Answer:
(a) Weight of a bolt = 12 grms and weight of a nut is 7 gms. Weight of one bolt and one nut = 12 + 7 = 19 gms
(b) 69 gms
(c) 209 gms
Step-by-step explanation:
This problem relates to solution of 2 unknowns using simultaneous equations.
Let B be the weight of a single bolt and N be the weight of a single nut
Since 8 bolts and 6 nuts weigh 138 grams we get one of the equations as
8B + 6N = 138 (1)
The second equation in the unknowns is
3B + 5N = 71 (2)
To solve, we eliminate one of the unknowns by making its coefficients the same and subtracting one from the other
Multiplying (1) by 5 ===> 40B + 30N = 690 (3)
Multiplying (2) by 6 ===> 18B + 30N = 426 (4)
(3) - (4) eliminates the N variables and yields
22B = 264 ==> B = 264/22 ==> B = 12
So the weight of a single bolt is 12 grams
We can find the weight of a single nut by substituting this value of B into any of the equations (1), (2), (3) or (4) and solving for N. Let's use equation (2)
3(12) + 5N = 71 ==> 36 + 5N = 71 ==> 5N = 71-36 = 35 ==> N= 7
So the weight of a single nut is 7 grams
Weight of one bolt and one nut is the sum of the above individual weights = 12 + 7 = 19 gms
To solve (b) and (c) we could set up two other equations and plug in values for B and N
(b) 4B + 3N = 4.12 + 3.7 = 69
However, an alternate way is to perceive that 4B + 3N is exactly half of 8B + 6N so the value of that must be 138/2 = 69
(c) If we add both equations (1) and (2), we get
11B + 11N = 138 + 71 = 209 which is the equation for the total weight of 11 bolts and 11 nuts
, that the ladder is making with the ground, we are going to use a trig function that relates the adjacent side of our angle with the hypotenuse. That trig function is Cosine. 
