Answer:
1 second
Step-by-step explanation:
Using -b/2a I can get the maximum height of the ball. Assuming that there is no outside forces, there should be an equal time of going up and coming down, so if max height is at 1/2 seconds, then it should take 1 full second for the ball to hit the ground.
For b. They are all correct, you can justify because all pairs are different versions of the same angle
The angle bisector QS is constructed using arcs of the same width
intersecting above the segment joining equidistant point from <em>Q</em>.
<h3>Correct Response;</h3>
- <u>c. ∠AQS ≅ ∠BQS when AS = BS and AQ = BQ </u>
<h3>Reasons why the selected option is correct;</h3>
The steps to construct an angle bisector are as follows;
- Draw an arc from the vertex of the angle, <em>Q</em>, intersecting the rays forming the angle, QP and QR, at points A and B respectively.
- From points <em>A</em>, and <em>B</em>, draw arcs having same radii to intersect between the rays QP and QR at point <em>S</em>.
- Join the point of intersection of the small arcs at <em>S</em> to <em>Q</em> to bisect the angle PQR.
The reason why Ben uses the same width to draw arcs from <em>A</em> and <em>B</em> is as follows;
The point <em>A</em> and <em>B</em> are equidistant from point <em>Q</em>, therefore, point <em>Q</em> is point of intersection of arcs of radius AQ = BQ drawn from <em>A</em> and <em>B</em>.
Similarly point <em>S</em> is the point of intersection of arcs AS = BS from points <em>A</em> and <em>B</em>.
Which gives that the line QS is the perpendicular bisector of the segment AB, where ΔABQ is an isosceles triangle, therefore, QS bisects vertex angle ∠PQR.
Therefore, the correct option is the option c.;
- <u>c. ∠AQS ≅ ∠BQS when AS = BS and AQ = BQ</u>
Learn more about angle bisector here:
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