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3 years ago

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6 plates and 5 cups cost 32.10 dollars. 7 plates and 6 cups cost 37.70 dollars. Each plate costs the same amount as the other pl

ates, and each cup costs the same amount as the other cups. What is the cost of 1 plate? What is the cost of 1 cup?

2 answers:

slavikrds [6]3 years ago

8 0

Answer:

Plates are 2.80 Dollars and cups are 1.40 dollars

Step-by-step explanation:

Cups = c

Plates = p

6p + 5c = 23.80 ==> multiply by 6 ==> 36p + 30c = 142.80

7p + 6c = 28.00 ==> multiply by 5 ==> 35p + 30c = 140.00

Now use algebra to have the 30c be on one side and the rest on the other:

36p + 30c = 142.80 | -36p

30c = 142.80 - 36p

35p + 30c = 140.00 | -35p

30c = 140.00 - 35p

Now set them equal to each other about the 30c:

142.80 - 36p = 140.00 - 35p

Use algebra to solve for p:

142.80 - 36p = 140.00 - 35p | +36p

142.80 = 140.00 + p | - 140

2.80 = p

Go back to one of the "30c" equations and plug in the value for p:

30c = 140 - 35p

30c = 140 - 35(2.80)

30c = 140 - 98

30c = 42 | /30

c = 42/30

c = 1.4

MrRa [10]3 years ago

3 0

Answer:

the cost of 1 plate $4.1

the cost of 1 cup $1.5

Step-by-step explanation:

6p + 5c = 32.10

7p + 6c= 37.70

Multiply the first by 6

Multiply the second by -5

and then ADD both

36p +30c = 192.6

-35p-30c = -188.5 (30c is cancelled)

p = 4.1

4.1(6) +5c = 32.10

24.6 +5c = 32.10

5c = 7.5

c= 1.5

1.5(5) =7.5

24.6 + 7.5 = 32.10

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The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
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When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

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$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

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First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

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$\mathbf{n = 10}$

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$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

<u>(b) The mathematics book, not together</u>

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

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$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

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We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

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We have:

$\mathbf{Mathematics = 2}$

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$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

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6 0

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