Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

giving medals. please help>

OpenStudy (anonymous):

Question 1 (Multiple Choice Worth 1 points) (07.01) Classify the following expression by degree and term: x3y + 5xyz 3rd degree trinomial 4th degree binomial 3rd degree binomial 5th degree binomial Question 2 (Multiple Choice Worth 1 points) (07.01) Which of the following shows 2x3y - 3x2 + 7y2x - 12y written in standard form? 2x3y + 7y2x - 3x2 - 12y -12y + 7y2x - 3x2 + 2x3y 7y2x - 12y + 2x3y - 3x2 -12y - 3x2 + 7y2x + 2x3y Question 3 (Multiple Choice Worth 2 points) (07.01) A company distributes free boxes of paperclips to all the workers in x offices. Each office has 8 more departments than the number of offices. The number of workers in each department is 4 less than the number of departments in each office. Each worker is given 3 boxes of paperclips. Which expression shows the number of departments in each office? x x + 8 x + 4 3x(x + 8)(x + 4) Question 4 (Multiple Choice Worth 1 points) (07.01) Which statement best demonstrates why the following is a non-example of a polynomial? 33 y squared all over x squared - 62y2xz - 35z2y2 The expression has a variable raised to a negative exponent. The expression has a negative coefficient. The expression has a variable raised to a fraction. The expression has a variable in the denominator of a fraction.

OpenStudy (anonymous):

1) Well, first we need to find the degree of the polynomial. \[x^3y = x^3y^1\] 3+1 = 4 The degree of the polynomial is 4. How many terms are there? 2, so it is a binomial.

OpenStudy (anonymous):

Wait- was that x to the third power times y or x times 3y? If it was the latter the answer above is wrong.

OpenStudy (anonymous):

@dominiquebee72, is @MustachioedPenguin's answer sufficient?

OpenStudy (anonymous):

Alright, let's do this thing, then. For y(x^3) + 5xyz, it looks like the first term; y(x^3); is the one with the greatest exponent sum. That is, I'm not assuming that 5 is an exponent here. If the 3 in y(x^3) is an exponent, then this whole problem: y(x^3) + 5xyz; is what you'd call a 4th-degree binomial. (That's what @MustachioedPenguin was getting at.) 'Bi' means 2, and we have 2 terms in this problem: y(x^3), and 5xyz. We say this is a 4th-degree binomial because when you have (x^3 + y^1) you get a degree measure of 4 when you sum up these exponents; 3+1 = 4. With the second question, don't lose heart! Just add up all the like-terms you can. So the original problem is: 2x3y - 3x2 + 7y2x - 12y. Are there any terms in here that we can add together? Don't look like it t'me. So 2x3y + 7y2x - 3x2 - 12y would have to be the answer. I'll post this right now and you can dwell on all of this for a bit while I finish up your problems, 'kay?

OpenStudy (anonymous):

For question 3, I can only see one possible conclusion. After reading the problem carefully, I noted that "Each office has 8 more departments than the number of offices." The number of offices is a given: 'x' represents that. To add 8 to the number of offices would then represent the total number of departments that each office has. Therefore the answer should be x+8. BTW, to help yourself visualize this stuff a little better, substitute easy values like 10 into variables like 'x'. It will look a lot clearer then.

OpenStudy (anonymous):

For the last question, I'm assuming that "all over" means "divided by". In that case, we've got a serious issue. Remember, it's abso-friggin'-lutely impossible to have a polynomial term with a variable in the denominator! So I would say that this problem can't be a polynomial because: The expression has a variable in the denominator of a fraction.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!