Answer:
someone solve it please usuck
Step-by-step explanation:
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:
b because 23.59+ (-16.75) = 6.84
Step-by-step explanation:
(x - xc)² + (y - yc)² = r²
Where xc and yc are center coordinates
(x - (-4))² + (y - 3)² = 2²

x² + 2.x.4 + 4² + y² - 2.y.3 + 3² = 4
x² + 8x + 16 + y² - 6y + 9 = 4
x² + y² + 8x - 6y + 25 - 4 = 0
x² + y² + 8x - 6y + 21 = 0
Answer:
y=15
Step-by-step explanation:
The formula for direct variation is
y = kx
5 =k*2
Solve for k
Divide by 2
5/2 = k
y = 5/2 x
Now we substitute x=6
y = 5/2(6)
y = 15