It’s the first one I believe.
Sin2x = 2sinxcosx;
cos2x = (cosx)^2 - (sinx)^2;
tan2x = (sin2x)/(cos2x);
cosx = 5/13 from formula (sinx)^2 + (cosx)^2 = 1;
=> sin2x = 120/169;
.................................
A.


cost is $16.50
B.

rearrange to solve for t:

C. This is the same as part B, only the rate changed

they ask for the greatest number of minutes, so we round down (take the floor of the value).
the answer is 105 minutes.
x = 3.
Plug in 3 for x in the equation
14(3) + 64
Remember to follow PEMDAS. First, multiply 3 with 14
3 x 14 = 42
Finally, add 64
42 + 64 = 106
106 is your answer
hope this helps
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)