1. Binary:100111111 Octal:477 Hexadecimal:13F 2. a. 280 b. 544 c.2805 Bonus:1452

25

1. EADB EAEB 2.E to B EADB EAEB

The depth of the tree is 5, leaf nodes are A,E,I,S and U, amount of external nodes are 13, internal path length is 33, external path length is 55.

ABBCDA BBCDA ABBCDA EABBCDA ABBCDA DABBCDA ans, ABBCDA

~AB+C+~D 3 false outputs

~~A+(~~B x ~~C) +~~A x ~~~B =A+(B xC) + A x ~B =(A+BC)A~B =AA~B+BCA~B =A~b+0 =A~b

f(18,6) =f(17,8)-10 =f(16,10)-20 =f(15,12)-30 =f(14,14)-40 =f(13,16)-50 =f(12,18)-60 =f(11,20)-70 =f(10,22)-80 =f(9,24)-90 =f(6,31)-81 =f(3,38)-72 =f(0,45)-63 =45-63 =-18

1. Pre. +/*58^22*46 post. 58+/*22^46* 2. 0 3. 60 Bonus:X=-3

1.11011 2.11000 bonus:01000

My decimal number is 100 Binary 1100100+110101=10011001 Octal 100-161=-61 Decimal 100*21=2100 converted to hexadecimal=13 Hexadecimal=100

number 100 converted to binary 100/2=50 R=0(R=remainder), 50/2=25 R=0 ,25/2=12 R=1, 12/2=6 R=0, 6/2=3 R=0, 3/2= 1 R=1, 1/2=0 R=1. Binary is 1100100 number 100 converted to octal 100/8=13 R=2, 13/8=1 R=5, 1/8=0 R=1 octal is 152 number 100 converted to hexadecimal 100/16=6 R=4, 6/16=0 R=1 hexadecimal is 14