So for this, we can simplify √63 as such:
Using the simplified version of √63, we can solve it:
Zero is your final answer.
The second one is the right answer
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.
Answer: 4.5 looks the most reasonable to me.
Step-by-step explanation:
Answer:
The scientist needs 80 liters of 25% solution and 40 liters of 10% solution.
Step-by-step explanation:
If x is the liters of 25% solution, and y is the liters of 10% solution, then:
x + y = 120
0.25x + 0.10y = 0.20(120)
Solve the system of equations using substitution or elimination. Using substitution:
0.25x + 0.10(120 − x) = 0.20(120)
0.25x + 12 − 0.10x = 24
0.15x = 12
x = 80
y = 40
The scientist needs 80 liters of 25% solution and 40 liters of 10% solution.
