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3 years ago

FOR 100 POINTS!!!!!!!!!!!Solve the equation. 3(4x + 4) = 2(5x + 9) - 12

Mathematics

2 answers:

Tatiana [17]3 years ago

6 0

Answer:

Step-by-step explanation:

Isolate x on one side of the equation.

DISTRIBUTIVE PROPERTY

A(B+C)=AB+AC

A(B-C)=AB-AC

3(4x+4) (First, expand.)

3*4x=12x

3*4=12

12x+12

2(5x+9)-12

2*5x=10x

2*9=18

10x+18

18-12=6

12x+12=10x+6

12x+12-12=10x+6-12 (Subtract 12 from both sides.)

6-12 (Solve.)

12-6=6

12x=10x-6

12x-10x=10x-6-10x (Subtract 10x from both sides.)

12x-10x (Solve.)

12x-10x=2x

2x=-6

2x/2=-6/2 (Then, divide by 2 from both sides.)

-6/2 (Solve.)

-6/2=-3

x=-3

As a result, the final answer is x=-3.

Pepsi [2]3 years ago

6 0

Answer:

x = -3

Step-by-step explanation:

Step 1: Distribute

3(4x + 4) = 2(5x + 9) - 12

(3 * 4x) + (3 * 4) = (2 * 5x) + (2 * 9) - 12

12x + 12 = 10x + 18 - 12

Step 2: Combine like terms

12x + 12 = 10x + 6

Step 3: Subtract 10x from both sides

12x + 12 - 10x = 10x + 6 - 10x

2x + 12 = 6

Step 4: Subtract 12 from both sides

2x + 12 - 12 = 6 - 12

2x = -6

Step 5: Divide both sides by 2

2x / 2 = -6 / 2

x = -3

Answer: x = -3

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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:

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

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

$\mathbf{Total = 302400}$