Answer:
-1, 4+i, 4-i
Step-by-step explanation:
x^4- 6x^3 + 2x^2 + 26x + 17
Using the rational root theorem
we see if 1, -1, -17 or 17 are roots
Check and see if 1 is a root
1^4- 6(1^3) + 2(1^2) + 26(1) + 17=0
1-6+2+26+17 does not equal 0 1 is not a root
-1
1^4- 6(-1^3) + 2(1^2) + 26(-1) + 17=0
1 +6 +2 -26+17 = 0
-1 is a root
Factor out (x+1)
(x+1) ( x^3-7x^2+9x+17)
Using the rational root theorem again on x^3-7x^2+9x+17
Checking -1
-1 -7 -9 +17=0
-1 is a root
(x+1) (x+1) (x^2-8x+17)
Using the quadratic on the last
8 ±sqrt(8^2 - 4(1)17)
--------------------------------
2
gives imaginary roots
4±i
Answer:
the probably could be 0.8 at most since there are other colours
Answer:
This is a weird one but d=443/24
Step-by-step explanation:
0.22 because one side is 11 and the other is 11 so 11+11 is 22 the pursent of 22 is 0.22
Answer:
Step-by-step explanation:
If BOTH equations are in slope-intercept form then the-graphing-? method would be best, but the-substitution-? method would also be effective since both y's are already by itself.
If ONE of the equations is solved for x or y and the other equation is not, then the-substitution-? method is best.
If BOTH equations are lined up in standard form & the coefficients of x or y are opposites then the BEST method is definitely the-elimination--? method.
If BOTH equations are lined up in standard form the elimination method would be best. But if the coefficient of x or y is 1, then the-substitution--? method is also effective.
