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:
Step-by-step explanation:
Two similar triangles.
10/26 = y/13
y = 5
x² = 13² - y² = 144
x = 12
9.675
Move the decimal back one spot
Mark brainliest please
Used show you have more than what is needed. Example is you need 6 slices of pizza for all your friends but have 7 slices you have more than is needed.
Answer:
a)

b) 0.09
Step-by-step explanation:
We are given the following in the question:

where B(t) gives the brightness of the star at time t, where t is measured in days.
a) rate of change of the brightness after t days.

b) rate of increase after one day.
We put t = 1

The rate of increase after 1 day is 0.09