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damaskus [11]
3 years ago
8

Which set of ordered pairs could be generated by an exponential function? A)(negative 1, negative one-half), (0, 0), (1, one-hal

f), (2, 1) B) (–1, –1), (0, 0), (1, 1), (2, 8) C) (negative 1, one-half), (0, 1), (1, 2), (2, 4) D)(–1, 1), (0, 0), (1, 1), (2, 4)
Mathematics
2 answers:
Vadim26 [7]3 years ago
3 0

Answer:

<h2>C) (negative 1, one-half), (0, 1), (1, 2), (2, 4)</h2>

Step-by-step explanation:

The set C could be generated by an exponential function. The main reason is that exponential functions hava a restricted range, it can't have negative numbers or the number zero, because power can only be equal or greater than 1.

Additionally, for all exponentials, a null exponent gives 1 as an answer, so point (0, 1) is always present in an exponential function.

Therefore, the right answer is C.

Tamiku [17]3 years ago
3 0

Answer:

C

Step-by-step explanation:

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Determine the above sequence converges or diverges. If the sequence converges determine its limit​
marshall27 [118]

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}}_.

Therefore, the partial sum of this series converges to \displaystyle \left(- \frac{2}{3}\right).

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Step-by-step explanation:

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2. 1/3 of 1/2 of the total amount of stock is worth $2700. That is 1/6 of the stock, so the worth of all the stock before the sale must be 6 times $2700, or $16,200.

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