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http://www.mediafire.com/?y8981oxxjpi1eoo

 

Customer Question

http://www.mediafire.com/?y8981oxxjpi1eoo

 



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It is one question

Submitted: 587 days and 14 hours ago.
Category: Pre-Calculus
Value: AU$35.61
Status: CLOSED

Accepted Answer

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Expert:  Osbert12 replied 587 days and 13 hours ago.

Thank you for your question.

 

graphic

Osbert

Expert TypeMaster's Degree
Category: Pre-Calculus
Pos. Feedback: 100.0 %
Accepts: 74
Answered: 8/31/2011

Experience: Math Tutor

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Customer replied 587 days and 12 hours ago.

Can you please help me with another.

http://www.mediafire.com/download.php?4pqbnn9lch7yxjw

http://www.mediafire.com/download.php?ywjp8z4bwnuk3jd

Accepted Answer

Picture
Expert:  Osbert12 replied 587 days and 12 hours ago.

Thank you for your questions.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

graphic

 

(b) p Λ (q V r) becomes False Λ (False V True) which is False

(p Λ q) V r becomes (False Λ False) V True which is True

Since p Λ (q V r) is not equivalent to (p Λ q) V r in this situation, they are not equivalent.

 

 

Osbert

Expert TypeMaster's Degree
Category: Pre-Calculus
Pos. Feedback: 100.0 %
Accepts: 74
Answered: 8/31/2011

Experience: Math Tutor

Ask this Expert a Question >
Customer replied 587 days and 3 hours ago.

Can you help with this one too

http://www.mediafire.com/?1m9uo6scdhk1fqx

And whatever the amount is set I don't know but I am sure to put in a bonus.

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Expert:  Osbert12 replied 587 days and 3 hours ago.

Thank you for your questions.

 

(a) False

 

(b) Consider x = 1. Then 1 ≥ y + 1 , which means y ≤ 0, which contradicts the premise that y is a positive integer.

 

(c) x is a positive integer larger than 1, y is a positive integer

 

(d) There exists an x value for which there is not a y such that x ≤ y + 1, where x and y are positive integers.

 

Osbert

Customer replied 587 days and 2 hours ago.

This one also

http://www.mediafire.com/?71hsxvt42rcbfci

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Expert:  Osbert12 replied 587 days and 2 hours ago.

Thank you.

 

Case (i)

|x| = x

|y| = y

x + y ≥ 0 → |x + y| = x + y

|x + y| = |x| + |y|

 

Case (ii)

|x| = x

|y| = -y

If x + y ≥ 0 → |x + y| = x + y < |x| + |y| = x - y

If x + y < 0 → |x + y| = -x - y ≤ |x| + |y| = x - y

 

Case (iii)

Replace y with x and x with y. This is proved in case (ii).

 

Case (iv)

|x| = -x

|y| = -y

x + y = -(-x + -y)

|x + y| = -(x + y) = -x + -y = |x| + |y|

 

Osbert

Customer replied 587 days and 1 hours ago.

And another

http://www.mediafire.com/?beew4fywycrxbno

Picture
Expert:  Osbert12 replied 587 days and 1 hours ago.

Thank you.

 

For n = 1,

1/(1(2)) = 1/2 = 1 - 1/(1+1)

 

Assume the equation holds for n = k.

 

For n = k+1,

1 / 1(2) + 1 / 2(3) + 1 / (k-1)k + 1 / k(k+1) = 1 - 1/k + 1 / k(k+1) = 1 - (k+1) / k(k+1) + 1 / k(k+1) = 1 - k / k(k+1) = 1 - 1 / (k+1)

 

Osbert

Customer replied 587 days ago.

And one final one please

http://www.mediafire.com/download.php?00qjkxlezm8yt9d

Accepted Answer

Picture
Expert:  Osbert12 replied 586 days and 14 hours ago.

Thank you.

 

graphic

Osbert

Expert TypeMaster's Degree
Category: Pre-Calculus
Pos. Feedback: 100.0 %
Accepts: 74
Answered: 9/1/2011

Experience: Math Tutor

Ask this Expert a Question >
 
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