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What are the steps for using a compass and straightedge to construct a square? Drag and drop the steps in order from start to fi
nish.
1.Construct a line perpendicular to line m through point F. Label a point on this line as point G.
2.Use a straightedge to draw line m and label a point on the line as point F.
3.Without changing the compass width, place the compass point on point H and draw an arc in the interior of ∠HFK .
4.Use the straightedge to draw JH¯¯¯¯¯ and JK¯¯¯¯¯.
5.With the compass open to the desired side length of the square, place the compass point on point F and draw an arc on line m and an arc on FG←→ . Label the points of intersection as points H and K.
6.Keeping the same compass width, place the compass on point K and draw an arc in the interior of ∠HFK to intersect the previously drawn arc. Label the point of intersection as point J.
1.Construct a line perpendicular to line m through point F. Label a point on this line as point G.
2.Use a straightedge to draw line m and label a point on the line as point F.
3.Without changing the compass width, place the compass point on point H and draw an arc in the interior of ∠HFK .
4.Use the straightedge to draw JH¯¯¯¯¯ and JK¯¯¯¯¯.
5.With the compass open to the desired side length of the square, place the compass point on point F and draw an arc on line m and an arc on FG←→ . Label the points of intersection as points H and K.
6.Keeping the same compass width, place the compass on point K and draw an arc in the interior of ∠HFK to intersect the previously drawn arc. Label the point of intersection as point J.
2 answers:
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<h3>Answer:</h3>
- f(x) = -2x^3 +3x^2 +11x -6
- see attached
- an infinite number. Since the magnitude of the leading coefficient is not specified, it may be any negative number. (We have chosen the smallest magnitude integer that makes all coefficients be integers.)
1. When "a" is a root of a polynomial, (x -a) is a factor of it. For the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).
In order to make the leading coefficient be negative, we need to multiply this product by a negative number. Any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.
Then ...
... f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)
... = -(2x² +3x -2)(x -3)
... = -(2x³ -3x² -11x +6)
... f(x) = -2x³ +3x² +11x -6
2. A graph created by the Desmos on-line graphing calculator is shown, and the zeros are highlighted.
3. As indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. You want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.
Solution: We are given:
We need to find the test statistic in order to test the claim that the mean weight of all green candies is equal to 0.8535
We know that the test statistic is:
Therefore, the test statistic is
Hence the option -1.36 is correct.