<span> Exercise #1: Point H = (–2, 2) Point J = (–2, –3) Point K = (3, –3)
It would be very helpful if you could take a pencil and a piece of paper, and sketch a graph with these points on it. Then you'd immediately see what's going on.
Notice that points H and J have the same x-coordinate, but different y-coordinates, so they're on the same vertical line.
</span><span>Notice that points J and K have different x-coordinates but the same y-coordinate, so they're on the same horizontal line.
Notice that point-J is on both the horizontal line and the vertical line, so the lines meet there, and they're perpendicular. Point-J is one corner of the square.
H is another corner of the square. It's 5 units above J.
K is another corner of the square. It's 5 units to the right of J.
The fourth corner is (2, 3) ... 5 to the right of H, and 5 above K. ____________________________________
Exercise #2: </span><span>Point H = (6, 2) Point J = (–2, –4) Point K = (-2, y) .
</span><span>It would be very helpful if you could take a pencil and a piece of paper, and sketch a graph with these points on it. Then you'd immediately see what's going on.
</span><span>Notice that points J and K have the same x-coordinate, but different y-coordinates, so they're on the same vertical line.
We need K to connect to point-H in such a way that it's on the same horizontal line as H. Then the vertical and horizontal lines that meet at K will be perpendicular, and we'll have the right angle that we need there to make the right triangle. So K and H need to have the same y-coordinate. H is the point (6, 2). So K has to be up at (2, 2) . ____________________________________________
Exercise #3: </span> <span>Point H = (-6, 2) Point J = (–6, –1) Point K = (4, 2) . </span> <span>It would be very helpful if you could take a pencil and a piece of paper, and sketch a graph with these points on it. Then you'd immediately see what's going on.
This exercise is exactly the same as #1, except that it's a rectangle instead of a square. It's still make of horizontal and vertical lines, and that's all we need to know in order to solve it.</span><span>
Notice that points H and J have the same x-coordinate, but different y-coordinates, so they're on the same vertical line.
</span><span>Notice that points H and K have different x-coordinates but the same y-coordinate, so they're on the same horizontal line.
Notice that point-H is on both the horizontal line and the vertical line, so the lines meet there, and they're perpendicular. Point-H is one corner of the rectangle.
J is another corner of the rectangle. It's 3 units below H.
K is another corner of the square. It's 4 units to the right of H.
The fourth corner is (2, -1) ... 4 to the right of J, and 3 below K.
Angle A is 29 and Angle B is 61. It’s problem can be solved by written the equation X+X+32=90. This becomes 2X+32=90 subtract 32 from each side making 2X=58 divide by 2 X=29 and just plug x in to solve for both angle names.
anyway, without any further information about ground consistency, friction and stuff, we have to assume ideal circumstances.
so, the same energy that made the ball flying through the air makes it also bounce or roll across the ground carrying it exactly the same distance at the same time.
the distances with an ideal, reflective ground will be the same.
it is the same energy released with the same inertia.