Jack and Jill both start at point A. They each walk in a straight line at an angle of 105° to each other. After 45 minutes Jack has walked 4.5km and Jill has walked 6km. How far apart are they? Round to the nearest hundredth.

The distance between them is 8.38 km

Step-by-step explanation:

Let's suppose both of them start at the point (0km, 0km), and that Jack walks along the positive x-axis, then Jill walks at an angle of 105° measured from the positive x-axis.

After 45 minutes, jack has walked 4.5km, then if his initial position was (0 km, 0 km)

Then his new position will be (4.5km, 0km)

Jill has walked 6km, but we need to write this in rectangular components.

We can think in this as a triangle rectangle, then the components will be:

x-component = 6 km*cos(105°) = -1.55 km

y-component = 6km*sin(105°) = 5.80 km

Then the new position of Jill is ( -1.55km, 5.80km)

Now, we know that the distance between two points (a, b) and (c, d) is:

Distance = √( (a - c)^2 + (b - d)^2)

Then the distance between Jill and Jack will be equal to the distance between ( -1.55km, 5.80km) and (4.5km, 0km)

This is:

Distance = √( (-1.55km - 4.5km)^2 + (5.80km - 0km)^2)

Distance = 8.38 km

This means that they are 8.38km apart.

answer: yes, since the line has a steeper rate of change than the exponential the y-values will be greater and they will only intersect one time.