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 .
Answer: Variant C
Step-by-step explanation:
f(x) = 3x^2 − 4
For finding f(-2) just replace x with - 2 like this:
f(-2)=3*((-2))^2-4=3*4-4=12-4=8
Answer:
2:3 , you divide by common denominator
i guess as a decimal it might be
.50 to .75 but dont take my word for it
Step-by-step explanation:
10 into 5 is 2
15 into 5 is 3
Answer:
108000000x^2
Step-by-step explanation: