For what value of n belonging to N does the monomial 7/5 * x^n-2 * y^n-7 have a whole result?

A) n>3

B) n>7

C) n>4

D) n> or equal 7

E) 2<n<7

Hi everyone,

the answer for this question is D), i’m not sure how to solve it

For what value of n belonging to N does the monomial 7/5 * x^n-2 * y^n-7 have a whole result?

A) n>3

B) n>7

C) n>4

D) n> or equal 7

E) 2<n<7

Hi everyone,

the answer for this question is D), i’m not sure how to solve it

I would also like to know how to solve this one! If you find out, please let me know

Hi!

sorry i realize that i didn’t type the question correctly:

the equation is 7/5 * x^(n-2) * y^(n-7), (so n-2 is the exponent of x)

By whole result i think they mean that the monomial must be a positive value

To have this we cannot have a negative exponent

with n=8:

7/5 * x^(n-2) * y^(n-7) = 7/5 x⁶y¹ = 7/5 x⁶y

with n=7:

7/5 * x^(n-2) * y^(n-7) = 7/5 x⁵y⁰ = 7/5 x⁵

with any value lower than 7, one or both exponents will be negative and cannot be a solution

Do you agree with my reasoning?

Ohhhh I understand!! Yes I agree, it’s so logical and makes so much sense how you explained it! Thank you