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: 88º
Step-by-step explanation:
180-47=133-45=88
Answer:
D.
Step-by-step explanation:
The area is the product of 2 adjacent sides:
Area = (5x - 3y)(5x - 3y) as the sides of a square are all equal.
= 5x(5x - 3y) - 3y(5x - 3y)
= 25x^2 - 15xy - 15xy + 9y^2
= 25x^2 - 30xy + 9y^2.
Let the two numbers be x and y, then,
xy = -12 . . . (1)
x + y = -10 . . . (2)
From (2), x = -10 - y . . . (3)
Putting (3) into (1), gives
(-10 - y)y = -12
-10y - y^2 = -12
y^2 + 10y - 12 = 0
Therefore, the two numbers are
and
