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Alex787 [66]
3 years ago
12

What is the sin, cos, and tan of -495 degrees, and -20π/3  ??

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
7 0
Sin-495 = 0.98
cos-495 = 0.19
tan-495 = 4.95

sin<span>-20π/3</span> = 0
cos<span>-20π/3</span> = 0.3
tan<span>-20<span>π/3</span></span> = 0





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Step-by-step explanation:

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Given that P = (-7, 16) and Q = (-8, 7), find the component form and magnitude of vector QP .
Katyanochek1 [597]

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\vec {QP}=\binom{ - 8}{7} -  \binom { - 7}{15}

We subtract the corresponding components to obtain:

\vec {QP}=\binom{ - 8 -  - 7}{7 - 15}

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\vec {QP}=\binom{ - 1}{ - 8}

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|\vec {QP}| = \sqrt{ 1 +  64 }

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We add the corresponding components to get;

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