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A point P is 45km from Q on a bearing of 74'. How far is P north of Q
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A scuba diver dove from the surface of the ocean to an elevation of −59
9/10 feet at a rate of −15 feet per minute. After
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If 3sin tita - cos tita=0,how do I find the value of 3tan tita=1(0° less than or equal to tita less than or equal to 360°)
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Pls I need answers
A point p is 45km from Q on a bearing of 75'.How far is P north of Q
100%=98g/mol=1mol 2%=X mol Therefore X=(2×1)÷100 X=0.02