The answer is 1/3 of 12 its just easy math for me
Answer:
(a) 218.6 N
(b) 97.14 N
Step-by-step explanation:
When the system is in equilibrium, the net torque on the system is zero.
AC = 1.5 m, CD = 2.3 m, DB = 5 - 1.5 - 2.3 = 1.2 m
Let the centre of gravity of plank is at G.
AG = 2.5 m, CG = 2.5 - 1.5 = 1 m, GB = 2.5 m
(a) Let the reaction at C is R and at D is R'.
R + R' = 29 x 9.8 = 284.2 N ... (1)
Take the torque about C.
29 x 9.8 x CG = R' x GD
29 x 9.8 x 1 = R' x 1.3
R' = 218.6 N
(b) Take the torque about D.
6 x 9.8 x AD = R x CD
6 x 9.8 x (1.5 + 2.3) = R x 2.3
R = 97.14 N
4n/4n-4 × n-1/n+1
= 4n(n-1)/(4n-4)(n+1)
= (4n^2 - 4n)/(4n^2 + 4n - 4n-4)
= (4n^2 - 4n)/(4n^2 - 4)
= 4(n^2 - n)/4(n^2 - 1)
=(n^2 - n)/(n^2 - 1)
g(x) = x - 3 = 0
g(x) = x = 3
f(x) = 2x^3 + x - 4
f(3) = 2(3)^3 + 3 - 4
f(3) = 2(27) - 1
f(3) = 54 - 1
f(3) = 53
The remainder when f(x) is divided by x - 3 is <u>53</u>.
5 (the sides) x 0.8 = 4 meters
