Question

which statement is always true if the domain of the variables is the set of positive integers?
a) √x^2 + y^2 = x+y
b) xy=y
c) (x+y/z) = (x/z) + (y/z)
d) (x+y)^2 = x^2 +y^2

Answer 1

Is that the full question

Answer 2

\[\sqrt{x ^{2}+y ^{2}} = x+y\]

Answer 3

\[\frac{ x+y }{ z}=\frac{ x }{ z }+\frac{ y }{ z}\]

Answer 4

\[(x+y)^{2}=x ^{2}+y ^{2}\]

Answer 5

@lolacole12 yes this is the whole question

Answer 6

Are there any you can eliminate just by looking at them? Remember that the domain is all whole numbers greater than zero.

Answer 7

I would just say, "which statement is always true?"

Answer 8

I eliminated b

Answer 9

well, you might want exclude 0, but, you jus have 3 false statements and one true for all values (unless z=0).

Answer 10

Good, b is wrong !!

Answer 11

It says in the problem that its domain is all positive integers, so 0 is not in the domain. Think about A if x and y were two different numbers. Would it ever be true?

Answer 12

If x and y were 2 different numbers and x is always positive then can y be a negative number? If y is a negative number then this equation would be false right?

Answer 13

Well remember that it says the domain (all the values the variables can take) will only be positive integers, so it can't be negative. Could it be true with two positive numbers though?

Answer 14

it could be true with positive numbers but it might or might not be true with y as a negative number?

Answer 15

What two positive numbers are you thinking of that make it true?

Answer 16

2 and 3

Answer 17

Are you sure?
\[\sqrt{2^{2}+3^{2}}=\sqrt{4+9}=\sqrt{13}\]\[2+3=5\]

Answer 18

I was thinking \[\sqrt{2^{2}+3^{2}} = 2+3\]

Answer 19

Unfortunately, that is not correct math. Think of it like the inside of the radical is in parentheses, you have to do PEMDAS, so do the parentheses first. Now do you think that A can ever be true?