All categories

2 years ago

Which expression represents “12 less than the product of n and 5”?

Mathematics

1 answer:

pogonyaev2 years ago

8 0

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

$\blue{\textsf{\textbf{\underline{\underline{Question:-}}}}$

Which expression represents "12 less than the product of n and 5"?

$\blue{\textsf{\textbf{\underline{\underline{Answer:-}}}}$

12-5n

$\blue{\textsf{\textbf{\underline{\underline{How\:to\:Solve:-}}}}$

First of all, the "product" indicates that you multiply these numbers.

Here we are asked to multiply a number n times 5. This can be written as follows:

$\red{\textsf{\textbf{5n}}}$

Now, "12 less" indicates that we subtract 12 from something; in this case, this "something" is 5n:

$\red{\textsf{\textbf{\underline{\underline{12-5n}}}}$

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

You might be interested in

Use the distributive property to simplify the expression 13(n+4+7m)

bulgar [2K]

91m+13n+52 ....
hope this helps!!!

6 0

2 years ago

Read 2 more answers

Consider the graph.

Arlecino [84]

Answer:

C, f(x) = 2x + 6

Step-by-step explanation:

First, we need to plug in the values of the x coordinates and see if it matches with the y coordinate to determine if it is on the same line. Startin with 2x + 8, we have the point (1, 8) on the graph. Plugging in 1 gets you 10 for the y. This is wrong since 8 is the y coordinate. Moving on, we have 6.4(1.25)^x for the same point. Plugging in 1, we have 6.4 * 1.25 = 8, which is true. Moving on to the second point, (2, 10), we have 1.25 squared times 6.4. This is thus wrong. So, moving on to 2x + 6, we have the point (1, 8), and plugging in 1 for x, we have 8 as y. Since this satisfies the equation we move on to the next point, (2,10). Plugging in x, we have 2 * 2 + 6 = 10, which is also true. Moving on to our third point (3 , 12), we plug in 3 for x. We then get 3 * 2 + 6 = 12, which is correct. This, is our answer then.

5 0

3 years ago

Read 2 more answers

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:

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

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

$\mathbf{Total = 302400}$