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3 years ago

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How can I find the area of a rectangle with a rectangle the size of w- 2in and l- 5 in. I need two ways and i need a lot of easy

explaining because I need to draw how I did it.

1 answer:

Anvisha [2.4K]3 years ago

6 0

Here's one way: A=WxL So in this case, it's 2x5 You would draw the rectangle, label one side 2 in. and the other 5 in. Draw an arrow from each no./side and put x underneath the line so that its 2 —x— 5 (2, line, x, line, 5. With x under line)

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The sum of 3 consecutive odd numbers is 177. What is the value of the smallest number in the sequence

riadik2000 [5.3K]

Ok this is a very question that can solved very easily. So we know that there three numbers, and they are consecutive, this means that they are all one bigger than the number before so for example 1,2, and 3 or 20, 21, and 22. These numbers are called consecutive. Now let's just pretend that the answer to this question is the variable "x". Now x is the smallest number in all the three numbers.

Using our knowledge of consecutive numbers we can say that the next to numbers will be one more than the number before so:

(x+1) and (x+2)

So we have our three numbers: x , x+1 , and x+2. Now let's set these numbers equal to 177.

177 = x + x + 1 + x + 2 → Combine like terms.

177 = 3x + 3 → Subtract 3 from both sides.

174 = 3x → Divide both sides by 3

58 = x

So that means that our first and smallest number is going to be 58 our second number will be +1 so 59 and our third number will be +2 so 60.

So the three numbers will be 58, 59, and 60.

Hope this helped.

4 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

If you had a right triangle with legs measuring 40 yards and 100 yards long,

tia_tia [17]

9514 1404 393

Answer:

(c) 108 yd

Step-by-step explanation:

For hypotenuse c and legs a and b, the Pythagorean theorem tells you the relation is ...

c² = a² +b²

Solving for c, we have ...

c = √(a² +b²)

c = √(40² +100²) = √(1600 +10000) = √11600

c ≈ 107.70 . . . . yards

The length of the hypotenuse is about 108 yards.

6 0

3 years ago

If the dimensions of the rectangle are doubled, what is the area of the new rectangle? in terms of X?

miv72 [106K]

4(x+3)=4x+12 because when you have the ( ) you have to multiply the front # with the # inside it.

I hope that my image explains it I’m not really good at explaining things

4 0

3 years ago

What is the value of ? Pls help asapppppp

Serhud [2]

Answer:

115-50=65

Step-by-step explanation:

the sum of two interior angles = ext angle

3 0

3 years ago