<u>Explanation</u>:
(a)
![f=A+A B+A C\\=A[1+B+C]=A \quad[1+x=1]\\F=A](https://tex.z-dn.net/?f=f%3DA%2BA%20B%2BA%20C%5C%5C%3DA%5B1%2BB%2BC%5D%3DA%20%5Cquad%5B1%2Bx%3D1%5D%5C%5CF%3DA)
No gate is required to implement this function
(b)
Note: Refer the first image.
(c)
![\begin{aligned}f &=\overline{A+B}+A \bar{B}+B \bar{C} \\&=(\bar{A} \bar{B})+A \bar{B}+B \bar{C} \\&=\bar{B}[A+\bar{A}]+B \bar{C} \\& F=\bar{B}+B \bar{C} =\bar{B}+\bar{C}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%20%26%3D%5Coverline%7BA%2BB%7D%2BA%20%5Cbar%7BB%7D%2BB%20%5Cbar%7BC%7D%20%5C%5C%26%3D%28%5Cbar%7BA%7D%20%5Cbar%7BB%7D%29%2BA%20%5Cbar%7BB%7D%2BB%20%5Cbar%7BC%7D%20%5C%5C%26%3D%5Cbar%7BB%7D%5BA%2B%5Cbar%7BA%7D%5D%2BB%20%5Cbar%7BC%7D%20%5C%5C%26%20F%3D%5Cbar%7BB%7D%2BB%20%5Cbar%7BC%7D%20%3D%5Cbar%7BB%7D%2B%5Cbar%7BC%7D%5Cend%7Baligned%7D)
Note: Refer the second image
(d)
![\begin{aligned}f=& A B \bar{c}+\overline{A+\bar{c}} \\=& A B \bar{c}+\bar{A} \bar{c}=\bar{A} B \bar{c}+\bar{A} c \\f=& \bar{A}[c+B \bar{c}] . \\& f=\bar{A} B+\bar{A} c=\bar{A}(B+c)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%3D%26%20A%20B%20%5Cbar%7Bc%7D%2B%5Coverline%7BA%2B%5Cbar%7Bc%7D%7D%20%5C%5C%3D%26%20A%20B%20%5Cbar%7Bc%7D%2B%5Cbar%7BA%7D%20%5Cbar%7Bc%7D%3D%5Cbar%7BA%7D%20B%20%5Cbar%7Bc%7D%2B%5Cbar%7BA%7D%20c%20%5C%5Cf%3D%26%20%5Cbar%7BA%7D%5Bc%2BB%20%5Cbar%7Bc%7D%5D%20.%20%5C%5C%26%20f%3D%5Cbar%7BA%7D%20B%2B%5Cbar%7BA%7D%20c%3D%5Cbar%7BA%7D%28B%2Bc%29%5Cend%7Baligned%7D)
Note: Refer the third image
(e)
![\begin{aligned}f=& A \bar{B}+\bar{B} C+A \bar{B} \\&=\bar{B}[A+\bar{A}+c] \\&=\bar{B}[1+C]\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%3D%26%20A%20%5Cbar%7BB%7D%2B%5Cbar%7BB%7D%20C%2BA%20%5Cbar%7BB%7D%20%5C%5C%26%3D%5Cbar%7BB%7D%5BA%2B%5Cbar%7BA%7D%2Bc%5D%20%5C%5C%26%3D%5Cbar%7BB%7D%5B1%2BC%5D%5Cend%7Baligned%7D)
(f)
![\begin{aligned}f &=A B C+A B D+A B C \\&=A B[C+C]+A B D \\&=A B+A B D \\&=B[A+A D] \\&=B[A+D] \\\therefore & A=B[A+D]\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%20%26%3DA%20B%20C%2BA%20B%20D%2BA%20B%20C%20%5C%5C%26%3DA%20B%5BC%2BC%5D%2BA%20B%20D%20%5C%5C%26%3DA%20B%2BA%20B%20D%20%5C%5C%26%3DB%5BA%2BA%20D%5D%20%5C%5C%26%3DB%5BA%2BD%5D%20%5C%5C%5Ctherefore%20%26%20A%3DB%5BA%2BD%5D%5Cend%7Baligned%7D)
Note: Refer the fourth image
Answer:
I think a circular particular circuit element is given by it equals 5 200.8 e blah blah blah blah
Explanation:
Answer:
Written in C++
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
float degreeC, degreeF;
cout<<"Degree Fahrenheit: ";
cin>>degreeF;
degreeC = 5 * (degreeF - 32)/9;
cout<<"Degree Celsius: "<<degreeC<<" C";
return 0;
}
Explanation:
The question requests that input should be in degree Fahrenheit
Declare all necessary variables
float degreeC, degreeF;
Prompt user for input in degrees Fahrenheit as stated in the question
cout<<"Degree Fahrenheit: ";
Get User Input
cin>>degreeF;
Convert degree Fahrenheit to Celsius
degreeC = 5 * (degreeF - 32)/9;
Display output
cout<<"Degree Celsius: "<<degreeC<<" C";
Answer:
6.5 × 10¹⁵/ cm³
Explanation:
Thinking process:
The relation 
With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹
and ni = 1.5 × 10¹⁰
Temperature, T = 300 K
K = 1.38 × 10⁻²³
This generates N₀ = 1.654 × 10¹⁶ per cube
Now, there are 10¹⁶ per cubic centimeter
Hence, 
