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