Good evening,
Answer:
x > 5
x ≤ 12
Step-by-step explanation:
Visualize as if the inequality symbols are equal signs and then solve it as you would for a normal equation, note if you divide by two negative numbers at all you then switch the sign.
On a number line, an inequality sign without the “equal to” is plotted with an open circe.
An inequality sign, with the “equal to” is plotted with a closed circle.
3x - 7> 8
Add seven on both sides, we do this because we want to eliminate the 7 from one side.
3x > 15
Divide both sides by 3, we do this because you want to get rid of the 3 from x.
x > 5
As for the second inequality.
Divide both sides by -3, as I mentioned earlier switch the side since we are dividing by two negative numbers.
x ≤ 12
For the greater than inequality x > 5, plot a open circle on 5 and draw the line going to the right.
For the less than or equal to inequality x ≤ 12, plot a closed cirlce on 12 and draw the line going to the left.
A unique trick is to graph the line based on the direction (right or left) the inequality symbol is pointing to.
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
