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love history [14]
3 years ago
12

Help me please thank you

Mathematics
2 answers:
Alex3 years ago
6 0
The answer is D because you need the horizontal translations in the parenthesis to equal zero

tatiyna3 years ago
5 0
Mortals, begging for help............... *Disgusted face*

Your answer would be D

From: Dark Angel
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3 cm 10 cm 10 cm 3cm ​
erik [133]

Answer:

what is the question about the given numbers

8 0
2 years ago
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Which shows the graph of the solution set of 3y – 2x > –18?
nikdorinn [45]

Answer:

Graph of the inequality 3y-2x>-18 is given below.

Step-by-step explanation:

We are given the inequality, 3y-2x>-18

Now, using the 'Zero Test', which states that,

After substituting the point (0,0) in the inequality, if the result is true, then the solution region is towards the origin. If the result is false, then the solution region is away from the origin'.

So, after substituting (0,0) in 3y-2x>-18, we get,  

3\times 0-2\times 0>-18

i.e. 0 > -18, which is true.

Thus, the solution region is towards the origin.

Hence, the graph of the inequality 3y-2x>-18 is given below.

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 barometer falls from a weather balloon
8_murik_8 [283]

Answer:

Step-by-step explanation:

h(t) = -16t² + 14400

0= -16t² + 14400

(1/-16)×-14400=-16t²(1/-16)

900=t²

\sqrt{900} =\sqrt{t^{2}}

30=t

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2 years ago
One type of firework splits 5 times after it is fired into the air. The number of new bursts after each split can be modeled by
kherson [118]
Probably 5 because it only states 1 firework
5 0
3 years ago
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