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Ulleksa [173]
2 years ago
8

Helpppppp plzzzzzzzzzzzzzzzz

Mathematics
2 answers:
dem82 [27]2 years ago
4 0

Answer:

the picture isnt loading. it says error.

Step-by-step explanation:

Arisa [49]2 years ago
3 0

Answer:

complementary supplementary

1 and 2 9 and 10

4 and 5 10 and 11

neither

1 and 6

3 and 4 6 and 7

9 and 11

hope this helps

good day mate

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For questions 3 – 5, an airplane is heading south at an airspeed of 540 km/hr, but there is a wind blowing from the northeast at
GuDViN [60]

First, let's calculate the horizontal and vertical components of the wind speed (W) and the airplane speed (A), knowing that south is a bearing of 270° and northeast is a bearing of 45°:

\begin{gathered} W_x=W\cos45°\\ \\ W_x=50\cdot0.707\\ \\ W_x=35.35\\ \\ \\ \\ W_y=W\sin45°\\ \\ W_y=50\cdot0.707\\ \\ W_y=35.35 \end{gathered}\begin{gathered} A_x=A\cos270°\\ \\ A_x=540\cdot0\\ \\ A_x=0\\ \\ \\ \\ A_y=A\sin270°\\ \\ A_y=540\cdot(-1)\\ \\ A_y=-540 \end{gathered}

Now, let's add the components of the same direction:

\begin{gathered} V_x=W_x+A_x=35.35+0=35.35\\ \\ V_y=W_y+A_y=35.35-540=-504.65 \end{gathered}

To find the resultant bearing (theta), we can use the formula below:

\begin{gathered} \theta=\tan^{-1}(\frac{V_y}{V_x})\\ \\ \theta=\tan^{-1}(\frac{-504.65}{35.35})\\ \\ \theta=-86° \end{gathered}

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