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Zinaida [17]
2 years ago
10

Given the formula:

" title="x = \frac{1 + a}{1 - a} " alt="x = \frac{1 + a}{1 - a} " align="absmiddle" class="latex-formula">
Solve for "a" when x=12.​
Mathematics
1 answer:
shtirl [24]2 years ago
8 0

12 = \frac{1+a}{1-a} => 12(1-a)= 1+ a \\=> 12 -12a = 1 + a => 12a+1 = 12-1 \\=> 13a = 11 => a= \frac{11}{13}

ok done. Thank to me :>

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Fr33 points 38/827. 72/727
Virty [35]

Answer:

thx

Step-by-step explanation:

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2 years ago
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Marco had sales of $28,000. He makes a graduated commission of 4% on the first $10,000, 5% on the next $20,000 and 6% on anythin
Paha777 [63]

Answer:

$1,300

Step-by-step explanation:

commission on the first $10,000

= 4/100 * 10000

= $400

remaining amount

= $28,000 - $10,000

= $18,000

commission on remaining $18,000

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total commission

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= $1,300

7 0
2 years ago
What is the completely factored form of x3-64?
nataly862011 [7]
<span>a^3 - b^3 = (a - b) </span><span>(<span>a^2 + </span>ab + b^2)
</span><span>so
x^3 - 4^3 = (x - 4)(x^2 + 4x + 16)

hope it helps</span>
7 0
2 years ago
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At Pizza Pi, 32% of the pizzas made last week had extra cheese. If 16 pizzas had extra cheese, how many pizzas in all were made
igor_vitrenko [27]

Answer:

50

Step-by-step explanation:

you need to put  first the 32 on top of 100 since it always goes over 100 then you put 16 over a blank and you divide 32 by 16 which equals 2 so since you divided at the top you need to divide at the bottom so you divide 100 by 2 and you get 50 so 50 is your answer

7 0
3 years ago
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).

8 0
3 years ago
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