UGRhIGx3b29za256IGJrbiBwZGEgeWR3aGhhamNhIGx3Y2EgZW86IHludWxwaw
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Q: AAAANNNSSSWWWEEERRR TTTHHIIISSS QQQUUEESStion UUUUUUUsingggggg Boooooooth algooooorithm innnnnnn…
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Q: Translate the lines marked by to Hex . LDI R16, 10 LDI R29, 3* OR R16, R29. ASR R29. LDI R29, 3: OR…
A: LDI R16, 10LDI R29, 3 *OR R16, R29 *ASR R29 *
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UGRhIGx3b29za256IGJrbiBwZGEgeWR3aGhhamNhIGx3Y2EgZW86IHludWxwaw==
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- on Translate the lines marked by to Hex LDI R16, 10 LDI R29, 3* OR R16, R29* ASR R29* LDI R29, 3: OR R16, R29: ASR R29: 0x95AA 0x1805 0x957A 0x2BOD OxE0A1 OxE0C9 OxE045 0xE0D3 0x9540 0x908C OXOFOA Ox95D5 0xE057 OxE098 OxE075 0x9587 0x9583 OxEDE6 ONEOB4 09595 0x9553 0x2308 0x270E 0x230C 0x1B0B 0x95C7 0x95E0 0x0F07 0x2809 0x2704mov r0, # 6Ah mov psw, # 00h 018 ric a 180100564-871598836 598836 b190100564871598836. 71598830 The final value of b190100567 1598836 2871698836 The accumulator's final value is not twice the 22h. 598636 00100504-98158836-In the representation Complement to 1, we the number when Code the absolute value thus is negative: A = True B- False 7- The result in base 10' A - 44.25 B-42015 C- 51. 13 45. 75 8- The result AT 40. B + 61 q of the transformation (101001.10) 2 usi o the transformation (52) a in bose 10 is: C² 42 D-41 WWW 30 O
- AAAANNNSSSWWWEEERRR TTTHHIIISSS QQQUUEESStion UUUUUUUsingggggg Boooooooth algooooorithm innnnnnn an 6-bit sssssssystemmmmm, multiply (-6)10 by (3)10Convert the following numbers into base ten. (a) 32five (b) 22six (c) 410five (d) 3203five (e) 202five (f) 1222six (g) 202four (h) 404seven (i) 1000two (j) 11011two (k) 88sixteen (l) 707sixteen ?Complete the Lesson with Eight LEDs with 74HC595. This code uses a shift register to use a single data pin to light up 8 LEDs. Run the code with the loop() replaced with void loop () { } leds = 0b00001111; Serial.println(leds); 00 digitalWrite (latchPin, LOW); shiftOut (dataPin, clock Pin, MSBFIRST, leds); digitalWrite (latchPin, HIGH); delay (1000); Which LED lights are turned on? QA QB QC QD QE QF QG QH
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