Answer:
4
Step-by-step explanation:
two negative numbers multiplied together makes a positive number. so there could not be 1,3, or 5 negative numbers, but there could be 0,2, or 4 negative numbers.
Elimination
A, they are both linear
linear equations are sometimes given by y=mx+b, equation A, if graphed, produces a hyperbola which is not linear so statement A is false
B they areboth non-linear
equationb is linear so this is false
C equation a is non-linear and equation b is linear
true
C is the answer
now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to
undefined.
now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.
![\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\ -------------------------------\\\\ \cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}](https://tex.z-dn.net/?f=%5Cbf%202-x%5E%7B12%7D%3D0%5Cimplies%202%3Dx%5E%7B12%7D%5Cimplies%20%5Cpm%5Csqrt%5B12%5D%7B2%7D%3Dx%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7Bx%5E2-9%7D%7B2-x%5E%7B12%7D%7D%5Cqquad%20%5Cboxed%7Bx%3D%5Cpm%20%5Csqrt%5B12%5D%7B2%7D%7D%5Cqquad%20%5Ccfrac%7Bx%5E2-9%7D%7B2-%28%5Cpm%5Csqrt%5B12%5D%7B2%7D%29%5E%7B12%7D%7D%5Cimplies%20%5Ccfrac%7Bx%5E2-9%7D%7B2-%5Cboxed%7B2%7D%7D%5Cimplies%20%5Cstackrel%7Bund%20efined%7D%7B%5Ccfrac%7Bx%5E2-9%7D%7B0%7D%7D)
so, the domain is all real numbers EXCEPT that one.
There is no quotient: if x is equal to zero, then your problem would simplify to a number divided by zero. As it is impossible to divide by zero, there is no answer.
