Determine where the real zeros of f(x)= 2x^4 + x^2 - 3x + 3 are located a. between -1 & 0 c. between 1 & 2 b. between 0 & 1 d. no real zeros

The given polynomial is
f(x) = 2x⁴ + x² - 3x + 3
There are 2 changes of sign

From Descartes' Rule of signs, there are
(a) 2 real positive zeros, or
(b) a conjugate pair of 2 complex zeros.

f(-x) = 2x⁴ + x² + 3x + 3
There is no change in sign
There are no negative real zeros.

Let us test for real zeros in the given ranges:
  x     f(x)
-----  --------
   -1    9
-0.5   4.875
    0   3
 0.5   1.875
    1    3
 1.5    10.875
   2    33
There are no changes in the sign of f(x).
Therefore no real zeros exist in (-1,0), (1,2), (0,1).
The graph of the function confirms the conclusion.

d. No real zeros.

The equation is y=17(7)+600 and it equals 717

The answer is 270 cars.you can do this by making a proportional relationship. for instance, 200 is the whole we are using, so on a “number line” 200 would be over 100%. now, since he sells 35% more, add 35% to 100% to get 135% on the other side of the “number line” on the bottom half, leaving the top half empty.cross multiply 200 by 135 then divide by 100 to get your answer. i hope this , if it did, , if you don’t mind, giving me brainiest! 200/100% } 200 • 135 /100 ? /135%