Can someone answer this please? Thanks

Can y’all plz help me ?

crimeas [40]

Answer:

x=-5

Step-by-step explanation

The first box has eight x's and six 1's which means that it would be written as 8x+6

The second box has four x's and fourteen -1's which means that it would be written as 4x-14

You set them equal to one another

8x+6=4x-14

You subtract 4x from both sides

4x+6=-14

Subtract 6 on both sides

4x=-20

Divide 4 on both sides

x=-5

Hope it helps :))

4 0

3 years ago

Answer this question please; number 5... show all work thank you

My name is Ann [436]

Answer:

rate of the plane in still air is 33 miles per hour and the rate of the wind is 11 miles per hour

Step-by-step explanation:

We will make a table of the trip there and back using the formula distance = rate x time

d = r x t

there

back

The distance there and back is 264 miles, so we can split that in half and put each half under d:

d = r x t

there 132

back 132

It tells us that the trip there is with the wind and the trip back is against the wind:

d = r x t

there 132 = (r + w)

back 132 = (r - w)

Finally, the trip there took 3 hours and the trip back took 6:

d = r * t

there 132 = (r + w) * 3

back 132 = (r - w) * 6

There's the table. Using the distance formula we have 2 equations that result from that info:

132 = 3(r + w) and 132 = 6(r - w)

We are looking to solve for both r and w. We have 2 equations with 2 unknowns, so we will solve the first equation for r, sub that value for r into the second equation to solve for w:

132 = 3r + 3w and

132 - 3w = 3r so

44 - w = r. Subbing that into the second equation:

132 = 6(44 - w) - 6w and

132 = 264 - 6w - 6w and

-132 = -12w so

w = 11

Subbing w in to solve for r:

132 = 3r + 3(11) and

132 = 3r + 33 so

99 = 3r and

r = 33

7 0

3 years ago

The first three terms of a sequence are given. Round to the nearest thousandth (if

Mumz [18]

Answer:

$\frac{3}{16}$

Step-by-step explanation:

The common ratio is $\frac{1}{2}$

24, 12, 6, 3, $\frac{3}{2}$ , $\frac{3}{4}$ , $\frac{3}{8}$ , $\frac{3}{16}$

so divide each term by 2

plz mark m brainliest. :)

6 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:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(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:

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

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

$\mathbf{Total = 302400}$