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3 years ago

8-5x =2x + 20 Simplify I’ll give brainliest

Mathematics

2 answers:

kobusy [5.1K]3 years ago

6 0

Answer:

X= - 12/7 i hope it is helpful have a nice day if you want steps i will show you

umka2103 [35]3 years ago

3 0

Answer/Step-by-step explanation:

Rearrange terms

8 − 5 = 2 + 2 0 − 5 + 8 = 2 + 2 0

Subtract 8 from both sides of the equation
− 5 + 8 = 2 + 2 0 − 5 + 8 − 8 = 2 + 2 0 − 8

Simplify
Subtract the numbers

− 5 = 2 + 1 2

Subtract 2 2x 2x from both sides of the equation
− 5 = 2 + 1 2 − 5 − 2 = 2 + 1 2 − 2x

Simplify
Combine like terms

− 7 = 1 2

Divide both sides of the equation by the same term

− 7 = 1 2 − 7 − 7 = 1 2 − 7

Simplify Cancel terms that are in both the numerator and denominator Divide the numbers

= − 1 2 7

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Find h(-5) for the function h(x)= -5x+9. h(-5) =

Helga [31]

Answer:

h(-5) = 34

Step-by-step explanation:

h(x) = -5x + 9

h(-5) = -5(-5) + 9 = 25 + 9 = 34

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3 years ago

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Answer: $7200

Step-by-step explanation:

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2 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:

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

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

$\mathbf{Total = 302400}$