anonymous
  • anonymous
PLEASE HELP!!!! Suppose X and Y represent two different school populations where X > Y and X and Y must be greater than 0. Which of the following expressions is the largest? Explain why. X^2 + Y^2 X^2 − Y^2 2(X + Y) (X + Y)^2
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@pooja195 @goformit100
anonymous
  • anonymous
may i help u
anonymous
  • anonymous
yes please

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
ok the answer is the second equation
anonymous
  • anonymous
can you explain how you did that?
anonymous
  • anonymous
cuz like the equation x=y it is going to equal 2 always that is the rule got it
anonymous
  • anonymous
wait, what?
anonymous
  • anonymous
uggh the answer is 2nd one cuz of how the equation is set up
anonymous
  • anonymous
@tom982 ?
anonymous
  • anonymous
Firstly, begin by expanding the parenthesis: \[x^2+y^2\]\[x^2-y^2\]\[2(x+y)=2x+2y\]\[(x+y)^2=x^2+2xy+y^2\] Let's consider the first one. It's clearly bigger than the second one as y^2 is positive so we can rule out the second one. Now we have:\[x^2+y^2\]\[2x+2y\]\[(x+y)^2=x^2+2xy+y^2\]Compare the first one with the second one, we can see that the first one is only bigger when x or y is greater than 2 - we can't really tell much from this, so let's move on. Now consider the last one, it's clearly bigger than the first one as it has a 2xy as well, which is positive, so we can rule out the first. Leaving us with:\[2(x+y)\]\[(x+y)^2=x^2+2xy+y^2\]The last bit is a bit tricky as we need to compare 2(x+y) with (x+y)^2. Break this down into two cases and test them:\[2(x+y) \lt (x+y)^2\]\[2<(x+y)\]So we can see that (x+y)^2 is greater than 2(x+y) when (x+y)>2. Next case: \[2(x+y) \gt (x+y)^2\]\[2 \gt x+y\]So 2(x+y) is greater than (x+y)^2 only when (x+y) is less than 2 - this is to say that either x=0, y=1 or x=1, y=0 which can't be the case as the question says that the population is non-zero. Hence (x+y)^2 is the largest :)
anonymous
  • anonymous
thank you so much @tom982
anonymous
  • anonymous
No problem, hope it made sense!

Looking for something else?

Not the answer you are looking for? Search for more explanations.