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2 years ago

Help with any ! If you know all I’ll mark as brainlist

Mathematics

2 answers:

aleksklad [387]2 years ago

5 0

15. The answer is D

16.The answer is 32

17. The answer is 13,248

labwork [276]2 years ago

4 0

15. 52 sqr inches
16. 32 sqr ft
17. I’m not sure since it is a rhombus but it could be 92 ft sqr if it is the same as the others

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Can someone help me please:

elena-14-01-66 [18.8K]

$819.20 is the answer considering each time the units increase by fours, it adds $204.80

8 0

3 years ago

Adam is building a rectangular planter without a top the planter will be 7 feet wide 5 feet wide, 5 feet long, and 9 inches high

Tom [10]

Answer: The surface area of the planter will be 120.5 square feet.

In our shape, we will have 5 sides. The base and the four sides going up from each edge of the base.

The base will be 7 x 5 = 35 square feet.

The front and back side will each be 7 x 0.75 = 5.25 square feet.

The left and right side will each be 5 x 0.75 = 3.75 square feet.

If we add up the 5 faces we get:
35 + 5.25 + 5.25 + 3.75 + 3.75 = 120.5 square feet

8 0

3 years ago

240,000 written as a whole number multiplied by the power of ten

Stolb23 [73]

This question can be also asked as what is the scientific notation of the number 240,000. The scientific notation is written as a whole number multiplied by the power of ten. It should be written as follows:

2.4 x 10^5

Hope this helps. Have a nice day.

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:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(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:

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

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

$\mathbf{Total = 302400}$