Answer:
A = 3
B = 1
C = 2
D = 10
E = 2
F = Not listed
G = 1
Step-by-step explanation:
![\sqrt[3]{270x^5y^7} =\sqrt[3]{27*10x^5y^7}=3xy^2\sqrt[3]{10x^2y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B270x%5E5y%5E7%7D%20%3D%5Csqrt%5B3%5D%7B27%2A10x%5E5y%5E7%7D%3D3xy%5E2%5Csqrt%5B3%5D%7B10x%5E2y%7D)
Answer:Your left hand side evaluates to:
m+(−1)mn+(−1)m+(−1)mnp
and your right hand side evaluates to:
m+(−1)mn+(−1)m+np
After eliminating the common terms:
m+(−1)mn from both sides, we are left with showing:
(−1)m+(−1)mnp=(−1)m+np
If p=0, both sides are clearly equal, so assume p≠0, and we can (by cancellation) simply prove:
(−1)(−1)mn=(−1)n.
It should be clear that if m is even, we have equality (both sides are (−1)n), so we are down to the case where m is odd. In this case:
(−1)(−1)mn=(−1)−n=1(−1)n
Multiplying both sides by (−1)n then yields:
1=(−1)2n=[(−1)n]2 which is always true, no matter what n is
total charge = days late * fine
3.77 = 13*fine
divide by 13 on each side
3.77 /13 = fine
.29
The charge is $.29 per day
0.05 kilogramos.................
