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3 years ago

The following patterns are made using small squares

Mathematics

1 answer:

Alex787 [66]3 years ago

5 0

Answer:

stuck on that too

Step-by-step explanation:

I use mathswatch aswell

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Fluffy the cat is 3 years older than Scuffy the dog. Six less than five times Scruffy's age is one more than three times Fluffy'

lesya692 [45]

Answer: fluffy is 11 years old

Scuffy is 8 years old

Step-by-step explanation:

Let x represent the present age of Fluffy the cat.

Let y represent the present age of Scuffy the dog.

Fluffy the cat is 3 years older than Scuffy the dog. This means that

x = y + 3 - - - - - -- - - - - 1

Six less than five times Scruffy's age is one more than three times Fluffy's age. This means that

5y - 6 = 3x + 1

5y - 3x = 1 + 6

5y - 3x = 7 - - - - - - - - - -2

Substituting equation 1 into equation 2, it becomes

5y - 3(y + 3) = 7

5y - 3y - 9 = 7

5y - 3y = 7 + 9

2y = 16

y = 16/2 = 8

x = y + 3

x = 8 + 3 = 11

6 0

3 years ago

What is the slope of the line?

Art [367]

Answer: B. 1/5

Step-by-step explanation:

4 0

2 years ago

Strong positive strong negative weak positive weak negative

Nesterboy [21]

If you're looking for the trend name, I would say Strong Positive, or (a).

6 0

3 years ago

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

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
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

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

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

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:

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

This gives

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

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

(c) The novels must be together and the chemistry books, together

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:

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

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$