Question

Classify the following expression by degree and term:

x2y - 7xy + xyz + x
2nd degree polynomial
7th degree polynomial
2nd degree trinomial
3rd degree polynomial

Answer 1

I think B... Any suggestions? @arabpride , @pooja195 @Leader @BrooklynRoyals

Answer 2

2ed

Answer 3

So i was right?

Answer 4

No.

Answer 5

Can you properly post your question?

Answer 6

Oh... :'( What do you mean?

Answer 7

How can we help you if you don't take the time to make sure that your question came out correct.

Answer 8

Copy-paste is not affective I do not know whether your question has anything to the power in it.

Answer 9

effective*

Answer 10

Whats wrong with it? Im sorry....

Answer 11

x\(^2\)y - 7xy + xyz + x

Answer 12

Is there any powers in it?

Is it x^2y - 7xy + xyz = x

Answer 13

Because you will never see a coefficient of a term written out in the form `x2y`

Answer 14

xyz + x

Answer 15

I know @Jhannybean But you know if they want some help they should put some effort into the question :)

Answer 16

Oh... I didnt know sorry... :'(

Answer 17

Well copy-pasting changed the format.

Answer 18

I dont need your help anyways if it bothered you that much... @Leader

Answer 19

Well @Tootles143 it is good to know what the exact question is.

Answer 20

The only power that is in it is X^2y...

Answer 21

No problem.

But anyway, for your question, first off you know it's a polynomial because any function after the first exponent of 1 makes the function a polynomial.

Secondly, in order to classify what type of equation it is, you look at what is being added r subtracted, and how many terms are being multiplied together.

Answer 22

So itss....
X^2y-7xy+xyz+x... There you go.

Answer 23

i.e

binomial - adding or subtracting a monomial (1 term function)

trinomial - adding or subtracting 3 monomials, or a combination of a binomial and monomial,

polynomial - equations that include binomials (2 term functions) , monomials and trinomials.

Answer 24

I g2g... Sorry it is late where i am at... Help if youu want.... I will get into trouble if i stay any longer...

Answer 25

Use what @jhannybean gave and you have your answer :)

Answer 26

Okay... Thanks to you both! :) @Jhannybean and @Leader

Answer 27

Np :)