What direction for negative and positive go in

  opposite directions

Step-by-step explanation:

Conventionally, positive is to the right or up. The actual direction depends on the desired coordinate system, and may vary from one context to another.

In spherical coordinates, positive distance measure is away from center. Positive angles of one sort are CCW from the "x direction" as seen from the "north pole" (z-direction), and positive angles of the other sort are away from the axis pointing to the "north pole."

In mechanics, reference points and directions are often chosen to simplify the problem as much as possible. The point of application of an unknown force, for example, may be chosen as a reference point simply so its moment about that point is zero and can be ignored.

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In navigation on the Earth's surface, bearing angles are generally positive measured clockwise from North. In Algebra, angles are usually positive measured counterclockwise from the +x axis (East on a map conventionally oriented).

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In conventional 2-dimensional algebra problem coordinates, positive x is to the right, and positive y is up. Angles are positive counterclockwise from the +x axis and negative clockwise from that axis. However, you can define reference points and positive directions any way you like, as long as you make your coordinate system known to anyone interested in your work. Generally, negative coordinates are in the direction opposite positive ones.

594 square inches

step-by-step explanation:

vote !

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}