I NEED THE ANSWER NOW PLEASE IM DOING USA TEST PREP PLEASE NOW!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
x = 0 and y = 3
Step-by-step explanation:
Using elimination method, we can solve for x and y in the simultaneous equation
y = 2x + 3 .................................eqn 1
y = -3x + 3 .................................eqn 2
multiply equation 1 by 3 and equation 2 by 2, then we will have
(y = 2x + 3) × 3
(y = -3x + 3) × 2
3y = 6x + 9 ...............................eqn 3
2y = -6x + 6 ................................eqn 4
Add eqn 3 and 4 together and we will eliminate x and have
5y = 15
Divide both sides by 5, and we will have
y = 3
To solve for x, substitute 3 for y in eqn 1
y = 2x + 3
3 = 2x + 3
collect like terms
3 - 3 = 2x
0 = 2x
Therefore x = 0
x <-3 includes the numbers-4, -5, -6..........
x> 5 includes the numbers 6, 7 , 8.....
here we have to find intersection of these two.
By intersection we mean the values of x that occur in both the sets,
here there is no number or value of x that occur in both the sets.
So answer is empty set
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
