Given :
A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj .
To Find :
A. The vector position of the particle at any time t .
B. The acceleration of the particle at any time t .
Solution :
A )
Position of vector v is given by :

B )
Acceleration a is given by :

Hence , this is the required solution .
There least common factor would be 5 since they are all divisible by 5.
The answer to the question
1) 5.1
2)7.9
3) 4
4)20
5)8.9
6) 4.2
7)5.2
8)15
I'am somewhat confident that some are right because iam not sure your suppose to round the nearst tenth
Answer:
7
n
Step-by-step explanation:
28
4n
divide by 4 on both sides
7
n