All categories

3 years ago

Hellllppppppppppp meeeee plzzzzzzz!!!!!!!!!!!!!’

Mathematics

2 answers:

adell [148]3 years ago

6 0

Answer:

Step-by-step explanation:

The third answer expresses one or more themes

MAXImum [283]3 years ago

6 0

Answer:

B I believe and if thats not correct C

Step-by-step explanation:

You might be interested in

HELP!! <br> 9. What is the area of triangle AOB?

Akimi4 [234]

Angles ∠ACD and ∠CAB are congruent because they are alternate angles. Then the area of the triangle AOB will be 22.53 square cm.

<h3>What is the area of the right-angle triangle?</h3>

The area of the right-angle triangle is given as

A = 1/2 x B x H

Where B is the base and H is the height of the right triangle.

We know that angles ∠ACD and ∠CAB are congruent because they are alternate angles.

α₁ = 40°

AO = OC = 7.8 cm

Then the area of the triangle will be

Area = 1/2 x 7.8 x 7.8 x tan40°

Area = 22.53 square cm

More about the area of the right-angle triangle link is given below.

brainly.com/question/16653962

#SPJ1

5 0

1 year ago

12 Two plumbers received a job. At first, one of the plumbers worked alone for 1 hour, and then they worked together for the nex

Degger [83]

The answer should be 12

4 0

2 years ago

Help me pleasee<br> Questions r in picture

jok3333 [9.3K]

<h2>Answer:</h2>

$\large \bf\implies\frac{12}{61}$

<h2>Step-by-step explanation:</h2>

<h2>Given :</h2>

$\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...$

nth term

<h2>Solution :</h2>

We have to add 1 in numerator and -10 in denominator because

$\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...[Given]$

$\frac{8 \: + \: 1}{101 \: - \: 10} = \frac{9}{91} \\\\ \frac{9 + 1}{91 - 10} = \frac{10}{81} \\ \\ \frac{10 + 1}{81 - 10} = \frac{11}{71} \\ \\ \frac{11 + 1}{71 - 10} = \frac{12}{61} ...$

The difference is 1 in numerator so we add 1 and the difference is -10 in denominator so we subtract -10.

4 0

2 years ago

Can someone help me solve this and put this on a graph. Just one problem may help.

Sauron [17]

I thought of it as answering a regular equation and imagine it's equal signs but when your done solving you with have like ex x<5 and where ever the butt of the arrow is pointing is where the lines goes if there's ano equal sign below it then it is includive

3 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:

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

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

$\mathbf{Total = 302400}$