Find a polynomial of degree 3 with a real coefficient that satisfies the given condition. Zeros: -2, 1, 0; f(2)=32
1 answer:
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Consider two points in a two dimensional coordinate plane
As you know mid point of a line joining two points is the coordinate of the middle point of only that portion of line joining between two points.
[tex]\text { Mid point of } (x_{1},y_{1})\text { and } (x_{2},y_{2}) =(\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2})[/tex]
As we know distance formula gives the Distance between line joining two points in two dimensional coordinate system.It is always positive.
[tex]\text { Distance formula between two points } (x_{1},y_{1})\text { and } (x_{2},y_{2}) =\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
Distance formula gives the distance between two points, whereas Mid point formula gives coordinate of the middle point.
Answer:
-5 ≤ x
Step-by-step explanation:
2(x -3) -5x ≤ 9 . . . . . given
2x -6 -5x ≤ 9 . . . . . . eliminate parentheses
-3x -6 ≤ 9 . . . . . . . . combine like terms
-6 ≤ 9 +3x . . . . . . . add 3x
-15 ≤ 3x . . . . . . . . subtract 9
-5 ≤ x . . . . . . . . . divide by 3
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The point x=-5 is included in the solution set, so there is a sold dot at that point. The rest of the solution set is values of x that are greater than that value, so the number line is shaded to the right of that point. Your number line should show an arrow at the right end of the graph to indicate it extends that direction indefinitely. (My graphing program doesn't draw arrows easily.)
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<em>Additional comment</em>
Perhaps the most straightforward way to solve this would have been to add 6 instead of 3x. That would give ...
-3x ≤ 15
Then dividing by -3 would give x. That operation would reverse the inequality symbol (because the divisor is negative). Then you would have ...
x ≥ -5
Either way works. We judge it to be less confusing when we don't have to change the direction of the inequality symbol.
