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
brainly.com/question/8783264
Answer:
- 3 \frac{4}{5} + 1 \frac{2}{5}
= -frac{19}{5} +frac{7}{5}
= frac{-19+7}{5}
= frac{-12}{5}
= -2frac{2}{5}
5 \frac{3}{5} - 7
= -2+\frac{3}{5}
= frac{-10+3}{5}
= frac{-7}{5}
= -1\frac{2}{5}
Step-by-step explanation:
Answer:
Option C.
Step-by-step explanation:
Answer:
326
Step-by-step explanation:
l x w
7x8
25x6
4x30
326
Answer: (w² + 10)inches²
Step-by-step explanation:
Since the width of the rectangular blanket is given as w and the length of the blanket is 10inches greater than it's width, therefore the length will be:
= 10 + w.
Therefore,
length = 10 + w
width = w
Area = length × width
Area = (10 + w) × w
Area = 10w + w²
Therefore, the area of the blanket will be (w² + 10)inches².
