Mathematics, 05.08.2019 00:30 asialoggin21

If x+3 and x-2 are the first two terms of a geometric series, find the values of x for which the series converges

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Answer: when $x \ \textgreater \ -\frac{1}{2}$ , the series will converge.sa1======since this is a geometric ratio, figure out the common ratio.the common ratio is the ratio between a term and the term before it. divide x-2 by the term before it, which is x+3. $r=\dfrac{x-2}{x+3}$ for a geometric series to be convergent, the common ratio r must be between -1 and 1, or $|r| \ \textless \ 1$ . $\begin{aligned} |r| & \ \textless \ 1 \\ \left|\dfrac{x-2}{x+3}\right| & \ \textless \ 1 \\ \dfrac{|x-2|}{|x+3|} & \ \textless \ 1 \end{aligned}$ we can multiply both sides by |x+3| because |x+3| is always positive. there are no inequality flips. $\begin{aligned} |x-2| \ \textless \ |x+3| \end{aligned}$ these absolute values represented in piecewise format: $\begin{aligned} |x-2| & = \left\{ \begin{array}{rl} x-2 & \text{if } x -2 \ge 0 \\ -(x-2) & \text{if } x -2 \ \textless \ 0 \end{array}\right. \\ & = \left\{ \begin{array}{rl} x-2 & \text{if } x \ge 2 \\ -(x-2) & \text{if } x \ \textless \ 2\end{array}\right. \end{aligned}$ and $\begin{aligned} |x+3| & = \left\{ \begin{array}{rl} x+3 & \text{if } x +3 \ \textgreater \ 0 \\ -(x+3) & \text{if } x +3 \ \textless \ 0\end{array}\right. \\ & = \left\{ \begin{array}{rl} x+3 & \text{if } x \ \textgreater \ -3 \\ -(x+3) & \text{if } x \ \textless \ -3\end{array}\right. \end{aligned}$ consider all possibilities. the left of x=-3 (and at x=3), between x=-3 and x=2, and the right of x=2 (and at x=2) are areas to check for since they change the signs of the piecewise expressions.case 1: when $x \le -3$ , then $|x+3| = -(x+3)$ and $|x-2| = -(x-2)$ therefore $\begin{aligned} |x-2| & \ \textless \ |x+3| \\ -(x-2) & \ \textless \ -(x+3) \\ -x + 2 & \ \textless \ -x -3 \\ 2 & \ \textless \ -3 \end{aligned}$ the last step was achieved by adding x to both sides.this is always a false statement. we obtained no solution here.case 2: when $-3\ \textless \ x\ \textless \ 2$ $|x+3| = x+3$ and $|x-2| = -(x-2)$ therefore $\begin{aligned} |x-2| & \ \textless \ |x+3| \\ -(x-2) & \ \textless \ x + 3 \\ -x + 2 & \ \textless \ x + 3 \\ -2x + 2 & \ \textless \ 3 \\ -2x & \ \textless \ 1 \\ x & \ \textgreater \ -\tfrac{1}{2} \end{aligned}$ but this case is only for $-3 \ \textless \ x \ \textless \ 2$ the intersection between $x \ \textgreater \ -\frac{1}{2}$ and $-3 \ \textless \ x \ \textless \ 2$ is $-\frac{1}{2} \ \textless \ x \ \textless \ 2$ so this is one of the solutions.case 3: when $x\ge 2$ , then $|x+3| = x+3$ and $|x-2| = x-2$ therefore $\begin{aligned} |x-2| & \ \textless \ |x+3| \\ x-2& \ \textless \ x+3 \\ -2 & \ \textless \ 3 \end{aligned}$ this is always a true statement.therefore, $x \ge 2$ is another solution.======the solution from case 2 is $-\frac{1}{2} \ \textless \ x \ \textless \ 2$ the solution from case 3 is $x \ge 2$ both of these solutions will work, so we can just combine them.joining these solutions together, we can simply say that $x \ \textgreater \ -\frac{1}{2}$ . this will cover the solutions from cases 2 and 3.so when $x \ \textgreater \ -\frac{1}{2}$ , the series will converge.

Answer from: Quest

the answer is 200

step-by-step explanation:

this is a problem about poisson probabilities. i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.

here's how you can see a poisson distribution coming. when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution. this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot. there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything. the form of the distribution is

probability of n chips = {[e^(-x)]*(x^n)} / (n! )

Answer from: Quest

idc

step-by-step explanation:

idk

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If x+3 and x-2 are the first two terms of a geometric series, find the values of x for which the ser...

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