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telo118 [61]
2 years ago
6

1 (a) (i) Yasmin and Zak share an amount of money in the ratio 21 : 19.

Mathematics
1 answer:
cupoosta [38]2 years ago
4 0

Answer: 1(a) $120

(ii a) $34

Step-by-step explanation:

21-19 = 2

$6/2=3

21x 3( one ratio =$3) =63

19x3 = 57

63+57=120 - (1 A ans)

(ii a) 40x0.15 (percentage decrease)= 6

40-6 = $34

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278 divided by 59 is 4.712, which can be rounded to 5
the answer is d. 5
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3 years ago
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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
2 years ago
f left parenthesis x right parenthesis equals 25 comma 000 left parenthesis 1 plus.025 right parenthesis to the power of x?
Oduvanchick [21]

Answer:

f(x) = 25,000 ( 1 + 0.025 )ˣ

Step-by-step explanation:

Given word problem becomes

f(x) = 25,000 ( 1 + 0.025 )ˣ

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2 years ago
0.07 is closest to 0,1, or 1/2?
gulaghasi [49]
0 is the right answer.
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3 years ago
Is 42.9 a rational number
Elan Coil [88]

Answer:

Yes.

Step-by-step explanation:

What is a rational number?

A whole number, fraction, terminating or recurring decimal.

42.9 = 429/10 which is a fraction.

So it's rational.

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3 years ago
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