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RoseWind [281]
3 years ago
13

help need to answer asap!!! a basketball team played 64 games. They won 20 more than they played; how many did they win?; how ma

ny did they loose?
Mathematics
1 answer:
melisa1 [442]3 years ago
7 0
20 more than what they played or 20 more than what they lost?
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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
Z to the power of 2 =1
stiv31 [10]

That would be written as:

z^{2} = 1


And the step-by-step equation would be:

z^{2} = 1 → we take the 2 to the right side of the equation and do the IO*

z = \sqrt{1} →we solve for \sqrt{1}

z = 1 → final answer


*Ps - IO is not an existing term and stands for inverse operation. In this case, because when we take to 2 to the right side of the equation (the 2 is a power) it'll have to turn into a square root (because exponents and roots are inverse operations)



Hope it helped,



BioTeacher101

3 0
3 years ago
I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS PLEASE AND PLEASE EXPLAIN WHY THAT IS THE ANSWER
BartSMP [9]
\bold{FULL ANSWERS:}

3 0
2 years ago
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10x - 80 = 16x - 12 <br> what is x?
AnnyKZ [126]
I just did it on paper but hope that helps:)

7 0
3 years ago
Can someone answer this!?
Karolina [17]

Answer:

x = 58

Step-by-step explanation:

Hope this helps

enjoy

4 0
3 years ago
Read 2 more answers
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