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Inga [223]
3 years ago
14

can someone please solve this? I clicked on homework help and it wasn't much help. I'm trying to turn in as much missing assignm

ents as I can before school ends on Thursday. Thank you!

Mathematics
2 answers:
MA_775_DIABLO [31]3 years ago
6 0

Answer:

  • One proportion to solve is 8/18 = x/8
  • Exact answer as a fraction: x = 32/9
  • Approximate answer in decimal form: x = 3.556

In reality, the 5's go on forever, but we have to round off somewhere.

========================================================

Explanation:

Break up the triangles as your teacher recommends. See the diagram below as to what I mean.

The smaller triangle has A = 8 as the horizontal side, and B = x as the vertical side

The larger triangle has C = 10+8 = 18 as the horizontal aside, and D = 8 as the vertical side.

We can form the proportion A/C = B/D.

A/C connects the horizontal sides (small/large) while B/D has the vertical sides tied together. The order of division is the same small over large. You could do large over small, but make sure you keep both sides consistent.

-----------------

Let's solve for x

A/C = B/D

8/18 = x/8

8*8 = 18*x .... cross multiply

64 = 18x

18x = 64

x = 64/18 .... divide both sides by 18

x = 32/9  exactly in fraction form

x = 3.556 approximately in decimal form

-------------------

Another proportion you can solve is A/B = C/D and you should get the same x value. It's the same as A/C = B/D because the B and C swap places. There are many other approaches you could take using different proportions.

Nesterboy [21]3 years ago
6 0
They are similar triangles. Meaning they are the same triangle just different sizes(smaller version of the big one) That means all the proportions are the same so to figure out x you would put it in proportion that would look like x/8= 8/18 (8+10)and you would cross multiply to solve it. So x would equal 3.5 or 32/9
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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)
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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)

[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.

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