1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Juliette [100K]
3 years ago
10

Pick the set of numbers which contain ALL perfect squares:

Mathematics
1 answer:
Ede4ka [16]3 years ago
6 0
Well, a way to do this problem would be to find the set of numbers that all of rational square roots. A list of perfect squares would be 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81, 10^2=100.
You might be interested in
The rate of growth of profit​ (in millions) from an invention is approximated by Upper P prime (x )equals x e Superscript negati
Alecsey [184]

Answer:

The function is  P(x) =  \frac{1}{2} e^{-x^2} +0.016

Step-by-step explanation:

From the question we are told that

    The rate of growth  is  P'(x) =  xe^{x} - x^2

     The total profit is P(2)  =$25,000

      The time taken to make the profit is x =  2 \ years

         

From the question

     P'(x) =  xe^{-x^2}  is the rate of growth

  Now here x represent the time taken

Now the total profit is mathematically represented as

       P(x) =  \int\limits {P'(x)} \,  =   \int\limits {xe^{-x^2}} \,

So using substitution method

   We have that

                      u =  - x^2

                      du =  2xdx

  So  

        p(x) =  \int\limits {\frac{1}{2} e^{-u}} \, du

       p(x) =  {\frac{1}{2} [ e^{-u}} +c ]

       p(x) =  {\frac{1}{2}  e^{-x^2}} + \frac{1}{2} c      recall  u =  - x^2   and  let  \frac{1}{2} c =  Z

       

At  x =  2 years  

     P(x)  =$25,000

So

       Since the profit rate is in million

    P(x)  =$25,000 = \frac{25000}{1000000} =$0.025 millon dollars

So  

       0.025 =  {\frac{1}{2}  e^{-2^2}} + Z  

=>    Z = 0.025 - {\frac{1}{2}  e^{-2^2}}  

       Z = 0.016  

So the profit function becomes

         P(x) =  \frac{1}{2} e^{-x^2} +0.016

     

       

5 0
3 years ago
Taylor says ABCD and EFGH are congruent because he can map ABCD to EFGH by multiplying each side length by 1.5 and translating t
Elina [12.6K]

Answer:

The E came before H

Step-by-step explanation:

8 0
2 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
Use mental math to find the sum. <br><br> 150 + 20 + 25
stiv31 [10]

Answer:


50 + 20 = 70
70 + 25 = 95
150 + 95 = 245

In conclusion, 150 + 20 + 25 equals to a total amount of 215.


Step-by-step explanation:

Have a great rest of your day
#TheWizzer

6 0
2 years ago
Read 2 more answers
The "Mr. Brust Dance School of Fine Ballet" charges $24 per dass and a one-time registration fee $150 A student paid a total of
irinina [24]

Answer- 28 classes

Step-by-step explanation:

822-150= 672  672/24 = 28

5 0
3 years ago
Read 2 more answers
Other questions:
  • Complete the square x^2+7x+11=0
    11·1 answer
  • A ray is a defined term because it ?
    15·1 answer
  • The third angle in an isosceles triangle is 8% more than twice as large as each of two base angles. Find the measure of each ang
    13·1 answer
  • What is the value of sinY?
    12·1 answer
  • What is (−3.2)⋅1.7 multipluied
    7·2 answers
  • If f(x)=2x^2-x-6 and g(x)=x^2-4, find f(x)/g(x)<br> 2x+3/x-2<br> 2x-3/x+2<br> 2x+3/x+2<br> 2x-3/x-2
    15·1 answer
  • At these numbers functions: (1, 2), (2, 3), (3, 4), (4, 3)
    7·1 answer
  • A shop produces 39 wetsuits every 2 weeks. how long will it take the shop to produce 429 wetsuits
    14·2 answers
  • 2. Which of these situations can be represented by the
    11·2 answers
  • A man made a loss of 15% by selling an article for $595. Find the cost price of the article​
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!