The number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
- Since one student runs at a speed of 180 feet per minutes, in time t minutes, he moves a distance of d = 180t.
- Also, the second student runs at a speed of 160 feet per minute, in time t minutes, he moves a distance of d' = 160t.
Since they are initially 10 feet apart, their total distance apart after t minutes is D = d + 10 + d'
D = 180t + 10 + 160t
D = 340t + 10
<h3>Number of minutes before they are 1870 feet</h3>
Making t subject of the formula, we have
t = (D - 10)/340
Since they are 1870 feet apart after t minutes, D = 1870 feet.
t = (D - 10)/340
t = (1870 - 10)/340
t = 1860/340
t = 5.47 minutes
So, the number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
Learn more about minutes of distance apart here
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Dear!
In my best knowledge,your ans is b.
Answer:
A, c(excluding tip, if the tip is included then c is not the answer)
Step-by-step explanation:
I solved the Equations:
A)1/8(x+16)=76/8----60 (this is the correct answer!)
B)1/8(x+16=76)------ 592 (NOT the correct answer
c)x=60 Represents total food bill excluding tip
d) x=60 does represent each friends share of the food bill
E) not correct
27 hours in the maximum hours
Answer:
It would be the 4th one and the last one I think
Step-by-step explanation:
