Hello from MrBillDoesMath!
Answer:
11/ 1009 (.0109 approximately)
Discussion:
99/9081 = => as 9081 = 9 * 1009 = 3 * 3 * 1009
99/ ( 3 * 3 * 1009) =
(99/(3*3)) * (1/1009) =
(99/9) * (1/1009) =
11/ 1009 =
.0109 approximately
Thank you,
MrB
B. x > 70
is the correct answer
Deijah has $6.25.
Now, this 6.25 is comprised of 0.05s and 0.25s.
We know that there are 12 0.25s.
We now want to know how many remaining 0.05s there are.
Again, we know that the number of 0.05s he has, which is 12, multiplied by 0.05, plus the number of 0.25s he has, multiplied by 0.25, equals 6.25.
Thus, the answer is A, 0.25 x 12 + 0.05 x n = 6.25.
.17 because there is a zero behind the 7! .170 and .165. .17 is bigger
(1) R+G+B=50
(2) R=B+6
(3) G=B-4 so substitute G in (1)
R+B-4+B=50 substitute (2) in here
B+6+B-4+B=50 group like terms and solve
3B+2=50
3B=50-2
B=48/3=16 substitute in (2) and (3)
R=16+6=22
G=16-4=12
Check
22+12+16=50✅
