There can't be more than seven $5 bills, because they would be worth $33.
So, let's try all other cases and see which fits the requests:
- If there are six $5 bills, they are worth $30. You need three more $1 bills to reach $33. So, you have a total of 9 bills, which is not what we want.
- If there are five $5 bills, they are worth $25. You need eight more $1 bills to reach $33. So, you have a total of 13 bills, which is not what we want.
- If there are four $5 bills, they are worth $20. You need thirteen more $1 bills to reach $33. So, you have a total of 17 bills, which is not what we want.
- If there are three $5 bills, they are worth $15. You need eighteen more $1 bills to reach $33. So, you have a total of 21 bills, which is not what we want.
- If there are two $5 bills, they are worth $10. You need twenty-three more $1 bills to reach $33. So, you have a total of 25 bills, which is not what we want.
- If there is one $5, it is worth $5. You need twenty-eight more $1 bills to reach $33. So, you have a total of 29 bills, which is not what we want.
So, there's no way you can have 31 bills worth $1 and $5 that are worth $33 in total.
68.5 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111 is the answer fam
3(4m-2)-2(m+5)
Distribute.
12m-6-2m-10
12m-2m= 10m
-6-10= -16
10m-16
I hope this helps!
~kaikers
You would first do $120 (0.6) = $72
$72 + $120 = $192
Hope this helped! Good luck!!! :)
Radius = 15/2 = 7.5
Volume of cone: pi(r)^2*h/2
= pi(7.5)^2*23/2
pi(56.25) * 23/2
(3.14)(56.25)(23)/2 = 2,031 inches
