Answer:
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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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Answer:
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Answer:
Numbers that aren't: 35 and 43. It can literally be any number.
Perfect cubes: 8,64,125,216. You can find perfect cubes by cubing any number. For example 6^3 (6 to the power of 3) is 216. A perfect cube.
Answer:
140,120
Step-by-step explanation:
sum of angles in a quadrilateral =360
sum of the 1st two angles=60+40=100
360-100=260
the ratio of the remaining angles 7:6
therefore:
=140
for the second angle
260-140=120
