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
Answer:
This is actually not an identity.
Answer:
4 hair bow
Step-by-step explanation:
if 1/4 can make 1 hair bows you can make 4 hair bows
, it's simple 4/1=4
Answer:
20
Step-by-step explanation:
Let's start by rewriting the second equation in terms of "x":
Subtract y from both sides:
Now, substitute "5-y" for "x" in the first equation:
Note that:
Cancel out like terms:
Subtract 25 from both sides:
Divide both sides by -10
Now, substitute this value back into either of the equations to solve for x.
Add 15/2 to both sides:
Now, find the difference:
Note:xy means x times y and x(y) means the same thing
so
first we get rid of square root then
make the equation equal to zero becaues if
xy=0 then x or/and y=0
squareroot(y-1)+3=y
isolate the squareroot
subtrac 3 from boht sides
squareroot(y-1)=y-3
square both sides (since they are equal, you should be able to square both sides and still make it true)
(squareroot(y-1))^2=(y-3)^2
(y-1)=(y-3)(y-3)
y-1=y^2-6y+9
subtrac y from both sides
-1=y^2-7y+9
add 1 to both sides
0=y^2-7y+10
find what two number multiply to make 10 and add to get -7
the answer is -2 and -5
0=(y-5)(y-2)
therfore
y-5=0
and/or
y-2=0
therefor
y=5 or/and 2 might work
let's try out 2
square root(2-1)+3=2
square root(1)+3=2
1+3=2
false
so 2 doesn't work
let's try 5
squareroot(5-1)+3=5
squareroot(4)+3=5
2+3=5
5=5
true
y=5
