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3 years ago

Please help me this is very important its a test due in 30 minutes

Mathematics

1 answer:

Mashcka [7]3 years ago

5 0

A
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The function g(x)=(x-2^2. The function f(x)=g(x) +3

OlgaM077 [116]

The parent function is:
y = x ^ 2
Applying the following function transformation we have:
Horizontal translations:
Suppose that h> 0
To graph y = f (x-h), move the graph of h units to the right.
We have then:
g (x) = (x-2) ^ 2
Then, we have the following function transformation:
Vertical translations
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
We have then that the original function is:
g (x) = (x-2) ^ 2
Applying the transformation we have
f (x) = g (x) +3
f (x) = (x-2) ^ 2 + 3
Answer:
the function f(x) moves horizontally 2 units rigth.
The function f (x) is shifted vertically 3 units up.

4 0

3 years ago

REAL LIFE A cylindrical hazardous waste container has a diameter of 1.5 feet and a height of 1.6 feet. Abo

aivan3 [116]

Answer:

It can hold up to $1.256ft^3$ of waste

Step-by-step explanation:

Given

$D = 1.5ft$ -- Diameter

$h = 1.6ft$ -- height

Required

Determine the amount of waste it can hold

This simply means that we calculate the volume of the cylindrical container.

And this is calculated using:

$Volume = \pi r^2h$

Where

$\pi = 3.14$

and

$r = \frac{D}{2}$

$r = \frac{1.5}{2}$

$r = 0.75$

So, we have:

$Volume = 3.14 * 0.5^2 * 1.6$

$Volume = 1.256ft^3$

Hence, it can hold up to $1.256ft^3$ of waste

3 0

3 years ago

1. Write a numerical expression for the phrase “twelve groups of the sum of fourteen and 10.”

Rom4ik [11]

The expression for five groups of a number added to seventeen.

Let the number be X
Five groups of the number will be 5X
The required expression will be
5(X) +17

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:

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

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

$\mathbf{Total = 302400}$