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Goryan [66]
3 years ago
5

Someone please please help with this I would really appreciate it !

Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0

Answer:

I think this is how the response is meant to be, do you need workings to solve AB aswell?

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A recipe for 11 pizza crust calls for 3838 tablespoon of salt. How much salt is needed to make 33 pizza crusts?
max2010maxim [7]
Its 11,514 tablespoons of salt
8 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
6. Riley let his friend borrow $12,750. He wants to be paid back in 4¾ years and is going to charge his friend a 5.5% interest r
Talja [164]

Answer:

ivcfvbnjbvgfcddf,;lhuybgtfrsxdcfvnjmkl,;kuytresdfghjiuytfrd

Step-by-step explanation:,vbnnbvgbhjnkmjhuygtfrdesdrtfyuhijohuygtfrd5eswed

7 0
3 years ago
What is the value of 3.45 + 2.6 + 32.007?​
bulgar [2K]
38.057 is the total
4 0
2 years ago
Read 2 more answers
Shaquil needs to borrow $400. The loan company charges him $80 in fees for the loan. Shaquil calculates that the annual percenta
lesya [120]

Answer:

(B)  

Step-by-step explanation:

PRINCIPAL AMOUNT(p) = $400

RATE (r)= 250% = 2.5

INTEREST = $80

t = time (in days) = t/365

By using the formula,

r = \frac{i}{pt}

2.5 = \frac{80*365}{400*t}

t = \frac{364*80}{400*2.5}

t = 29.2

t = 29 days (approx)

Hence option (B) is correct.


6 0
3 years ago
Read 2 more answers
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