PR = √((5+2)^2 + (3+3)^2)
= √85
You do SQ and let me know what you got
i have a question in math
(( distance between two points ))
A quardilateral has vertices P(3,5), Q(-4,3), R(-3,-2), and S(5,-4). find the lengths of the diagonals,to the nearest tenth .
please in detales .. thank u
4 answers
it is easy thank u
i will do it
:)
QS = �ã((-4+5)^2 + (3+4)^2)
=�ã50
i will do it
:)
QS = �ã((-4+5)^2 + (3+4)^2)
=�ã50
hi
i think theere is something wrong in the answer
PR = �ã((5+2)^2 + (3+3)^2)
= �ã85
because the rule is
�ã((x2-x1)^2 + (y2-y1)^2)
i think theere is something wrong in the answer
PR = �ã((5+2)^2 + (3+3)^2)
= �ã85
because the rule is
�ã((x2-x1)^2 + (y2-y1)^2)
3892