If a number is a multipule of 6 then it is a multiplule of 2 AND 3
if a number is a multiplule of 2 then the last digit should be divisible by 2 (ie, 0,2,4,6,8)
if a number is divisible by 3, then the sum of its digits is divisble by 3
try each
A. 333
last number is not divisible by 2 so not divisible by 6
B. 882
2 is divisible by 2
8+8+2=18
18/3=6
divisible by 6 yes
C. 424
4 is disibielbe by 2
4+2+4=10
10/3=not a whole number so not divisible by 6
D. 106
6 is divisible by 2
1+0+6=7
7/3= not a whole number so not divibislb eby 6
answer is B
Answer:
u didn't show any pictures so I can't help u. can u show. pictures or something
Answer:
To prove:
X+Y.Z=(X+Y).(X+Z)
Taking R.H.S
= (X+Y).(X+Z)
By distributive law
= X.X+X.Z+X.Y+Y.Z --- (1)
From Boolean algebra
X.X = X
X.Y+X.Z = X.(Y+Z)
Using these in (1)
=X+X(Y+Z)+Y.Z
=X(1+(Y+Z)+Y.Z --- (2)
As we know (1+X) = 1
Then (2) becomes
=X.1+Y.Z
=X+Y.Z
Which is equal to R.H.S
Hence proved,
X+Y.Z=(X+Y).(X+Z)
8:15 or 8 to 15 would be the ratio because you compare 8 quarters to 15 coins.
