![\bf sin(x)[csc(x)-sin(x)]~~=~~cos^2(x) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(x)\left[\cfrac{1}{sin(x)}-\cfrac{sin(x)}{1} \right]\implies \underline{sin(x)}\left[\cfrac{1-sin^2(x)}{\underline{sin(x)}} \right] \\\\\\ 1-sin^2(x)\implies cos^2(x)](https://tex.z-dn.net/?f=%5Cbf%20sin%28x%29%5Bcsc%28x%29-sin%28x%29%5D~~%3D~~cos%5E2%28x%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20sin%28x%29%5Cleft%5B%5Ccfrac%7B1%7D%7Bsin%28x%29%7D-%5Ccfrac%7Bsin%28x%29%7D%7B1%7D%20%5Cright%5D%5Cimplies%20%5Cunderline%7Bsin%28x%29%7D%5Cleft%5B%5Ccfrac%7B1-sin%5E2%28x%29%7D%7B%5Cunderline%7Bsin%28x%29%7D%7D%20%5Cright%5D%20%5C%5C%5C%5C%5C%5C%201-sin%5E2%28x%29%5Cimplies%20cos%5E2%28x%29)
recall again, sin²(θ) + cos²(θ) = 1.
Either way. The probability of hitting the circle is:
P(C)=Area of circle divided by area of square
P(W)=(area of square minus area of circle divided by area of square
P(C)=(πr^2)/s^2
P(W)=(s^2-πr^2)/s^2
...
Okay with know dimensions, r=1 (because r=d/2 and d=2 so r=1), s=11 we have:
P(inside circle)=π/121 (≈0.0259 or 2.6%)
P(outside circel)=(121-π)/121 (≈0.9744 or 97.4%)
Answer:
34.3948
Step-by-step explanation:

Answer:
x axis? is that what your asking
Step-by-step explanation: