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.
It either 7|x^3|y^4 or 7x^3|y^4| I'm pretty sure it's the first one right?
Work out what 4th root 2401 is and the rest is probably more straightforward.
Well, remember that \(|y^4| = y^4\) since anything to an even power is going to be positive.
This isn't the case for \(|x^3|\) which is why the absolute value bars must stay.
@Trillography Given what I said, you can figure it out with certainty, right?
Dude, both of my answers that I think are right have A.V. bars ...
Was going to say 3 or 4 but now I've got it.
Cheers wio. ;D
-x * -x * -x does not equal -x * -x Good luck. BTW are we answering your monday maths homework? You've asked a few questions.
No, your definitely not. I've already finished all my hw. I'm working on a p. exam. I've asked more than a few and about 80% of them didn't get answers.
Ok 'x' flips its sign, potentially, if it starts out negative. As does the 'y' term. However, the 'y' term is multiplied an even number of times - whereas the 'x' term is multiplied an odd number of times.
So: -1*-1=1 -1*-1*-1=-1 -1*-1-*1-*1=1 So by using the bars i.e. the absolute value we get answer (c.) that is 1 from zero. Be it +1 or -1