Find the quotient of the quantity negative 15 times x to the 2nd power times y to the 6th power plus 50 times x to the 4th power times y to the 3rd power minus 20 times x to the 2nd power times y all over 5 times x to the 2nd power times y.

So you have: ((-15(x^2)(y^6))+(50(x^4)(y^3))+(20x^2y))/((5x^2)(y))

which simplifies to

(-3y^5)+(10(x^2)(y^2))+4

as there are no like terms and x and y have not been given, this is as far as it can be simplified.

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.