Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

Choose the polynomial that is written in standard form. 2x2y2 + 3x4y + 10x6 4x4y2 + 6x3y5 + 10x −3x8y2 + 9x2y + 10x2 −7x6y2 + x3y8 + 10x2

OpenStudy (anonymous):

standard form has decreasing degrees of x and increasing degrees of y

OpenStudy (anonymous):

so the one that has that property is your answer

OpenStudy (anonymous):

What one do you think?

OpenStudy (anonymous):

I don't know how to look at the decreasing degrees of x and increasing degrees of y

OpenStudy (anonymous):

look at the exponents of x..if the exponents are decreasing for every term, then that has decreasing degree

OpenStudy (anonymous):

for example \(x^3 + x^2 + x + 1\)

OpenStudy (anonymous):

so that's decreasing

OpenStudy (anonymous):

is it B?

OpenStudy (anonymous):

if there is no y...that means the degree of y for that term is 0

OpenStudy (anonymous):

theres a y in every single one

OpenStudy (anonymous):

is this B? 4x4y2 + 6x3y5 + 10x feels like something is missing

OpenStudy (anonymous):

the degrees I cant put them on here

OpenStudy (anonymous):

http://tutorial.math.lamar.edu/Classes/Alg/Polynomials.aspx#Pre_Poly_Ex2_a search that page for polynomials in two variables

OpenStudy (anonymous):

i guess you ad xs exponent and y's exponet in each term then take that sum and put it in decreasing order.

OpenStudy (anonymous):

|dw:1372894843615:dw| −3x8y2 + 9x2y + 10x2 is right

OpenStudy (anonymous):

The answer would be C Jaderbrown. Polynomials go from highest degree to smallest. To find the degree, of one of its terms you would do the following.. −3x8y2 + 9x2y + 10x2 \( -3x^8,^2 \) = degree 10 8+ 2 = 10

OpenStudy (anonymous):

That is \( -3x^8y^2 \) = 10 degree because 8+2 = 10, which when times you add exponents. You got it Jaderbrown? Highest degree to lowest.

OpenStudy (anonymous):

yes I get it thank you :)

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!