It's the zero in the tenth place
Bear in mind that, when it comes to trigonometric functions, the location of the exponent can be a bit misleading, however recall that sin²(θ) is really [ sin( θ )]²,
![\bf 2sin^2(2x)=2\implies sin^2(2x)=\cfrac{2}{2} \\\\\\ sin^2(2x)=1\implies [sin(2x)]^2=1\implies sin(2x)=\pm\sqrt{1} \\\\\\ sin(2x)=\pm 1\implies sin^{-1}[sin(2x)]=sin^{-1}(\pm 1)](https://tex.z-dn.net/?f=%5Cbf%202sin%5E2%282x%29%3D2%5Cimplies%20sin%5E2%282x%29%3D%5Ccfrac%7B2%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%5E2%282x%29%3D1%5Cimplies%20%5Bsin%282x%29%5D%5E2%3D1%5Cimplies%20sin%282x%29%3D%5Cpm%5Csqrt%7B1%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%282x%29%3D%5Cpm%201%5Cimplies%20sin%5E%7B-1%7D%5Bsin%282x%29%5D%3Dsin%5E%7B-1%7D%28%5Cpm%201%29)
Answer:

Step-by-step explanation:
we know that
In the right triangle ABC
The function sine of angle 68 degrees is equal to divide the opposite side SB by the hypotenuse AC
so

substitute the values and solve for AC



50 + x ≥ 268
X ≥ 218
STEP BY STEP WORK
50 + x ≥268
-50 -50
x ≥218
Jan needs to save at least $218 for camp
Hope I helped :)