Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out
Answer:
Distillation, heat
Explanation:
Here in this question, we simply want to look at the best options that could fit in the gaps.
We have a mixture of liquids having boiling points which is far from each other.
Whenever we have a mixture of liquids with boiling points far away from each other, the best technique to use in separating them is to use distillation. That is why we have that as the best fit for the first missing gap.
Now, to get the liquids to separate from each other, we shall be needing the heating mantle for the application of heat. This ensures that the mixture is vaporized. After vaporization, the condensing tube will help to condense the vapor of each of the liquids once we reach the boiling point of either of the two.
Kindly note that the liquid with the lower temperature will evaporate first and will be first obtained. In fact after reaching a little above the boiling point of the lower boiling liquid, we can be sure that what we have left in the mixture pot is the second other liquid with the higher boiling point.
Answer:
prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S
Explanation:
Given that S = { p q |p, q are prime numbers greater than 0}
E = {0, −2, 2, −4, 4, −6, 6, · · · }
To prove by constructing a bijection from S to E
detailed solution attached below
After the bijection :
<em>prove that | S | = | E |</em> : every element of S there is an Image on E , while not every element on E has an image on S
∴ we can say sets E and S are infinite sets
Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
Answer:
//Define the header file
#ifndef PLAYER_H
#define PLAYER_H
//header file.
#include <string>
//Use the standard namespace.
using namespace std;
//Define the class Player.
class Player
{
//Declare the required data members.
string name;
int score;
public:
//Declare the required
//member functions.
void setName(string par_name);
void setScore(int par_score);
string getName();
int getScore();
}
//End the definition
//of the header file.
#endif
Player.cpp:
//Include the "Player.h" header file,
#include "Player.h"
//Define the setName() function.
void Player::setName(string par_name)
{
name = par_name;
}
//Define the setScore() function.
void Player::setScore(int par_score)
{
score = par_score;
}
//Define the getName() function.
string Player::getName()
{
return name;
}
//Define the getScore() function.
int Player::getScore()
{
return score;
}
