Answer:
The answer to your question is L = 22; w = 15
Step-by-step explanation:
Data
length = L
width = w
Perimeter = 74
Condition
w = L/2 + 4
Formula
P = 2L + 2w
Substitution
74 = 2L + 2(L/2 + 4)
Simplification
74 = 2L + L + 8
74 = 3L + 8
74 - 8 = 3L
3L = 66
L = 66/3
L = 22
w = 22/2 + 4
w = 11 + 4
w = 15
The answer is
5mn(KUASA DUA) -6gh(KUASA DUA) +2
Answer a
Step-by-step explanation: 67 x 3
A. Meter because it is the only one that is metric (which is used internationally, the I in ISU or International System of Units) the rest are usually only used here in the state’s and honestly aren’t as exact either.
9514 1404 393
Answer:
a) E = 6500 -50d
b) 5000 kWh
c) the excess will last only 130 days, not enough for 5 months
Step-by-step explanation:
<u>Given</u>:
starting excess (E): 6500 kWh
usage: 50 kWh/day (d)
<u>Find</u>:
a) E(d)
b) E(30)
c) E(150)
<u>Solution</u>:
a) The exces is linearly decreasing with the number of days, so we have ...
E(d) = 6500 -50d
__
b) After 30 days, the excess remaining is ...
E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days
__
c) After 150 days, the excess remaining would be ...
E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system
The supply is not enough to last for 5 months.