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rodikova [14]
3 years ago
14

Find the product. Simplify your answer.

Mathematics
2 answers:
statuscvo [17]3 years ago
7 0

Answer:

1/5

Step-by-step explanation:

Factor x² - 5x - 6

x² - 5x - 6 = (x + 1)(x - 6)

\frac{x-6}{(x+1)(x-6)}

Cancel the common factor.

\frac{1}{x+1}

\frac{1}{x+1} · \frac{x+1}{5}

\frac{1*(x+1)}{5*(x+1)}

Cancel the common factor.

1/5

tangare [24]3 years ago
3 0

Answer:

1/5

Step-by-step explanation:

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Its 36 because 6x2 is 12 and 12 x 3 equals 36
5 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}

<u />

<u>(a) The books in any order</u>

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}

<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}

This gives

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

\mathbf{Maths\ not\ together = 2903040}

<u>(c) The novels must be together and the chemistry books, together</u>

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}

<u>(d) The mathematics must be together and the chemistry books, not together</u>

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}

Read more about permutations at:

brainly.com/question/1216161

8 0
2 years ago
A used car is on sale for $3,600. Eric offered the owner of the car 4/5 of the asking price. How much was Eric's offer?
Andrei [34K]
2880. What you do is you set this problem up as a proportion. So you do 4/5= x/3600. You then cross multiply and divide.
8 0
4 years ago
Read 2 more answers
Which property was used to simplify the expression?<br><br> (y^5)^2 = y^10
motikmotik
<span>(y^5)^2 = y^10

This is power of a power property</span>
8 0
3 years ago
Select two ratios that are equivalent to 7:10. Choose 2 answers: (Choice A) 21:30, (Choice B) 17:20, (Choice C) 10:13, (Choice D
snow_lady [41]

Answer:

A and E

Step-by-step explanation:

3 0
3 years ago
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