Which coordinate (g, z) lies in the solution set for the following system of inequalities? 3g + 4z < 24 3g - 4z > 12 select one or more: a. (4, 2) b. (6, 1) c. (-2, 8) d. (3, 9)

B. (6, 1)

Step-by-step explanation:

A graph of the solution space shows that the point (6, 1) is the only one that satisfies both inequalities.

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Verification

3(6) +4(1) = 18 +4 = 22 < 24

3(6) -4(1) = 18 -4 = 14 > 12

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Notes on the Solution

The graphing calculator used here only recognizes variables x and y. Since g is the first variable in the ordered pair (g, z) we use x to represent g in the equations that are graphed. Similarly, y is used to represent z in the graphed equations.

You graph an inequality by graphing it as though it were an equation, then choosing the appropriate half-plane (one side of the line or the other) to shade as the solution space. Where the shadings from the two inequalities overlap, both are satisfied by points in that area.

In the first inequality, x and y values that are less than those on the line will satisfy the inequality, so the shading is below the line.

In the second inequality, x values greater than those on the line will satisfy the inequality, so the shading is to the right of the line.

In both cases, the line is graphed as a dashed line. This is because the points on the line do not satisfy the < or > conditions, so are not part of the solution set. (If one or both conditions were ≤ or ≥, then the corresponding line would be solid, and the points on the line would be part of the solution.)

.mp4 files, because, video

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}