Two geometry questions in the attached photos - If I have a line, do I have a plane? & If I have three points, do I always have a plane?

18) A: only if you have an additional point not on the line.

19) D: No, because all the points may be on the same line.

Step-by-step explanation:

There are basically 4 ways we can determine a plane and they are;

- When we have three non - collinear points: What this means is that if we have three points that are not on one line, then it means that only one particular plane can pass through those points.

- When we have a line and a point that's not on the line

- When we have two lines that intersect each other.

- When we have two lines that are parallel to each other.

From the four conditions above, we can now answer the question;

18) You can only have a plane if there is an additional point not on the line.

Option A

19) Having just three points may not always be sufficient to have a plane because it's possible that the three points may be on the same line which is not a condition for a plane. Thus, option D is correct.

hi,

3. 12.15, you are correct.

4.2.01, your first digit is correct,

hope this !

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}