Answer:
Yes.
Step-by-step explanation:
Yes, because we can replace x^3 by another variable, say y:
x^9 - 5x^3 + 6 = 0
Let x^3 = y, then we have x^9 = (x^3)^2 = y^2:
y^2 - 5y + 6 = 0 which is quadratic from.
The equation is not quadratic in form.
The variable part of the first term is not the square of the variable part of the
second term.
The result of squaring x cubed is x to the 6th power, not the 9th power.
x9(not equal to) (x^3)^2
Answer: 1/6 of the total sheet cake
Step-by-step explanation: 1/2 divided by 3 = 1/6
the answer is x<0
because
x-4+4<-4+4
