The point (1,0) lies on the curve y=sin(10π/x).

A) if Q is the point (x,sin(10π/x), find the slope of the secant line PQ.

Points are 2,1.5,1.4,1.3,1.2,1.1,0.5,0.6,0.7,0.8,0.9

Do slopes appear to be approaching a limit?

There is no 10π/x on the unit circle..... So would you put it in this format?
2, 10π/2?
For point 2 the y equals to 1.7321? How does this happen?

don't use the graph of the unit circle

instead look at the graph of the function sin((10ð/x)

(1,0) lies on the curve because sin((10ð) = 0

for x = 2, y = sin(5ð), which is 0 NOT 1.7321

you want the points

(2, sin10π/2),(1.5, sin10π/1.5) and so on

Do not expect the slope of the secant to approach a limit. It will oscillate between values approaching -1 and 1

No the answer is 1.7321 if x = 2 then y is 1.7321 that is the answer in the back

The sin function range is between -1 and +1

1.7321 is not the sin of any angle.

sin(10π/2) = sin(5π) = 0

the slope of the line joining (1,0) and (2,0) is 0

something is wrong with this picture.

The slope of the line from the point (1.5,0.866) to the point (2,0) is -1.7321.

But the value of the function y=sin(10Pi/x) at x=2 is y=0.

The sine function will always return 0 when given an integer number times Pi, and 10/2 is an integer.

To find the slope of the secant line PQ, we need to select another point on the curve and calculate the rise over run between the two points.

Let's choose the point P(2, 1.7321) as the other point. To find this point, we can substitute x=2 into the equation y=sin(10π/x):

y = sin(10π/2) = sin(5π) = 0

Therefore, the point P is (2, 0).

Now, we can calculate the slope of the secant line PQ:

Slope = (Change in y)/(Change in x)
= (0 - 0)/(2 - 1)
= 0/1
= 0

So, the slope of the secant line PQ is 0.

Regarding the second part of your question, the given curve y = sin(10π/x) does not directly relate to the unit circle. The values of y for different x values are calculated using the sine function, which is a trigonometric function. The point (2, 1.7321) is the result of evaluating the sine function at x=2, which gives the y-coordinate on the curve. It is not related to the unit circle directly.