All categories

3 years ago

How do I find the answer to 10=7-m

Mathematics

1 answer:

andrey2020 [161]3 years ago

3 0

Answer:

M=-3

Step-by-step explanation:

10=7-m

Subtract 7 from both sides

3=-m

Divide by negative

-3=m

You might be interested in

What two numbers have a product of -2 and a sum of 7

Citrus2011 [14]

<em>Greetings from Brasil...</em>

According to the statement of the question, we can assemble the following system of equation:

X · Y = - 2 i

X + Y = 7 ii

isolating X from i and replacing in ii:

X · Y = - 2

X = - 2/Y

X + Y = 7

(- 2/Y) + Y = 7 <em>multiplying everything by Y</em>

(- 2Y/Y) + Y·Y = 7·Y

- 2 + Y² = 7X <em> rearranging everything</em>

Y² - 7X - 2 = 0 <em>2nd degree equation</em>

Δ = b² - 4·a·c

Δ = (- 7)² - 4·1·(- 2)

Δ = 49 + 8

Δ = 57

X = (- b ± √Δ)/2a

X' = (- (- 7) ± √57)/2·1

X' = (7 + √57)/2

X' = (7 - √57)/2

So, the numbers are:

and

7 0

3 years ago

What is not a statistical question?

Aleksandr [31]

Answer:

Step-by-step explanation:

Because there is only one table and you are asking for only one answer

6 0

2 years ago

What is the surface area of the triangular prism? A. 13 in.2 B. 36 in.2 C. 78 in.2 D. 84 in.2

NISA [10]

Answer :36in2

example:
my math teacher told me

4 0

2 years ago

What is the product? 3x^5(2x^2+4+1)

const2013 [10]

Answer:

$6x^{7} +15x^{5}$

Step-by-step explanation:

$3x^{5}$ ( $2x^{2} +4+1$ )

Multiply $3x^{5}$ by each term in the parentheses.

$6x^{7}$ + $12x^{5} +3x^{5}$

Now, all you have to do is simplify.

$6x^{7} +15x^{5}$

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:

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

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

$\mathbf{Total = 302400}$