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BartSMP [9]
3 years ago
14

You expect to receive $5,000 at graduation one year from now. Your plan is to invest this money at 6.5 percent, compounded annua

lly, until you have $50,000. At that time, you plan to travel around the world. How long from now will it be until you can begin your travels?
Mathematics
1 answer:
frez [133]3 years ago
6 0

Answer:

Approx 37 years.

Step-by-step explanation:

The compound interest formula is :

A=P(1+\frac{r}{n})^{nt}

Here, A = $50000

P = $5000

r = 6.5% or 0.065

n = 1

t = ?

Putting these values in formula we get;

50000=5000(1+\frac{0.065}{1})^{1t}

=> 50000=5000(1.065)^{t}

=> 10=(1.065)^{t}

Taking log on both sides;

ln(10)=ln(1.065)^t

=> t=\frac{ln(10)}{ln(1.065)}

We get t = 36.56 rounding to 37 years.

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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.

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Which relationship describes angles 1 and 2?
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We need to see the angels do u have a picture??

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I think it (3) one I’m not sure
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G(x)=4x^2<br><br> what would be the table to graph?
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Answer:

no you will not use a table

The answer is

g(x) = 4x

g{x} = 4x {  \: }^{2}  = 8x

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If the population has gone up 20,000 in 6yr what percent is that
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