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
Given that
z₁ = 15 (cos(90°) + i sin(90°))
z₂ = 3 (cos(10°) + i sin(80°))
we get the quotient z₁/z₂ by dividing the moduli and subtracting the arguments:
z₁/z₂ = 15/3 (cos(90° - 10°) + i sin(90° - 10°))
z₁/z₂ = 5 (cos(80°) + i sin(80°))
so that z₁ is scaled by a factor of 1/3 and is rotated 10° clockwise.
Answer:
56 is the answer to your question
7 + 3(4) - 2
7 + 12 - 2
19 - 2
= 17
Answer:
B
Step-by-step explanation:
n is how many you buy times the price of $0.85 and then you subtract a coupon for $0.99 for the entire purchase.
