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3 years ago

PlZ help I need an answer I don't need to waste points so plz Help!!!!!!!!!

Mathematics

2 answers:

avanturin [10]3 years ago

8 0

Answer:

option A.

Step-by-step explanation:

the rising rate isn't constant in exponential and B. makes no sense at all C. is constant D. curves incorrectly so the only reasonable answer is A.

nadya68 [22]3 years ago

5 0

Answer:

The first obe

Step-by-step explanation:

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Round 18,792 to the nearest thousand.

Mariulka [41]

Answer: 18,800

Step-by-step explanation: u only need 9 more numbers for 18,792 just to hit 18,800.

7 0

2 years ago

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What is the value of this expression if h = 8,j = -1, and k = -12

Rashid [163]

Answer:

Step-by-step explanation:

The given expression is :

$\dfrac{j^3k}{h^0}$

We need to find the value of this expression when h = 8, j = -1 and k = -12.

Put all the values in the given expression.

$\dfrac{j^3k}{h^0}\\\\=\dfrac{(-1)^3\times (-12)}{(8)^0}\\\\=\dfrac{-1\times -12}{1}\\\\=12$

So, the value of the given expression is 12.

7 0

3 years ago

The figure shows two intersecting lines and the measures of the resulting angles. Write an equation to help you solve for x. The

Svetlanka [38]

Given:

The given figure shows the intersection of the two lines.

The angles formed by the intersection of the two lines are (3x - 8)° and (2x + 12)°

We need to determine the equation to solve for x and to find the value of x.

Equation to solve for x:

Since, the two angles (3x - 8)° and (2x + 12)° are vertically opposite angles and the vertical angles are always equal.

Hence, we have;

$3x-8=2x+12$

Thus, the equation to solve for x is $3x-8=2x+12$

Value of x:

The value of x can be determined by solving the equation $3x-8=2x+12$

Thus, we have;

$x-8=12$

$x=20$

Thus, the value of x is 20.

3 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

3 years ago

How many fiths are in 1 whole

sattari [20]

Five fifths are in one whole.
Don't believe me?
So you just do 1÷5
which is 5!
~JZ
Hope it helps you :)!

6 0

3 years ago

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