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3 years ago

6

What is the first step in simplifying

1 answer:

seraphim [82]3 years ago

7 0

Distribute the negative b and negative one to he terms in the parentheses
In an expression, if there area brackets, you always have to expand THEN solve it

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Plz help me with this Im timed will mark as brainliest

viktelen [127]

Well what i got was 79.3 by finding the average by adding it all together then divided it by how many numbers

4 0

3 years ago

Read 2 more answers

Which equation represents the line that passes through (-6, 7) and (-3, 6)?

Degger [83]

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

$y = mx + b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following points through which the line passes:

$(x_ {1}, y_ {1}): (- 6,7)\\(x_ {2}, y_ {2}): (- 3,6)$

We find the slope of the line:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {6-7} {- 3 - (- 6)} = \frac {-1} {-3 + 6} = \frac {-1} {3} = -\frac {1} {3}$

Thus, the equation of the line is of the form:

$y = - \frac {1} {3} x + b$

We substitute one of the points and find b:

$6 = - \frac {1} {3} (- 3) + b\\6 = 1 + b\\b = 5$

Finally, the equation is:

$y = - \frac {1} {3} x + 5$

Answer:

$y = - \frac {1} {3} x + 5$

8 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 cooler has 12 apple juices and 15 grape juices. 9 of the apple juices are sugar free and 5 of the grape juices are sugar free.

antoniya [11.8K]

To me: given that the juice is sugar free means that I can ignore the 3 sugar apple juices and the 10 sugr grape juices.

Therefore the field of drinks is 14 sugar free drinks and the changes that it is apple juice, is 9:14 or 9/14.

If, on the other hand, the problem is saying that the juice must be sugar free and also apple, that would be 9/27 or 1/3

8 0

3 years ago

Read 2 more answers

How do you simplify this problem

vampirchik [111]

Reduce the fraction with 2
$\frac{2 \times x {}^{ - 2} y {}^{3}z {}^{ - 1} }{xy} \\ \\ \frac{3y {}^{2}z {}^{ - 1} }{x {}^{3} } \\ \\ \frac{3y {}^{2} }{x {}^{3} z}$

7 0

3 years ago