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Alexxandr [17]
3 years ago
15

5 more than the product of 9 and 3

Mathematics
2 answers:
dolphi86 [110]3 years ago
7 0
(9)(3) + 5

27 + 5

32



Your answer is 32.
frosja888 [35]3 years ago
7 0
If first you take 9x3, you get 27. 9x3=27. Then you add 5 more 27+5=32
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For x, y ∈ R we write x ∼ y if x − y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1)
vodomira [7]

Answer:

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

  • The relation being reflexive (∀a∈R, a∼a)
  • The relation being symmetric (∀a,b∈R, a∼b⇒b∼a)
  • The relation being transitive (∀a,b,c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it´s known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, -k ∈ Z

b∼a, ∀a,b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a,b,c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k,l ∈ Z

(a-b)+(b-c)=k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a,b,c ∈ R

∼ is Transitive by definition

We´ve shown that ∼ is an equivalence relation on R.

B. Now we have to show that there´s a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F(x)=[x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x,y∈[0,1), F(x)=F(y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F(x)=b):

F is injective:

let x,y ∈ [0,1) and suppose F(x)=F(y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x,y ∈ [0,1), then k must be 0. If it isn´t, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x,y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let´s find x such as x ∈ [0,1) and F(x)=[b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let´s show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F(r)=[b]

∼ is surjective

Then F maps [0,1) into C, i.e [0,1) is a set of representatives for the set of the equivalence classes.

4 0
3 years ago
Plz help this is complicated
abruzzese [7]

Answer:

Graph attached.

Step-by-step explanation:

In the question the given parameters are as followed.

y=7cos(\frac{x}{2})+1

1). Amplitude a = 7

2). Period = \frac{2\pi }{\frac{1}{2}}=2\pi \times 2=4\pi

3). Vertical shift = 1 means mid line of the graph will be y =1 and maximum of the graph is 8 and minimum is -6.

7 0
3 years ago
Just need the equation
Anna007 [38]

a. $ x = -2\\b. $ x = \frac{3}{2} \\c. $ x = 0\\d. $ x = -\frac{1}{5}

Each of the equations can be solved as shown below:

a. 3x + 7 = - x - 1

3x + 7 + x  = - x - 1 + x (<em>addition property of equality</em>)

4x + 7  = - 1\\4x + 7 - 7 = -1-7 (<em>Subtraction property of equality</em>)

4x = -8\\\frac{4x}{4} = \frac{-8}{4}<em> (Division property of equality)</em>

<em />x = -2

b. 1 - 2x + 5 = 4x - 3\\

<em>Add like terms </em>

<em />6 - 2x = 4x - 3\\\\6 - 2x - 6 = 4x - 3 - 6 (<em>Subtraction property of equality</em>)

- 2x = 4x - 9\\-2x - 4x = 4x -9 - 4x<em> (Subtraction property of equality)</em>

<em />-6x = -9 \\\frac{-6x}{-6} = \frac{-9}{-6}<em> (Division property of equality)</em>

<em />x = \frac{3}{2}

c. 4x - 2 + x =-2 + 2x\\

5x - 2 = -2 + 2x\\5x - 2x = -2 + 2\\3x = 0\\ x = \frac{0}{3} \\x = 0

d. 3x - 4 + 1 = - 2x -5 + 5x\\

3x - 3 = - 7x -5\\3x + 7x = 3 - 5\\10x = -2\\x = \frac{-2}{10} \\x = -\frac{1}{5}

<em>The value of x in each </em><em>equation</em><em> are:</em>

<em />a. $ x = -2\\b. $ x = \frac{3}{2} \\c. $ x = 0\\d. $ x = -\frac{1}{5}

Learn more here:

brainly.com/question/1527981

5 0
3 years ago
What is 2,527 divided by 8 equal step by step?
aksik [14]

Answer: 2527 divided by 8 equals

315 with a remainder of 7

Step-by-step explanation:

7 0
3 years ago
ΔABC with vertices A(-3, 0), B(-2, 3), C(-1, 1) is rotated 180° clockwise about the origin. It is then reflected across the line
Darina [25.2K]

A: because "reflection across the y axis then across the x axis is the only option that does not work"

3 0
3 years ago
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