will medal and fan!!! A base of a parallelogram is on the x-axis and the origin is located at the left endpoint of that base. Three consecutive vertices are (h, j), (0, 0), and (k, 0), where h > 0. How many units to the right of a vertical line through (k, 0) must the fourth vertex be? h 2h k + h

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will medal and fan!!! A base of a parallelogram is on the x-axis and the origin is located at the left endpoint of that base. Three consecutive vertices are (h, j), (0, 0), and (k, 0), where h > 0. How many units to the right of a vertical line through (k, 0) must the fourth vertex be? h 2h k + h

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The x-axis and y-axis are lines of symmetry for a square. If one of its vertices has coordinates ( a, b), then the vertice opposite it is ( a, -b) (- a, -b) (- a, b)
The base of that parallelogram is from (0,0) to (k,0)
ok..

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so what do i do now
(h, j ) is the upper left point from consecutive terms To get the upper right point, you stay at the same y-coordinate, and move right the same length as the base, k units
i still nit understanding
not
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would it be 2h? just asking
The base is k units long... So the top must also be k units long the y coordinate stays the same value
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so its H+k?
yep
yay!!
thank you!!!!!!!! what about the second one?
oh, wait
i think its A but im not to sure,,
it is h, sorry, they want to know the horizontal distance from (k , 0) to ( k + h , j )
ohhhhhhhhhhh okay
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:)
is the second one A? thats what i got
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oh .. i get it.........thx!!!!
welcome

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