2. A spy uses a telescope to track a rocket launched vertically from a launch pad 6 km away. The rocket travels upwards at a velocity of 21.6 km per minute. Find the rate of change of the distance between the spy and the rocket when the rocket is 8 km in the air. 17.28 km/mim 3. An observer stands 40 feet away from the Ferry Building’s clock tower. Kells Phelan ‘18 (an apprentice steeplejack) is climbing to the top of the flagpole on top of the Ferry Building to inspect it. He maintains a velocity of 2 ft/sec. (Assume the scenario can be modeled as a right triangle) How fast is the observer to Kells’ distance changing when he is 280 feet above the Embarcadero? 1.98 ft/s

The rate of change in distance between the spy and the rocket is 10.8 km/min,

or about 180 m/sec.

Step-by-step explanation:

At "Time=0" the rocket is 6km from the spy.

The rocket travels vertically 8km in about 0.37 minutes, calculated by 8km÷21.6km/min.

At that point, the rocket is 10 km from the spy., calculated by 6²+8²=10²

The difference is 4 kiilometers over 0.37minutes, about 22.2 seconds

Divide Distance by Time to get Rate:

4km/0.37min = 10.8 km/min

That is about 180 meters per second.

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

add an image, and i will answer

step-by-step explanation: