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Juliette [100K]
2 years ago
15

write an Equation for the angle relationship shown in the figure and solve for x. Find the measures of RQS and TQU?

Mathematics
1 answer:
Dvinal [7]2 years ago
4 0

Answer:

∠RQS = 21°

∠TQU = 28°

Step-by-step explanation:

90° + 221° + 3x + 4x = 360°

combine like terms:

7x + 311° = 360°

subtract 311° from each side of the equation:

7x = 49°

divide both sides by 7:

x = 7°

∠RQS = 3x = 3(7°) = 21°

∠TQU = 4x = 4(7°) = 28°

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Step-by-step explanation:

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Determine the above sequence converges or diverges. If the sequence converges determine its limit​
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Answer:

This series is convergent. The partial sums of this series converge to \displaystyle \frac{2}{3}.

Step-by-step explanation:

The nth partial sum of a series is the sum of its first n\!\! terms. In symbols, if a_n denote the n\!th term of the original series, the \! nth partial sum of this series would be:

\begin{aligned} S_n &= \sum\limits_{k = 1}^{n} a_k \\ &=  a_1 + a_2 + \cdots + a_{k}\end{aligned}.

A series is convergent if the limit of its partial sums, \displaystyle \lim\limits_{n \to \infty} S_{n}, exists (should be a finite number.)

In this question, the nth term of this original series is:

\displaystyle a_{n} = \frac{{(-1)}^{n+1}}{{2}^{n}}.

The first thing to notice is the {(-1)}^{n+1} in the expression for the nth term of this series. Because of this expression, signs of consecutive terms of this series would alternate between positive and negative. This series is considered an alternating series.

One useful property of alternating series is that it would be relatively easy to find out if the series is convergent (in other words, whether \displaystyle \lim\limits_{n \to \infty} S_{n} exists.)

If \lbrace a_n \rbrace is an alternating series (signs of consecutive terms alternate,) it would be convergent (that is: the partial sum limit \displaystyle \lim\limits_{n \to \infty} S_{n} exists) as long as \lim\limits_{n \to \infty} |a_{n}| = 0.

For the alternating series in this question, indeed:

\begin{aligned}\lim\limits_{n \to \infty} |a_n| &= \lim\limits_{n \to \infty} \left|\frac{{(-1)}^{n+1}}{{2}^{n}}\right| = \lim\limits_{n \to \infty} {\left(\frac{1}{2}\right)}^{n} =0\end{aligned}.

Therefore, this series is indeed convergent. However, this conclusion doesn't give the exact value of \displaystyle \lim\limits_{n \to \infty} S_{n}. The exact value of that limit needs to be found in other ways.

Notice that \lbrace a_n \rbrace is a geometric series with the first term is a_0 = (-1) while the common ratio is r = (- 1/ 2). Apply the formula for the sum of geometric series to find an expression for S_n:

\begin{aligned}S_n &= \frac{a_0 \cdot \left(1 - r^{n}\right)}{1 - r} \\ &= \frac{\displaystyle (-1) \cdot \left(1 - {(-1 / 2)}^{n}\right)}{1 - (-1/2)} \\ &= \frac{-1 +  {(-1 / 2)}^{n}}{3/2} = -\frac{2}{3} + \frac{2}{3} \cdot {\left(-\frac{1}{2}\right)}^{n}\end{aligned}.

Evaluate the limit \displaystyle \lim\limits_{n \to \infty} S_{n}:

\begin{aligned} \lim\limits_{n \to \infty} S_{n} &= \lim\limits_{n \to \infty} \left(-\frac{2}{3} + \frac{2}{3} \cdot {\left(-\frac{1}{2}\right)}^{n}\right) \\ &= -\frac{2}{3} + \frac{2}{3} \cdot \underbrace{\lim\limits_{n \to \infty} \left[{\left(-\frac{1}{2}\right)}^{n} \right] }_{0}= -\frac{2}{3}\end{aligned}}_.

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Answer:

See below

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