Answer: 10:55
Step-by-step explanation:
Taking statement at face value and the simplest scenario that commencing from 08:00am the buses take a route from depot that returns bus A to depot at 25min intervals while Bus B returns at 35min intervals.
The time the buses will be back at the depot simultaneously will be when:
N(a) * 25mins = N(b) * 35mins
Therefore, when N(b) * 35 is divisible by 25 where N(a) and N(b) are integers.
Multiples of 25 (Bus A) = 25, 50, 75, 100, 125, 150, 175, 200 etc
Multiples of 35 (Bus B) = 35, 70, 105, 140, 175, 210, 245 etc
This shows that after 7 circuits by BUS A and 5 circuits by Bus B, there will be an equal number which is 175 minutes.
So both buses are next at Depot together after 175minutes (2hr 55min) on the clock that is
at 08:00 + 2:55 = 10:55
Step-by-step explanation:
F(x) = ∫ₐˣ t⁷ dt
F(x) is the area under f(t) between t=a and t=x. When x=a, the width of the interval is 0, so the area is zero.
F(6) = 0, so a = 6.
F(x) = ∫₆ˣ t⁷ dt
F(6) = ∫₆⁶ t⁷ dt
F(6) = 0
Answer:
4
Step-by-step explanation:
2L=w
32=LxW
so you put the 2L as the w
which is 32=Lx2L
32=2L^2
16=L^2
4=L
*To find the width you input 4 as L*
So 8=W
The shortest side would be 4
