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Nezavi [6.7K]
3 years ago
11

A puzzle contains a triangular piece with a base of 3 in. and a height of 4 in. A manufacturer wants to make 80 puzzles. Find th

e amount of wood used if each puzzle contains 20 triangular pieces.
Mathematics
2 answers:
kari74 [83]3 years ago
8 0
Since the base is 3 inches and the height is 4 inches, as well as the area of a triangle being base*height/2, we get 3*4/2=6 in. for each triangle. Since we need 20 pieces for each puzzle, we multiply 6 by 20 to get 120. In addition, since 80 puzzles are needed, with 20 pieces for each puzzle, we have 120 inches for each puzzle and 120*80 = 9600 square inches of wood used
Mashcka [7]3 years ago
4 0

Answer:

9600 square inches.

Step-by-step explanation:

We have been given that a puzzle contains a triangular piece with a base of 3 in. and a height of 4 in.

The would used for each triangular piece would be equal to area of triangle.

\frac{1}{2}\times 3\text{ in}\times 4\text{ in}=3\text{ in}\times 2\text{ in}=6\text{ in}^2

Each puzzle contains 20 triangular pieces, so wood used for each piece of puzzle would be 20 times 6 square inches.

\text{Wood used for each puzzle}=20\times 6\text{ in}^2

\text{Wood used for each puzzle}=120\text{ in}^2

\text{Wood used for 80 puzzles}=80\times 120\text{ in}^2

\text{Wood used for 80 puzzles}=9600\text{ in}^2

Therefore, 9600 square inches of wood is needed to make 80 puzzles.

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<h3>Answer:  10</h3>

===========================================================

Explanation:

Even though your teacher doesn't want you to list the items of the set, it helps to do so.

We'll be working with these two sets

A = {b, d, f, h, j, I, n, p, r, t}

C =  {d, h, I, p, t}

When we union them together, we combine the two sets together. Think of it like throwing all the letters in one bin rather than two bins.

A u C = {b, d, f, h, j, I, n, p, r, t,   d, h, I, p, t  }

The stuff that isn't bolded is set A, while the stuff that is bolded is set C

After we toss out the duplicates, we end up with this

A u C = {b, d, f, h, j, I, n, p, r, t}

But wait, that's just set A. Notice how everything in set C can be found in set A. This indicates set C is a subset of set A.

That's why all of the stuff in bold was tossed out (because they were duplicates of stuff already mentioned).

Once we determine what set A u C looks like, we count out the number of items in that set to determine the final answer.

There are 10 items in {b, d, f, h, j, I, n, p, r, t} which means 10 is the final answer.

----------------------------

An alternative method is to use the formula below

n(A u C) = n(A) + n(C) - n(A and C)

n(A u C) = 10 + 5 - 5

n(A u C) = 10

The notation n(A and C) counts how many items are found in both sets A and C at the same time. But as mentioned earlier, this is identical to just counting how many items are in set C. So we'll have n(C) cancel out with itself.

In short, n(A u C) = n(A) = 10

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