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Serjik [45]
3 years ago
13

Which table represents a proportional relationships

Mathematics
2 answers:
Soloha48 [4]3 years ago
6 0
I think The last one in right

y |2| 6 |10 |14
x |6 |18 |30 |42
skad [1K]3 years ago
6 0

Box 3 because it has a proportion of two on both sides...also i had to dead with i-ready as well i hated it

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The ratio of Mary’s downloaded iPad apps to Katie’s is 12 to 8. If Katie has 96 apps how many does Mary have?
sladkih [1.3K]
8×12=96. Katie
12×12=144. Mary
you have to Divide Katie's 96 by 8 to get 12
then take Mary's 12 and multiply it by 12 to get the answer of 144 apps that Mary owns

3 0
2 years ago
one moving company charges $800 plus $16 per hour another moving company charges $720 plus $21 an hour at what number of hours w
Gekata [30.6K]

800 + 16x = 720 + 21x. Cancel out the 720 to get 80 + 16x = 21x. Subtract 16x from both sides to get 80 = 5x. Divide by 5 to get 16 = x, or x = 16. 16 hours. The charge is 800 + 16(16), = 720 + 21(16) which = 800 + 256 = 720 + 336 which = 1,056 = 1,056. Hope this helped!

6 0
3 years ago
PLEASE HELP MEEEEE<br><br><br> &lt;3
Svetradugi [14.3K]

Answer:

The second choice is correct.

Step-by-step explanation:

The softball cannot be anything under 180 grams because that would go past the ten gram limit. As well as the ball cannot go any higher than 200 grams.

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
A pyramid has a rectangular base with dimensions of 3 feet by 7 feet. The height of the rectangular pyramid is 6 feet. Water fil
Alja [10]

Answer:

Step-by-step explanation:

Michael and Juanita are interested in the amount of sand required to build a sand pyramid of the size that they see while at a sand sculpture exhibit.  They carefully measure the width of the square base of the sand pyramid which is 4 feet and the height of the pyramid which is 5 feet.  How much sand is required to build this sand pyramid?

3 0
3 years ago
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