1st digit: 9 possibilities (can’t have 0)
2nd digit: 10 possibilities
3rd digit:10
4th digit:10
5th digit:10
6th digit: 4 possibilities (it can only be even)
9•10•10•10•10•4 =360,000 possibilities
Answer:
No it's 2 Lol not 929 UwU
X= 30
I don’t know if this is correct or not
There are two equations and two variables. You can solve for each variable using either the elimination method or the substitution method. Here, I believe the elimination method would be best:
5a = -4b + 5
3a = -2b + 3
You can multiply the second equation by two, so the 4b and -4b in both equations will cancel each other out when you add them. So:
5a = -4b +5
6a = 4b +6
Add both equations together.
11a = 11
a = 1
Plug in the a value into any of the previous equations:
5(1) = -4b + 5
-4b = 0
b = 0
So, since we know that b = 0, 6b is also 0.
Answer:
I think you would do $9.70 times 2. Afterward, you would divide 9.70 by 5 which gives you 1.94. Finally, you add them together, leaving you with. 11.64. I hope I helped with this.
:D
