14. trapezoid
- the bases are parallel
15. rhombus
- if you add 57 and 33 you will get 90 degrees. that means all the other angles the diagonals create intersecting each other are 90 degrees, and one of the properties of a rhombus is they have perpendicular diagonals.
16. kite
- two adjacent sides are congruent, and the two angles drawn are congruent
The problem statement gives a relation between the amount removed from one bag and the amount removed from the other. It asks for the amount remaining in each bag. Thus, there are several choices for variables in this problem, some choices resulting in more complicated equations than others.
Let's do it this way: let x represent the amount remaining in bag 1. Then the amount removed from bag 1 is (100-x). The amount remaining in bag 2 is 2x, so the amount removed from that bag is (100-2x). The problem statement tells us the relationship between amounts removed:
... (100 -x) = 3(100 -2x)
... 100 -x -3(100 -2x) = 0 . . . . . . subtract the right side
... 5x -200 = 0 . . . . . . . . . . . . . . eliminate parentheses and collect terms
... x -40 = 0 . . . . . . . . . . . . . . . . .divide by 5
... x = 40 . . . . . . . . . . . . . . . . . . . add 40
- 40 kg is left in the first bag
- 80 kg is left in the second bag
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<u>Check</u>
The amount removed from the first bag is 60 kg. The amount removed from the second is 20 kg. The amount removed from the first bag is 3 times the amount removed from the second bag, as described.
The sides of a rhombus are equal length, so you have
... 5x + 20 = 6x + 10
... 10 = x . . . . . . . . subtract 5x+10
Then the side lengths are
... 5·10 +10 = 70
