C. for(int i=0; i<3; i++) { tone(7, 500); delay(1000); }
D. tone(7, 1000); delay(200); tone(7, 1000); delay(200); tone(7, 1000);
Solution
Step 1: Understand the beep pattern needed
Three quick beeps require playing tone briefly, stopping it, then repeating.
Step 2: Analyze each option
for(int i=0; i<3; i++) { tone(7, 1000, 200); delay(300); noTone(7); } uses a loop to play 200 ms tone, waits 300 ms, then stops tone before next beep. This creates three distinct beeps. tone(7, 1000, 600); delay(1800); noTone(7); plays one long beep. for(int i=0; i<3; i++) { tone(7, 500); delay(1000); } plays tones without stopping, causing continuous sound. tone(7, 1000); delay(200); tone(7, 1000); delay(200); tone(7, 1000); plays tones without stopping, overlapping sounds.