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svetoff [14.1K]
3 years ago
7

Each of the 10 students in the baking club made 2 chocolate cakes for a fundraiser they all used the same recipe, using C cups o

f flour in total. Write an expression that represents the amount of flour for one cake. Use C as your variable.
Mathematics
1 answer:
IrinaK [193]3 years ago
6 0

Answer:

C / 20

Step-by-step explanation:

Number of students in the baking club = 10

Each student made 2 chocolate cakes

Total cakes = number of students × number of cakes per student

= 10 × 2

= 20

Total cups of flour used by 10 students = C cups

Amount of flour for one cake = Total cups of flour / Total cakes

= C / 20

The expression that represents the amount of flour for one cake = C / 20

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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)

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x ∈ [y]

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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.

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vladimir1956 [14]
Um.. Both of those equations are the exact same. Divide both sides of the first equation by 6. You get:

\frac{30x  - 42y}{6} = \frac{-6}{6} \\ 5x - 7y = -1

That is the exact same as the second equation. This system has an infinite number of solutions. 5x - 7y = -1 is a line, so basically every point on that line is a solution to the system.

For example, x = 0 and y = \frac{1}{7} would work, but so would x = 1 and y = \frac{6}{7}
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A toy box has 6 faces.There are square there are 12 edges and 8 vertices.Identify the shape of the toy box.
xxTIMURxx [149]

Answer:

A cube or a cuboid.

Step-by-step explanation:

A cube or a cuboid is a three dimensional shape that has

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The difference between a cube and cuboid is that a cube has all edges equal which makes every face a square while a cuboid has two square faces and four rectangular faces.

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LekaFEV [45]

Answer:

Well, we have to remember that we must use PEMDAS, or the order of operations, to solve the expression.

<em>Therefore, we must first either divide or multiply anything that is inside of the parentheses. (In this case divide)</em>

12/3=4

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<em>Now solve what remains in the parentheses.</em>

4+6=10

10-4

<em>Finally, subtract 4 from 10 and get your answer:</em>

6


4 0
2 years ago
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