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irina [24]
2 years ago
14

A library buys 36 English books, 48 Science books and 72 Mathematics books. The thickness of each book is the same. Now, the lib

rarian wants the books to be placed in stacks, such that each stack has books of the same subject, and the height of each stack is the same. Also, the librarian wants as few stacks as possible. How many stacks of books will there be?
Mathematics
2 answers:
Shtirlitz [24]2 years ago
7 0

Answer:

12

Step-by-step explanation:

First, find the HCF/GCF of 36, 48, and 72. The HCF is 12. So, there will be 12 stacks of books. This is a simple example of HCF. Please mark me as brainliest. Thanks

fomenos2 years ago
5 0

Answer:

13

Step-by-step explanation:

The GCF of 36, 48, and 72 is 12 so there will be 36 / 12 = 3 stacks of English books, 48 / 12 = 4 stacks of science books and 72 / 12 = 6 stacks of math books for a total of 3 + 4 + 6 = 13 stacks.

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Joanne has a goal of buying a townhome and saved for 5 years. She expected to have the down payment of $15,000 saved by now but
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Answer:


Step-by-step explanation:

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Louis is climbing steps to the top of a monument. After climbing 15 steps, Louis stops to tie his shoe. If there is a total of 7
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4 0
2 years ago
19. Bella is putting down patches of sod
Fofino [41]

Answer:

The dimensions of the two different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

The perimeter of the two different rectangular regions are;

1st Arrangement:

P₁ = 18 yards

2nd Arrangement:

P₂ = 24 yards

Step-by-step explanation:

Bella is putting down patches of sod to start a new lawn.

She has 20 square yards of sod.

We are asked to provide the dimensions of two different rectangular regions that she can cover with the sod.

Recall that a rectangle has an area given by

Area = W*L

Where W is the width of the rectangle and and L is the length of the rectangle.

Since Bella has 20 square yards of sod,

20 = W*L

There are more than two such possible rectangular arrangements.

Out of them, two different possible arrangements are;

1st Arrangement:

20 = (4)*(5) = (5)*(4)

Width is 4 yards and length is 5 yards or width is 5 yards and length is 4 yards

2nd Arrangement:

20 = (2)*(10) = (10)*(2)

Width is 2 yards and length is 10 yards or width is 10 yards and length is 2 yards

Therefore, the dimensions of two  different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

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P₁ = 2(4 + 5)

P₁ = 2(9)

P₁ = 18 yards

The perimeter of the 2nd arrangement is

P₂ = 2(2 + 10)

P₂ = 2(12)

P₂ = 24 yards

So the perimeter of the 1st arrangement is 18 yards and the perimeter of the 2nd arrangement is 24 yards.

Note:

Another possible arrangement is,

20 = (1)*(20) = (20)*(1)

Width is 1 yard and length is 20 yards or width is 20 yards and length is 1 yard.

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