Answer:
1a) 93ten
1b) 135eight
2a) 565ten
2b) 769ten
Step-by-step explanation:
1a)
1011101
7 digits long the first digit (the one that is most far right) is the ones digit or 2^0 digit and so the last digit (the most far left) is 2^6.
So to put the number in base 10 we do:
1(2^6)+0(2^5)+1(2^4)+1(2^3)+1(2^2)+0(2^1)+1(2^0)
=64+0+16+8+4+0+1
=64+16+8+4+1
=(64+16)+(8+4+1)
=80+13
=93
1b)
93
What is the hightest power of 8 in 93?
8^x=93
8^1=8
8^2=64
8^3 would be a factor of 8 too much
So how many 8^2 's are in 93?
64 goes into 93 how many times?
It's once with a remainder of 93-64=29.
So now 29...what is the highest power of 8 that goes into 29? 8^1=8.
How many times? 3(8)=24 with remainder 5.
So 93=1(8^2)+3(8^1)+5(8^0).
93=135eight
2a) 2341six
2(6^3)+3(6^2)+4(6^1)+1(6^0)
2(216)+3(36)+4(6)+1
432+108+24+1
540+25
565
2b) 30001four
3(4^4)+0(4^3)+0(4^2)+0(4^1)+1(4^0)
3(256)+0+0+0+1
768+1
769
