Find a polynomial f(x) of degree 3 that has the following zeros. 0, 4, -2
2 answers:
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All the points that are 6 units from (-1, 1) are those on the circle
(x+1)^2 +(y-1)^2 = 36
For y=0, the two points of interest satisfy
(x+1)^2 +1 = 36
(x+1)^2 = 35 . . . . . . subtract 1
x+1 = ±√35
x = -1±√35
The points you seek are (-1-√35, 0) and (-1+√35, 0), about (-6.916, 0) and (4.916, 0).
(x+1)^2 +(y-1)^2 = 36
For y=0, the two points of interest satisfy
(x+1)^2 +1 = 36
(x+1)^2 = 35 . . . . . . subtract 1
x+1 = ±√35
x = -1±√35
The points you seek are (-1-√35, 0) and (-1+√35, 0), about (-6.916, 0) and (4.916, 0).
Answer:
Step-by-step explanation:
In every right triangle, a leg is given by the multiplication of the sine of its opposite angle and the hypothenuse.
In this case, the opposite angle of x is 15°, and the hypothenuse is 300. So, we have
