The question is in the photo

9514 1404 393

  108.8 km/h at 84.3°

Step-by-step explanation:

The law of cosines can be used to find the resultant ground speed. In the attached diagram, the length of interest is OR. It will be found as ...

  OR² = OP² +PR² -2·OP·PR·cos(P)

  OR² = 130² +26² -2×130×26×cos(32°) ≈ 11843.19

  OR = √11843.19 ≈ 108.8

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The angle POR can be found from the law of sines.

  sin(POR)/PR = sin(OPR)/OR

  sin(POR) = PR/OR×sin(32°) ≈ 0.12660

  ∠POR ≈ arcsin(0.12660) ≈ 7.27°

Then the bearing of the ground track of the airplane is 77° +7.27° = 84.27°.

The airplane is traveling at about 108.8 km/h on a bearing of 84.3°.

answer: 48

just remember to use pemdas:

parentheses

expressions

multiplication

division

addition

subtraction

step-by-step explanation:

what are the statements? and what do you mean by "216 2 1 6"?