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Gala2k [10]
3 years ago
8

A car manufacturer builds trucks and cars both in red, blue, and green colors. Let B represent the set of all red vehicles produ

ced by the manufacturer and let C represent the set of all trucks produced by the manufacturer. If all of the vehicles below are produced by the manufacturer, which could be in the set (B∩C')?
Mathematics
1 answer:
telo118 [61]3 years ago
5 0

Answer:

Step-by-step explanation:

The universal set, U is a set of all vehicles produced by the manufacturer.

U = red, blue and green vehicles(both cars and trucks)

B = all red vehicles produced by the manufacturer(both cars and trucks)

C = red, blue and green trucks

C ' = all elements that are not in C but are in the universal set

C ' = red, blue and green cars

B intersection C ' = all elements that are common to set B and set C '. Therefore,

B intersection C ' = red cars

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Given that 8 bolts and 6 nuts weigh 138grams and 3 bolts and 5 nuts weigh 71 grams. Find the weight of;(a)one bolt and one nut (
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(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

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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

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