explain the lines of code
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A: Introduction: I've listed the benefits and drawbacks of imperative, functional, and declarative…
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A: Introduction: Define "strongly typed checking" and "loosely typed checking" in the introduction.
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A: A network must satisfy the following three requirements to be effective and successful:
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A: Describe the advantages and disadvantages of the different Wi-Fi security measures.
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A: Here is the python code. see below step for output.
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A: Waterfall model used to be the most used model in earlier times of software development.
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A: The answer is
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A: A method that calls itself is known as a recursive method.
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Q: 8: 8 642 7: 1 3 5 7 6: 6 4 2 5: 1 35
A: I have completed the code below:
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A: digitizer is technology that convert analog to digital explain it
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Q: epts of integrity.
A: Still, you probably remember receiving some information about your academy's academic honesty policy…
Q: Parallel prefix computation on the q-cube Processor x, 0≤x N₂(x) then u[x] := u[x] Ⓡy endfor
A: explaination of code is given in next step:-
Q: What is the main distinction between procedural and object-oriented programming?
A: Programming is a a technique to create set of instructions executed by computer system to achieve…
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A: Given f(x,y)=y(x+y) X=4 and y= 3
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Q: What are the three requirements for a network to be effective and successful? Give a brief…
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- EXPLAIN LINE BY LINE THIS ARDUINO CODE Please explain each and every line in detail Simple Cylon Cylon Eye sweep using 5 LEDs */ unsigned char upDown=1; unsigned char cylon=0; void setup() { DDRB = B00011111; } void loop() { if(upDown==1){ cylon++; if(cylon>=4) upDown=0; } else { cylon--; if(cylon==0) upDown=1; } PORTB = 1 << cylon; delay(150); }int X[900]; int Y[600]; int sum, sum1, sum2, sum3; //parallelism : dividing outer loop in three parts //i = 1 to 300 for(i=1;i<=300;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 301 to 600 for(i=301;i<=600;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 601 to 900 for(i=601;i<=900;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } sum = sum1 + sum2 + sum3;} another way to solve the question that send in the pic#include Servo Main Servo int trigpin = echo pin = 8) opm distance i i Note This Code for Smart Ductoin using Arduino, Sensor & Motor → please give Suplanation about this code int float durahimn float cm Void Setup() Servo Main, allach (7) i Pin Mode (trigpin, output) i Pin Mode (echopin, imput); f void loop() digital write (trigping low) i delay (2); digital write (trigpin, High) ; delay Microsecond (10): digital write / trigpin, low) i duration= Pulse cm = (duration / 5882); distance = Cmi. if (distance (30) { Servo Main write (180), delay (3000) i f else & Servo Mam write (o); delay (50); f In (echopin, High) i
- Q1: Correct the following code, then list the functions of this code. const int Sensorl = A1 // analog pin const int Batt = A0; // analog pin const byte interruptpin = 2; void setup() { Serial.begin(6900): pinMode(interruptPin, INPUT); } void loop) { attachInterrupt(2,tilt sensor.RISING): } void send() { int valuel = analogRead(Sensol): float millivolts = (valuel / 1024.0) * 5000; celsius = millivolts / 10; Serial.printin(celseus); delay(1000); int valuel = analogRead(Batt) float Battery = (value2 / 1024.0) * 500; Battery-Battery/1000; Serial.printin(Battery); delay(1000); detachinterrupt(0); }convert c code to mips these 2 functions are linked with each other. #define MAX_BOARD_SIZE 12 // Players #define PLAYER_EMPTY 0 #define PLAYER_BLACK 1 #define PLAYER_WHITE 2 int board_size; int current_player = PLAYER_BLACK; char board[MAX_BOARD_SIZE][MAX_BOARD_SIZE]; int main(void); void announce_winner(void); unsigned int count_discs(int player); void announce_winner(void) { int black_count = count_discs(PLAYER_BLACK); int white_count = count_discs(PLAYER_WHITE); if (white_count > black_count) { printf("The game is a win for WHITE!\n"); white_count += count_discs(PLAYER_EMPTY); } else if (black_count > white_count) { printf("The game is a win for BLACK!\n"); black_count += count_discs(PLAYER_EMPTY); } else { printf("The game is a tie! Wow!\n"); } printf("Score for black: %d, for white: %d.\n", black_count, white_count); } unsigned int count_discs(int player) { int count = 0; for (int row = 0; row < board_size; ++row) { for (int col = 0; col < board_size; ++col) { if…Match the C-function on the left to the Intel assemble function on the right. W: cmpl $4 movl %edi , %edi jmp .L4(,%rdi,8) %edi .L3: movl $17, %eax ret .15: movl $3, %eax int A ( int x , int y) { int a ; if ( x == 0 ) else i f ( x == 1 ) a = 3 ; else i f ( x == 2 ) a = 2 0 ; else i f ( x == 3 ) a = 2 ; else i f ( x == 4 ) a = 1 ; ret .L6: a = 17; movl $20, %eax ret .L7: movl $2, %eax ret else a = 0; .L8: return a ; movl $1, %eax .L2: ret . section .rodata . L4: .quad .L3 .quad .L5 .quad .L6 .quad .L7 .quad .L8 X: testl %edi, %edi je cmpl je cmpl je стр1 je cmpl .L16 $1, %edi .L17 $2, %edi .L18 $3, %edi int B (int x, int y) { int a; switch (x) { .L19 $4, %edi %al movzbl %al, %eax case 0: a = 17; break; sete break; case 1: a = 3; case 2: a = 20; break; case 3: a = 2; break; case 4: a = 1; a = 0; } return a; ret .L16: break; movl $17, %eax ret .L17: movl $3, %eax } ret .L18: movl $20, %eax ret .L19: movl ret $2, %eax
- int x1 = 66; int y1 = 39; int d; _asm { } mov EAX, X1; mov EBX, y1; push EAX; push EBX; pop ECX mov d, ECX; What is d in decimal format?3. Show the stack with all activation record instances, including static and dynamic chains, when execution reaches position 1 in the following skeletal program. Assume bigsub is at level 1. function bigsub() { function a(flag) { function b() { *** a(false); } // end of b *** *** if (flag) b(); else c(); } // end of a function c() { function d() { <--- *** } // end of d d(); } // end of c *** 2 a(true); } // end of bigsub The calling sequence for this program for execution to reach dis bigsub calls a a calls b 12T2 read_item (X); read_item (Y); Z = Y - X write_item (Z); T1 read_item (X); read_item (Y); Y = Y + X write_item (Y); Suppose: TS(T1) = 3 TS(T2) Using Basic Timestamp Ordering to show the execute T1 and T2
- #include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…#include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…int func(int a, int b) { return (aSEE MORE QUESTIONS