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borishaifa [10]
3 years ago
14

I dont remember how to rewrite formulas properly

Mathematics
1 answer:
Anastaziya [24]3 years ago
6 0

Answer:

h= (V/πr²)(3)

Step-by-step explanation:

You might be interested in
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
A recipe for banana bread requires 3 cups of bananas for every 1 1/2 cups of sugar used. At this rate how many cups of sugar sho
mr Goodwill [35]

Answer:

The answer is 1 cup of sugar

Step-by-step explanation:

x = 2 x 1.5/3

x = 3/3

x = 1 cup of sugar

hope it helps!

4 0
3 years ago
Pls<br> Help me <br> I need it
ad-work [718]

Answer:

{2}^{7}

Step-by-step explanation:

{2}^{7}  = 128

{7}^{2}  = 49

So,

128 > 49 \\  {2}^{7}  \:  \:  \:  \:    >  {7}^{2}

Therefore 2^7 is greater

4 0
3 years ago
Suppose A and B represent two different school populations where A &gt; B and A and B must be greater than 0. Which of the follo
Ulleksa [173]
We can reject the last one: subtracting a non-zero value will result in a smaller value.

So now we have:

<span>2(A + B)
(A + B)2
A2 + B2

If all of them are mulptiplications, then they are all equivalent!

I mean by this, if what you meant is this:

</span>
<span>2*(A + B)
(A + B)*2
A*2 + B*2

If there is no sign, then the multiplication sign is implicit,


and all of these expressions say exactly the same: two of A and two of B.
</span>
6 0
3 years ago
Which is equivalent to 5+2(3x-4)=12x-11? <br>C. 9x=17 <br>B. 6x=8 <br>D. 9x=-39​
Rzqust [24]
5+2(3x-4)=12x-11 is 5+6x -8 = 12x -11, 5-8+11= 12x-6x, “8= 6x” so option B is correct
4 0
3 years ago
Read 2 more answers
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