Alright so I'm coming up with this on the fly; you have the first six letters (a,b,c,d,e,f) and 0-9 and your ten numbers. calculate the amount of possible combinations for the letters by simply writing them down.
ab, ac, ad, ae, a f- five
bc, bd , be, bf- four
cd, ce, cf- three
de, df- two
ef,- one.
adding these all together gets a total of 15 for the letters. now the numbers
01, 02, 03, 04, 05, 06, 07, 08, 09- nine
12, 13, 14, 15, 16, 17, 18, 19- eight
23, 24, 25, 26, 27, 28, 29- seven
34, 35, 36, 37, 38, 39- six
45, 46, 47, 48, 49- five
56, 57, 58, 59- four
67, 68, 69- three
78, 79- two
89- one
added together with a total of 45 combinations.
alright so, 45 different number combinations and 15 letter combinations. multiplying 15 by 45 should tell you the total possible combinations for a two letter and two number serial-number
Is that supposed to be x squared, or x times 2 in the beginning of the first one?
Hello good person. The answer is C, (4,7). If you replace the x's in both equations (4) and the y's i both equations (7) you would get your answer for both.
3 + 4abs(x/2 + 3) = 11 Subtract 3
4 abs(x/2 + 3) = 11 - 3
4 abs(x/2 + 3) = 8 Divide by 4
abs(x/2 + 3) = 8/4
abs(x/2 + 3) = 2
Solution 1
x/2 + 3 = 2 Subtract 3
x/2 = 2 - 3
x/2 = - 1 Multiply by 2
x/2 *2 = - 2
x = - 2
Solution 2
x/2 + 3 = - 2
x/2 = - 2 - 3
x/2 = - 5 Multiply by 2
x = -5 * 2
x = -10
Solutions
x = - 2 <<<< answer 1
x = -10 <<<< answer 2
