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2 years ago

What are the x and the y in this function

Mathematics

1 answer:

Marianna [84]2 years ago

7 0

Answer:

X = 3

Y = all real numbers

Step-by-step explanation:

x is the input

y is the output

The equation would be x = 3

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PLEASE HELP ME!!! PLEASE TRY TO DO ALL OF THEM THX! :)

kakasveta [241]

Hello there!

Number 1 is C (11, 4) Image is below

I don't know number 2, so so sorry!

Number 3 is A. 36 The formula for perimeter of a rectangle is (2 * base) + (2 * height)
The base = 13
The height = 5

(5 * 2) + (13 * 2)

10 + 26 = 36

Number 4 is 23 units.

I hope I was of assistance!

~ Fire

5 0

3 years ago

Help! I don't know how to do this! I will mark Brainliest!

gogolik [260]

Answer:

- 3y² - 7y + 6

Step-by-step explanation:

Given

(- 3y + 2)(y + 3)

Each term in the second factor is multiplied by each term in the first factor, that is

- 3y (y + 3) + 2(y + 3) ← distribute both parenthesis

= - 3y² - 9y + 2y + 6 ← collect like terms

= - 3y² - 7y + 6

4 0

3 years ago

Read 2 more answers

Which expression is equivalent to 2^4 ⋅ 2^−7?

IRISSAK [1]

the answer for this is:

subtract 7 from 4 to get -3

you then have 2^-3

remove the negative exponent by rewriting as 1/(2^3)

4 0

3 years ago

Read 2 more answers

Exercise 7.3.5 The following is a Markov (migration) matrix for three locations        1 5 1 5 2 5 2 5 2 5 1 5 2 5 2 5 2

fredd [130]

Answer:

Both get the same results that is,

$\left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

Step-by-step explanation:

Given :

$\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

and initial population,

$\bf P=\left[\begin{array}{ccc}130\\300\\70\end{array}\right]$

a) - After two times, we will find in each position.

$P_2=[P].[M]^2=[P].[M].[M]$

$M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

$=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]$

$\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right]$