A very small dinosaur , the microraptor was only 1.3 feet long. one of the largest dinosaures the diplodocus was about 91 feet long. how many times as long as the microraptor was the diplodocus ? =78 feet
I think the answer is B. 150 i may be wrong though
Answer:
d = 90°
e = 41°
f = 139°
Step-by-step explanation:
d + 90° = 180° (Angles in linear pair)
-> d = 180° - 90°
-> d = 90°
e = 41° (vertical angles)
f + 41° = 180° (Angles in linear pair)
-> f = 180° - 41°
-> f = 139°
Answer:
Let x rep the lenght of the shorter one
Then the longer one is 2x+1
Therefore
(2x+1) + x = 16
We now solve for x
2x + 1 + x = 16
Group and evaluate like terms
3x +1 = 16
3x = 16 -1
× = 15/3
x = 5
So the shorter one is 5 ft
The longer one is 2(5)+1= 11
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