<span>-3(1+6r)=14-r
-3 - 18r = 14 - r ...expand by using distributive property
-3 -17r = 14 ...add (r) to both sides
-17r = 17 ...add (3) to both sides
r = -1 ....divide both sides by (-17)</span>
Evaluate.
1934917632
this is evaluation
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.
Answer:
(-1,1,3)
Step-by-step explanation:
midpoint = (x1+x2/2, y1+y2/2, z1+z2/2)
midpoint = ( (-6+4)/2, (3+-1)/2, (4+2)/2 )
midpoint = (-2/2, 2/2, 6/2)
midpoint = (-1, 1, 3)
Answer:
x = 19
Step-by-step explanation:
Solve for x:
3 x + 43 = 100
Subtract 43 from both sides:
3 x + (43 - 43) = 100 - 43
43 - 43 = 0:
3 x = 100 - 43
100 - 43 = 57:
3 x = 57
Divide both sides of 3 x = 57 by 3:
(3 x)/3 = 57/3
3/3 = 1:
x = 57/3
The gcd of 57 and 3 is 3, so 57/3 = (3×19)/(3×1) = 3/3×19 = 19:
Answer: x = 19
