put 2 in each
Bubble sort is easy to program, slower, iterative. Compares neighboring numbers swaps it if required and continues this procedure until there are no more swaps Quick Sort is little difficult to program, Fastest, Recursive. Pivot number is selected, other numbers are compared with it and shifted to the right of number or left depending upon criteria again this method is applied to the left and right list generated to the pivot point number. Select pivot point among that list.
2 4 6
F. Numbers in hecadecimal start 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, 12, ...
In series like so ---6 ohms ---- 12 ohms --- , the total resistance is just 6 ohms + 12 ohms.assuming you mean in parallel like this:_|---6 ohms-----|-|~|-_|---12 ohms---|then the resistance of this can be calculated like so:1/6 ohms + 1/12 ohms = 1/R (where R is the resistance of the circuit as a whole)2/12 ohms + 1/12 ohms = 1/R3/12 ohms = 1/R1/4 ohms = 1/Rso R = 4 ohmsA few notes, if the resistors are in parallel the total resistance will always be less than or equal to the lowest resistance in parallel (i.e 6 ohms in parallel with 12 ohms will have resistance less than 6 ohms).Also if two resistances in parallel are the same, then the resistance is half of the resistance of both resistors (i.e. 1/2 ohms + 1/2 ohms = 1/R; 1 = 1/R, R=1 ohm which is half of 2 ohms).This process can be extended to 2 or more resistors in parallel.i.e if we had a 6 ohm, 6 ohm and 12 ohm resistor in parallel we could go1/6 ohms + 1/6 ohms + 1/12 ohms = 1/R(1/6 ohms + 1/6 ohms) + 1/12 ohms = 1/R1/3 ohms + 1/12 ohms = 1/R4/12ohms + 1/12 ohms = 1/R5/12 ohms = 1/Rso R = 12/5 ohms or 2.4 ohms
No. 12 volts peak to peak would be 6 in the positive polarity and 6 negative polarity. Simply saying 12 volts AC would be 12 volts in each polarity or 24 volts peak to peak
They are all positive numbers.
There are 6 groups.
6 different groupings can be made from 12: 12 groups of 1, 6 groups of 2, 4 groups of 3, 3 groups of 4, 2 groups of 6, 1 group of 12.
Use the definition of median: sort the numbers from smallest to largest, then choose the number in the middle.
2 groups
12C6 = (12 x 11 x 10 x 9 x 8 x 7)/ (6 x 5 x 4 x 3 x 2 x 1) = 924
The sum of the six numbers/6 = 12 The sum of the six numbers = 72 (72-6)/6=11
If the six numbers are all different, then the answer is 6C4 = 6*5/(2*1) = 15
5
The GCF of 12 and 42 is 6
12 and 252 is a pair of numbers that fit those criteria.
Yes, they can.