2(3x - 1) ≥ 4x - 6 Remove the brackets on the left.
6x - 2 ≥ 4x - 6 Subtract 4x from both sides.
6x - 4x - 2 ≥ - 6 combine like terms on the left.
2x - 2 ≥ - 6 Add 2 to both sides
2x ≥ - 6+ 2
2x ≥ -4 Divide by 2
x ≥ - 4/2
x ≥ - 2 answer <<<<<<<<
Check
2(3*-2 - 1) ≥ 4x - 6
2(-6 - 1) ≥ 4(-2) - 6
2(-7) ≥- 8 - 6
- 14 = - 14 and this checks.
Answer:
See explanation
Explanation:
From the formula;
0.693/t1/2 = 2.303/t log (Ao/A)
t1/2 = half life of Sodium-24
Ao = initial activity of Sodium-24
A= activity of Sodium-24 at time = t
So,
0.693/15 = 2.303/15 log (800/A)
0.0462 = 0.1535 log (800/A)
0.0462/0.1535 = log (800/A)
0.3 = log (800/A)
Antilog(0.3) = (800/A)
1.995 = (800/A)
A = 800/1.995
A = 401 Bq
ii) 0.693/15 = 2.303/30 log (800/A)
0.0462 = 0.0768 log (800/A)
0.0462/0.0768 = log (800/A)
0.6 = log (800/A)
Antilog (0.6) = (800/A)
3.98 = (800/A)
A = 800/3.98
A = 201 Bq
iii)
0.693/15 = 2.303/45 log (800/A)
0.0462 = 0.0512 log (800/A)
0.0462/0.0512 = log (800/A)
0.9 = log (800/A)
Antilog (0.9) = (800/A)
7.94 = (800/A)
A = 800/7.94
A= 100.8 Bq
iv)
0.693/15 = 2.303/60 log (800/A)
0.0462 = 0.038 log (800/A)
0.0462/0.038 = log (800/A)
1.216 = log (800/A)
Antilog(1.216) = (800/A)
16.44 = (800/A)
A = 800/16.44
A = 48.66 Bq
The maximum magnitude of the net force on the box is 20 N, which is only possible if the boy and the girl pull the box together in the same direction, horizontally and parallel to the ground.
The minimum magnitude of the net force on the box is 0 N, which will occur when the boy and the girl pull the box together in the parallel but opposite direction.
If either of them pulls at an angle from the horizontal, then the magnitude of the net force will be between 0 N and 20 N.
Answer:
303.9481875 N
Explanation:
t = Time taken = 2 seconds
F = Force
r = Radius = 1.5 m
I = Moment of Inertia
= Angular Acceleration
Torque
Angular velocity
Angular acceleration
The magnitude of the force to stop the merry-go-round is 303.9481875 N
