i%5E2%29 Multiply


%28-1%29%2F%28-1%29 Replace i%5E2 with -1


Note: i=sqrt%28-1%29. So i%5E2=%28sqrt%28-1%29%29%5E2=-1


1 Reduce


So %282i%29%281%2F%282i%29%29=1


Quadratic_Equations/189008: how would you solve for 8x squared – 12x = 11
1 solutions

Answer 141753 by jim_thompson5910(28598) About Me  on 2009-03-30 21:56:45 (Show Source):
You can put this solution on YOUR website!
8x%5E2-12x=11 Start with the given equation.


8x%5E2-12x-11=0 Subtract 11 from both sides.


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=8, b=-12, and c=-11


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%28-12%29+%2B-+sqrt%28+%28-12%29%5E2-4%288%29%28-11%29+%29%29%2F%282%288%29%29 Plug in a=8, b=-12, and c=-11


x+=+%2812+%2B-+sqrt%28+%28-12%29%5E2-4%288%29%28-11%29+%29%29%2F%282%288%29%29 Negate -12 to get 12.


x+=+%2812+%2B-+sqrt%28+144-4%288%29%28-11%29+%29%29%2F%282%288%29%29 Square -12 to get 144.


x+=+%2812+%2B-+sqrt%28+144--352+%29%29%2F%282%288%29%29 Multiply 4%288%29%28-11%29 to get -352


x+=+%2812+%2B-+sqrt%28+144%2B352+%29%29%2F%282%288%29%29 Rewrite sqrt%28144--352%29 as sqrt%28144%2B352%29


x+=+%2812+%2B-+sqrt%28+496+%29%29%2F%282%288%29%29 Add 144 to 352 to get 496


x+=+%2812+%2B-+sqrt%28+496+%29%29%2F%2816%29 Multiply 2 and 8 to get 16.


x+=+%2812+%2B-+4%2Asqrt%2831%29%29%2F%2816%29 Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


x+=+%2812%2B4%2Asqrt%2831%29%29%2F%2816%29 or x+=+%2812-4%2Asqrt%2831%29%29%2F%2816%29 Break up the expression.


x+=+%283%2Bsqrt%2831%29%29%2F%284%29 or x+=+%283-sqrt%2831%29%29%2F%284%29 Reduce


So the answers are x+=+%283%2Bsqrt%2831%29%29%2F%284%29 or x+=+%283-sqrt%2831%29%29%2F%284%29


Inequalities/188951: 4+2(3-2y)<_6y-20
1 solutions

Answer 141750 by jim_thompson5910(28598) About Me  on 2009-03-30 21:44:46 (Show Source):
You can put this solution on YOUR website!

4%2B2%283-2y%29%3C=6y-20 Start with the given inequality.


4%2B6-4y%3C=6y-20 Distribute.


-4y%2B10%3C=6y-20 Combine like terms on the left side.


-4y%3C=6y-20-10 Subtract 10 from both sides.


-4y-6y%3C=-20-10 Subtract 6y from both sides.


-10y%3C=-20-10 Combine like terms on the left side.


-10y%3C=-30 Combine like terms on the right side.


y%3E=%28-30%29%2F%28-10%29 Divide both sides by -10 to isolate y. note: Remember, the inequality sign flips when we divide both sides by a negative number.


y%3E=3 Reduce.


----------------------------------------------------------------------

Answer:

So the answer is y%3E=3


logarithm/189006: This question is from textbook saxon algebra 2
Use the properties of logarithms to evaluate ln(8e)^4. Round to the nearest tenth.
4*ln(8e)
4*ln(8e)=4*(ln8+lne)
4(ln8+lne)=?
What do I do next?

1 solutions

Answer 141749 by jim_thompson5910(28598) About Me  on 2009-03-30 21:41:12 (Show Source):
You can put this solution on YOUR website!
ln%28%288e%29%5E4%29 Start with the given expression.


4%2Aln%288e%29 Pull down the exponent using the identity ln%28x%5Ey%29=y%2Aln%28x%29%29


4%2A%28ln%288%29%2Bln%28e%29%29 Break up the log using the identity ln%28A%2AB%29=ln%28A%29%2Aln%28B%29


4%2A%28ln%288%29%2B1%29 Evaluate the natural log of "e" to get 1.


Note: since the natural log is really a log with base "e", this means that


y=ln%28e%29 ---> e%5Ey=e ----> y=1


4%2A%282.0794%2B1%29 Evaluate the natural log of 8 to get approximately 2.0794


4%2A%283.0794%29 Add


12.3176 Multiply


12.3 Round to the nearest tenth



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) About Me  on 2009-03-30 21:33:55 (Show Source):
You can put this solution on YOUR website!
A=Pe%5E%28rt%29 Start with the given equation.


A=%282000%29e%5E%28%280.05%29%284%29%29 Plug in P=2000, r=0.05 and t=4


A=2000e%5E%280.2%29 Multiply 0.05 and 4 to get 0.2


A=2000%281.2214%29 Raise "e" (which is roughly 2.718...) to the 0.2 power to get approximately 1.2214


A=2442.8 Multiply (this is also an approximate value)



So the account will have about $2,442.80 in 4 years.


Rational-functions/188989: This question is from textbook algebra2
Find (f+g)(x), (f-g)(x), (f*g)(x)and (f/g)(x) for each f(x) and g(x)
2.f(x)=8x^2
G(x)=1/x^2
1 solutions

Answer 141745 by jim_thompson5910(28598) About Me  on 2009-03-30 21:28:41 (Show Source):
You can put this solution on YOUR website!
Note: for every example, x%3C%3E0 (to avoid division by zero)


%28f%2Bg%29%28x%29 Start with the given expression.


f%28x%29%2Bg%28x%29 Expand


8x%5E2%2B1%2Fx%5E2 Plug in f%28x%29=8x%5E2 and g%28x%29=1%2Fx%5E2


8x%5E2%28x%5E2%2Fx%5E2%29%2B1%2Fx%5E2 Multiply the first term by x%5E2%2Fx%5E2 (to get the denominators the same.


8x%5E4%2Fx%5E2%2B1%2Fx%5E2 Multiply


%288x%5E4%2B1%29%2Fx%5E2 Combine the fractions.


So %28f%2Bg%29%28x%29=%288x%5E4%2B1%29%2Fx%5E2


-------------------------------------------------------

%28f-g%29%28x%29 Start with the given expression.


f%28x%29-g%28x%29 Expand


8x%5E2-1%2Fx%5E2 Plug in f%28x%29=8x%5E2 and g%28x%29=1%2Fx%5E2


8x%5E2%28x%5E2%2Fx%5E2%29-1%2Fx%5E2 Multiply the first term by x%5E2%2Fx%5E2 (to get the denominators the same.


8x%5E4%2Fx%5E2-1%2Fx%5E2 Multiply


%288x%5E4-1%29%2Fx%5E2 Combine the fractions.


So %28f-g%29%28x%29=%288x%5E4-1%29%2Fx%5E2

--------------------------------------------------------

%28f%2Ag%29%28x%29 Start with the given expression.


f%28x%29%2Ag%28x%29 Expand


%288x%5E2%29%281%2Fx%5E2%29 Plug in f%28x%29=8x%5E2 and g%28x%29=1%2Fx%5E2


8x%5E2%2Fx%5E2 Combine the fractions.


8 Simplify (the x%5E2 terms cancel out)


So %28f%2Ag%29%28x%29=8


--------------------------------------------------------

%28f%2Fg%29%28x%29 Start with the given expression.


f%28x%29%2Fg%28x%29 Expand


%288x%5E2%29%2F%281%2Fx%5E2%29 Plug in f%28x%29=8x%5E2 and g%28x%29=1%2Fx%5E2


%288x%5E2%29%28x%5E2%2F1%29 Multiply by the reciprocal of the second fraction.


%288x%5E2%2Ax%5E2%29%2F1 Combine the fractions.


8x%5E4%2F1 Multiply (add the exponents)


8x%5E4 Reduce


So %28f%2Fg%29%28x%29=8x%5E4


Exponents-negative-and-fractional/188984: how do i solve(5xE3yE4)E2
1 solutions

Answer 141742 by jim_thompson5910(28598) About Me  on 2009-03-30 21:16:46 (Show Source):
You can put this solution on YOUR website!
%285x%5E3y%5E4%29%5E2 Start with the given expression.


%285%5E1x%5E3y%5E4%29%5E2 Rewrite 5 as 5%5E1


5%5E%281%2A2%29x%5E%283%2A2%29y%5E%284%2A2%29 Multiply the outer exponent by EVERY exponent in the parenthesis.


5%5E2x%5E6y%5E8 Multiply


25x%5E6y%5E8 Square 5 to get 25


So %285x%5E3y%5E4%29%5E2=25x%5E6y%5E8


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: ~p
1 solutions

Answer 141620 by jim_thompson5910(28598) About Me  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) About Me  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) About Me  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 ~t
1 solutions

Answer 141609 by jim_thompson5910(28598) About Me  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