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Stels [109]
3 years ago
6

I don’t understand this question. Can anyone help? I need answers ASAP. Thanks for all the help

Mathematics
1 answer:
zavuch27 [327]3 years ago
7 0

There's some unknown (but derivable) system of equations being modeled by the two lines in the given graph. (But we don't care what equations make up these lines.)

There's no solution to this particular system because the two lines are parallel.

How do we know they're parallel? Parallel lines have the same slope, and we can easily calculate the slope of these lines.

The line on the left passes through the points (-1, 0) and (0, -2), so it has slope

(-2 - 0)/(0 - (-1)) = -2/1 = -2

The line on the right passes through (0, 2) and (1, 0), so its slope is

(0 - 2)/(1 - 0) = -2/1 = -2

The slopes are equal, so the lines are parallel.

Why does this mean there is no solution? Graphically, a solution to the system is represented by an intersection of the lines. Parallel lines never intersect, so there is no solution.

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Which of the following geometric series converges?
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All three series converge, so the answer is D.

The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.

Consider a geometric sequence with the first term <em>a</em> and common ratio |<em>r</em>| < 1. Then the <em>n</em>-th partial sum (the sum of the first <em>n</em> terms) of the sequence is

S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}

Multiply both sides by <em>r</em> :

rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n

Subtract the latter sum from the first, which eliminates all but the first and last terms:

S_n-rS_n=a-ar^n

Solve for S_n:

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Then as gets arbitrarily large, the term r^n will converge to 0, leaving us with

S=\displaystyle\lim_{n\to\infty}S_n=\frac a{1-r}

So the given series converge to

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(II) -1.1/(1 + 1/10) = -1

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The third story, to get area you have to multiply length times width

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