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3 years ago

Which faces have an area of 28in squared

Mathematics

1 answer:

jeyben [28]3 years ago

3 0

To find area, multiply the length by the width

So, the very bottom box with a width of 4 in and a length of 7 inches, that equals an area of 28 in

Also, the very middle box as well

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Which ordered pair is a solution of the equation

andrew11 [14]

(0,-7)
-7=11(0)+4
-7=0+4
-7 is not 4 -----not a solution

(-1,-7)
-7=11(-1)+4
-7=-11+4
-7=-7 ----solution

(1,-7)
-7=11(1)+4
-7=11+4
-7=15
-7 is not 15 ----not a solution

(2,26)
26=11(2)+4
26=22+4
26=26 ----soltion

(-1, -7) & (2, 26) are solutions to the equation

4 0

3 years ago

If f(x)=5x, what is f^-1(x)?

Georgia [21]

Answer:

first thing to do is get up from your sit and go cry because that question is to hard

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3 years ago

HELP NOW PLZ AND I WILL LOVE YOU FOREVER

mr_godi [17]

Answer:

12/4+13=16 2+22/4=7.5

Step-by-step explanation:

you answer is the last option im 100 percent sure.

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3 years ago

You and a friend are driving go-karts on two different tracks. As you drive on a straight section heading east, your friend pass

yKpoI14uk [10]

No they cross each other like a highway. You are on the underpass and your friend passes over you. As in, they are crossing a bridge & you are below them.

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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:

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

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

$\mathbf{Total = 302400}$