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BARSIC [14]
3 years ago
14

Describe the sequence of basic rigid motions that shoews S1 = S2

Mathematics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

S1 is congruent to S2

Step-by-step explanation:

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The following proportion could be used to change from gallons to which measure.
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Step-by-step explanation:

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2 years ago
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Could someone please help me:) I am stick and I am not sure what to do ​
Delicious77 [7]

Answer:

Part 5.1.1:

\displaystyle \cos 2A = \frac{7}{8}

Part 5.1.2:

\displaystyle \cos A = \frac{\sqrt{15}}{4}

Step-by-step explanation:

We are given that:

\displaystyle \sin 2A = \frac{\sqrt{15}}{8}

Part 5.1.1

Recall that:

\displaystyle \sin^2 \theta + \cos^2 \theta = 1

Let θ = 2<em>A</em>. Hence:

\displaystyle \sin ^2 2A + \cos ^2 2A = 1

Square the original equation:

\displaystyle \sin^2 2A = \frac{15}{64}

Hence:

\displaystyle \left(\frac{15}{64}\right) + \cos ^2 2A = 1

Subtract:

\displaystyle \cos ^2 2A = \frac{49}{64}

Take the square root of both sides:

\displaystyle \cos 2A = \pm\sqrt{\frac{49}{64}}

Since 0° ≤ 2<em>A</em> ≤ 90°, cos(2<em>A</em>) must be positive. Hence:

\displaystyle \cos 2A = \frac{7}{8}

Part 5.1.2

Recall that:

\displaystyle \begin{aligned}  \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ &=   1- 2\sin^2\theta \\ &= 2\cos^2\theta - 1\end{aligned}

We can use the third form. Substitute:

\displaystyle \left(\frac{7}{8}\right) = 2\cos^2 A - 1

Solve for cosine:

\displaystyle \begin{aligned} \frac{15}{8} &= 2\cos^2 A\\ \\ \cos^2 A &= \frac{15}{16} \\ \\ \cos A& = \pm\sqrt{\frac{15}{16}} \\ \\ \Rightarrow \cos A &= \frac{\sqrt{15}}{4}\end{aligned}

In conclusion:

\displaystyle \cos A = \frac{\sqrt{15}}{4}

(Note that since 0° ≤ 2<em>A</em> ≤ 90°, 0° ≤ <em>A</em> ≤ 45°. Hence, cos(<em>A</em>) must be positive.)

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3 years ago
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Hi! First, we multiply both sides by 5, -n/5*5=5*5=-n=25. Second, we divide both sides by -1, -n/-1=25/-1. Therefore, our answer is n=-25
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How much would $100 invested at 8% intrest compounded annually be worth after 15 years? round your answer to the nearest cent. d
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