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3 years ago

Find the value of f (2).

Mathematics

2 answers:

patriot [66]3 years ago

5 0

Answer:

y=-5

Step-by-step explanation:

You just have to see what y is when x is 2.

saw5 [17]3 years ago

4 0

Step-by-step explanation:

The value of f(2) is -5.

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2x = x +3 Help pls I can’t do this one

Fudgin [204]

Answer:

x=3

Step-by-step explanation:

Subract the x from the left side and add it to the right side and then you got your answer

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3 years ago

Una embarcación tenía 120 tripulantes al naufragar parece el 30% los sobrevivientes el 25% son casados¿ Cuántos de los sobrevivi

Wewaii [24]

Answer:

Solteros= 27

Step-by-step explanation:

Entiendo que de los 120 tripulantes, sobrevivió el 30%. Y de los sobrevivientes, 25% son casados.

Primero, debemos calcular la cantidad de sobrevivientes:

Sobrevivientes= 120*0.3= 36

Ahora la cantidad de casados, y por diferencia los solteros:

Casados= 36*0.25= 9

Solteros= 36 - 9

Solteros= 27

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3 years ago

Is perpendicular to line segment

Masja [62]

Answer:

AB is perpendicular to [GH] and GH is [A]

Step-by-step explanation:

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3 years ago

The following two parallelograms are drawn below. Find the measurement of FE or the variable “x”

labwork [276]

Answer:

x=15 in.

Step-by-step explanation:

If the parallelograms are equal (I assume they are), then the ratios are the same. 12 is 3/4 of 16. 3/4 of 20 is 15.

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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:

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

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

$\mathbf{Total = 302400}$