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Leona [35]
3 years ago
10

1.) A.) tan 50° B.) cot 50° C.) sin 50° D.) cos 50° E.) cot 50°

Mathematics
2 answers:
Elina [12.6K]3 years ago
5 0

Answer:

cot 50 is thr answer

Step-by-step explanation:

Elan Coil [88]3 years ago
4 0

Answer:

\cot 50 \textdegree

Step-by-step explanation:

\cos\theta=\frac{Adjacent}{Hypotenuse}\\ \\\sin\theta=\frac{Opposite}{hypotenuse}\\ \\\tan\theta=\frac{Opposite}{Adjacent}\\ \\\cot\theta=\frac{Adjacent}{Opposite}\\ \\\frac{\cos\theta}{\sin\theta}=\frac{\left(\frac{Adjacent}{Hypotenuse}\right)}{\left(\frac{Opposite}{Hypotenuse}\right)}=\frac{Adjacent}{Opposite}=\cot\theta\\ \\\Rightarrow \frac{\cos\theta}{\sin\theta}=\cot\theta\\ \\Substitute \ \theta=50\textdegree\\ \\\frac{\cos50\textdegree}{\sin50\textdegree}=\cot50\textdegree

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Wittaler [7]

We are given equation :x^6+6x^3+5=0

Use\:the\:rational\:root\:theorem

\mathrm{Therefore,\:we\:need\:to\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:5}{1}

-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1

\mathrm{Compute\:}\frac{x^6+6x^3+5}{x+1}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }x^5-x^4+x^3+5x^2-5x+5

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x^6\:+\:6x^3\:+\:5=\left(x+1\right)\left(x^5-x^4+x^3+5x^2-5x+5\right)

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5 0
3 years ago
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Answer:

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