, 24.01.2020chino54

# Arocket is dropping to the moons surface at a constant rate of 15 feet per second. after 2 minutes of descent the rocket is at a height of 11,400 feet above the moons surface. at what elevation in feet was the rocket when it started to drop to the moons surface?

2/4 is the answer

step-by-step explanation:

x= total pieces

both ate 9 pieces

mindy ate 3 so troy ate 6

6= (1/4)x

x =24

given : the shape of his satellite can be modeled by   where x and y are modeled in inches.

now,the given equation   is a equation of parabola.

here, coefficient of x is positive, hence the parabola opens rightwards.

on comparing this equation with standard equation , we get

in standard equation, coordinates of focus=(0,m)

thus for given equation coordinates of focus=(0,1.5)

step-by-step explanation:

13,200 feet

Step-by-step explanation:

So, the rocket starts descending at a given rate and time, so we first need to find the length it passed during that descent.

Length equals to speed multiplied with time:

s = v • t

since speed is given in feet per second, we need to convert minutes to seconds; if one minute has 60 seconds, then two minutes have 120 seconds.

s = 15 feet/s • 120 s

s = 1800 feet

So, the rocket descended 1800 feet, but it still has 11,400 feet to the Moon's surface. That means that the rocket was on 11,400 + 1,800 = 13,200 feet elevation when it started to drop.

Do you know the answer?

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