Another Math question help

Test Preparation
1

<p>I have 2 very difficult questions, which I dont know how to solve it.
Please Help!!!</p>

<p>If (a+b)^0,5= (a-b)^0,5 which of the following must be true?</p>

<p>A) b=0
B) a+b=0
C) a-b=0
D) a^2+b^2= 0
E) a^2-b^2= 0</p>

<hr>

<p>Set X has X members and Set Y has Y members. Set Z consists all members that are in either set X or set Y with the exception of the K common members (K>0). Which of the following represents the number of members in set Z?</p>

<p>A) X+Y+K
B) X+Y-K
C) X+Y+2K
D) X+Y-2K
E) 2X+2Y-2K</p>

<p>Trinya</p>

2

<p>first tell me if the answers are A) and B)</p>

3

<p>For the second problem, just think about the fact that the common numbers occur seperately in both sets. The answer is x+y-2k</p>

4

<p>Thnx, now try the first one</p>

5

<p>and please explain the second better, I dont understand "Common"</p>

6

<p>hmm Ill try the second one. What its saying is:
square root(a+b)=square root(a-b)
square both sides.
a+b=a-b
a+2b=a
2b=0
b=0
answer A Tell me if its right.
x^1/n=n root.
Example: 27^(1/3)=cube root of 27=3. x^.5 or 1/2 is just another way of saying square root of.</p>

<p>I remember the answer to the second one is X+y-2k. Make example sets say X=(3,4,6,7) and Y=(1,2,3,4,5) So Z would be the amount of numbers that only appear in one. That is 1,2,5,6,7=5 numbers. X= 4numbers Y=5 numbers K=2. So x+y-2k=4+5-4=5=z. You take off 2k because if a number appears in both sets, you got to get rid of that number from both sets, so you double k.</p>

<p>BTW, Before you ask about the problem please announce the source, because you may screw up a practice test for someone ( I happened to see these problems in the 8 reals book).</p>

7

<p>I am very sorry for the source, I forgot it.</p>

<p>But can you explain what "common" means?</p>

8

<p>It just means numbers that are in both sets.
Just phrase it differently. What do the sets have in common? The numbers 2,3,... or whatever.</p>

9

<p>1-a 2-d.........</p>