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Pachacha [2.7K]
2 years ago
6

Please please help me with this answer

Mathematics
1 answer:
umka21 [38]2 years ago
4 0
The length is 12m from the solution above

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Extra Credit: Graph the function r(x) = |x −5| + 4. Compare the graph to the graph of f(x) = |x|. Describe the domain and range.
Vlad [161]

I forgot. Ill edit this when i remember

3 0
3 years ago
SOMEONE PLEASE HELP ME ON QUESTIONS 17-19 I NEED ASAP
Lana71 [14]

Answer:

17. 10x+24 OR 108   18. 72   19. 8.4

Step-by-step explanation:

(10x+24)+72=180

10x+96=180

10x=84

x=8.4

10x+24

10(8.4)+24

84+24

108

8 0
3 years ago
I need to find a expression equivalent to this with the same numbers <br><br> 2•((4•x•3)•6)
Kryger [21]

Answer:

144 * x

Step-by-step explanation:

2*4*x*3*6

2*4*3*6*x

8*3*6*x

24*6*x

144*x

3 0
2 years ago
A segment has an endpoint at (2, 4) and a midpoint at (4. 2). What are the coordinates of the other endpoint (2, 10) B. (3, 1) C
stiks02 [169]

Answer:

b

Step-by-step explanation:

6 0
3 years ago
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
2 years ago
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