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krok68 [10]
4 years ago
12

Does anyone know how to do this?

Mathematics
1 answer:
Anton [14]4 years ago
7 0
For 1. −39.98
2. −23.64
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An angle is drawn in standard position passing through the unit circle at (0.643,0.766). The angle in standard position θ has a
Flura [38]
<h2>Hello!</h2>

The asnwer is: Cos50° ≈ 0.643

<h2>Why?</h2>

A unit circle is a cirgle with a radius equal of 1, knowing that,  also know the following:

The angle is drawn passing trough the unit circle at (0.643,0.766) it means that:

x=0.643\\y=0.766

So,

Cos50° ≈ 0.643

We can prove that by following the next steps:

- If it's a unit circle,here is a right triangle with hypotenuse of 1,

1^{2}=x^{2}+y^{2}

1^{2}=0.643^{2}+0.766^{2}

1^{2}=0.4134 +0.5867

1=1.0001=1

- We can determine the cosine of the angle by the following formula:

cos(\alpha)=\frac{x}{hypotenuse} \\cos(\alpha)^{-1}=cos(\frac{0.643}{1})^{-1} \\\alpha=49.98

Therefore,

Cos(α)=49.98°≈50°

Also, if there is a right triangle, according to the Pythagorean Thorem:

1^{2}=(Cos(\alpha))^{2}+(Sin(\alpha))^{2} \\Cos(50)=\sqrt{1-(Sin(50))^{2}}=0.6427

Hence,

Cos50° ≈ 0.643

Have a nice day!

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\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-2}x^2\stackrel{\stackrel{b}{\downarrow }}{+2}x\stackrel{\stackrel{c}{\downarrow }}{+3} \qquad \qquad  \left(-\cfrac{ b}{2 a}~~~~ ,~~~~  c-\cfrac{ b^2}{4 a}\right)


\bf \left(-\cfrac{2}{2(-2)}~~,~~3-\cfrac{2^2}{4(-2)}  \right)\implies \left( \cfrac{1}{2}~~,~~3+\cfrac{4}{8} \right)\implies \left(\cfrac{1}{2}~~,~~\cfrac{28}{8}  \right) \\\\\\ \left(\cfrac{1}{2}~~,~~\cfrac{7}{2}  \right)\implies \left(\frac{1}{2}~,~3\frac{1}{2}  \right)

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