Agirl wants to count the steps of a moving escalator which is going up. if she is going up on it, she counts 60 steps. if she is walking down, taking the same time per step, then she counts 90 steps. how many steps would she have to take in either direction, if the escalator were standing still?

Let

V the speed of the girl per step
v the speed of the escalator per step
L the length between ground and floor in number of steps

V*60+v*60=L => V+v=L/60
V*90-v*90=L => V-v=L/90

By adding the equations V+v=L/60 and V-v=L/90 we get:

2V=L(3+2)/180

V=5L/360

V = L/72

Time to climb (in steps) with v=0 (escalator standing still) is:

V*t=L

t = L/V

t = L/(L/72)

t =72 steps

The girl will count 72 steps.

this question relies purely on our observation of the graph and a knowledge of the behaviour of a graph of a hyperbola.

now, what we should realise is that there are two asymptotes - one at x = 0 and one at y = 0. what this means is that the graph will gradually approach, but never touch, these two lines. so, as x becomes more positive (ie. 'a very large positive number'), it will get closer to the line y = 0 (ie. the x-axis) - which means that the value of f(x) will gradually become smaller and closer to 0 - but will never reach it.

we can also observe this from the right half of the graph, where the graph gradually lowers closer and closer to the x-axis as x increases. what is important to note however is that it will never actually touch the x-axis (ie. f(x) will not equal 0) since there is an asymptote there. given this, we can see that option d, f(x) is a very small positive number, would be correct.

step-by-step explanation: