Okay this is how you count by 3's...
3,6,9,12,15,18,21,24,27,30,33,and so on...
JUST KEEP ADDING 3 EVERY TIME AND YOU WILL DO IT !!!!!!!!!!!!
It is only because we count in tens - and 5 10 and 2 are factors.
3s + 1
3S Facility simply means sales, services and spares
To find out how many 3s go into 140, divide 140 by 3. This gives you approximately 46.67. Since we're looking for whole 3s, you can fit 46 complete 3s into 140.
12r - 15s - r + 3s = 11r -12s
Every 30 numbers you end up in the 10s place when you count by 3s. 30,60,90
There are fewer than 6 dozen egs in a basket If you count them by 3s there are 2 left over Three are left if you count by 5s How many eggs are there?
It is only because we count in tens - and 5 10 and 2 are factors.
3s + 1
135=3s +15 120=3s 40=s
28 3s
3S Facility simply means sales, services and spares
To find out how many 3s go into 140, divide 140 by 3. This gives you approximately 46.67. Since we're looking for whole 3s, you can fit 46 complete 3s into 140.
-7 = 3s - 1 +1 +1 (add 1 to both sides to get the variable alone) ___ ____ -6 = 3s __ ___ (-6 and 3s both divided by 3) 3 3 -2 = s
Oh, dude, let me grab my calculator for this super challenging math problem. So, in 3333, there are... drumroll... four 3s! Shocking, right? It's like they planned it that way or something.
4 3s = 4*3 = 12, which is a rational number.
The easiest way is to "flip" the inequality symbol end divide by the negative number:Example:6 < 3 - 3s6 - 3 < 3 - 3s -33 < -3s Method a) Divide by negative coefficient and flip the inequality symbol3/-3 > -3s/-3-1 > s or s< -13 < -3s Method b) Full algorithm, eliminate -3s by adding 3s on both sides3 +3s < -3s + 3s3 + 3s < 03 - 3 + 3s < 0 -33s < -33s/3 < -3/3s < -1 Looks familiar? So basically if you perform the full algorithm (method b) you can understand why we flip the inequality symbol when we have to eliminate a negative coefficient but it is faster just to flip the symbol (method a)