The answer would be <span>-10.125</span>
Answer:
2/3a^5
Step-by-step explanation:
we know that 2/7x2/7x2/7 = 8/27
a^15 cube will be a^5
because we know (2^3)^2 = 2^6
so (a^5)^3 = a^15
2/3a^5
We have to assume that the speed before being stuck was sufficient to get to the destination on time had there been no delay. Call that speed "s" in km/h.
Since 200 km is "halfway", the total distance must be 400 km.
time = distance / speed
total time = (time for first half) + (delay) + (time for second half)
400/s = 200/s + 1 + 200/(s+10) . . . .times are in hours, distances in km
200/s = 1 + 200/(s+10) . . . . . . . . . . subtract 200/s
200(s+10) = s(s+10) +200s . . . . . . .multiply by s(s+10)
0 = s² +10s - 2000 . . . . . . . . . . . . . .subtract the left side
(s+50)(s-40) = 0
Solutions are s = -50, s = 40
The speed of the bus before the traffic holdup was 40 km/h.
The answer should be 4x+ 17 > x - 1
