If the 1500-lblb boom ABAB, the 220-lblb cage BCDBCD, and the 169-lblb man have centers of gravity located at points G1G1, G2G2 and G3G3, respectively, determine the resultant moment produced by all the weights about point A.

 (M)_a = -8171 lb-ft

Step-by-step explanation:

Step 1:

- We will first mark each weight from left most to right most.

               Point:                      Weight:                             Moment arm r_a

               G_3                          W_g3 = 169                     30*Cos(75) + 4.25

               G_2                          W_g2 = 220                    30*Cos(75) + 2.5

               G_1                           W_g1 = 1500                    10*cos(75)

Step 2:

- Set up a sum of moments about pivot point A, the expression would be as follows:

(M)_a = -W_g3*(30cos(75) + 4.25) - W_g2*(30*Cos(75) + 2.5) - W_g1*10*cos(75)

Step 3:

- Plug in the values and solve for (M)_b, as follows:

    (M)_a = -169*(30cos(75) + 4.25) - 220*(30*Cos(75) + 2.5) - 1500*10*cos(75)

    (M)_a = -2030.462559 -2258.205698 - 3882.285677

    (M)_a = -8171 lb-ft

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

dont know

step-by-step explanation: