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hjlf
3 years ago
10

Please help

Mathematics
2 answers:
vladimir1956 [14]3 years ago
8 0
It would be supplementary because the angles outside the triangle are connected though the line segment of the triangle.
Goshia [24]3 years ago
6 0
It forms Obtuse angle......
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Anon25 [30]

Answer:

19

Step-by-step explanation:

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A wallet is originally $11.00. With a 40% discount, what is the sales price? Show your work.
NeX [460]

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$6.6

Step-by-step explanation:

you want to find out what 40% of the original price is ($11)

100% = 11

10%= 1.1

to get 40% you times it by 4

40%= 4.4

you then have to take away the sale reduction from the original

11- 4.4 =$6.6

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X^(log_(\sqrt(x))(x-3))=4
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Answer:

X= -4

Step-by-step explanation:

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In the basket and 12 of them are oranges what percentage of the pieces of fruit in the basket or orange show your work in the bo
notsponge [240]

Answer:

60%

Step-by-step explanation:

There are 20 pieces of fruit in a basket and 12 of them are oranges what percentage of the pieces of fruit in the basket are orange show your work in the box and circle your answer

Total oranges in the basket = 20

Number of oranges = 12

what percentage of the pieces of fruit in the basket are orange

= Number of oranges / Total pieces of fruits × 100

= 12/20 × 100

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= 60%

Percentage of the pieces of fruit in the basket that are oranges = 60%

3 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
3 years ago
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