Tenesha is trying to draw a parallelogram and find its area using the two line segments shown. On a coordinate plane, 2 line segments are formed by points (1, 5), (5, 2), (12, 2). She begins by finding the missing vertex on the graph to be point (9, 5). Then she calculates the area of the parallelogram. What errors did she make? Check all that apply.

The missing vertex is not (9,5) it is rather (8,5)

Step-by-step explanation:

We cannot see Tenesha's solution steps, so we cannot identify her mistakes.

Moreover we cannot see the options.

However, the mistake we can spot with the available information is that, the fourth vertex of a parallelogram with vertex (1, 5), (5, 2), (12, 2) is not (9,5).

The two adjacent horizontal vertices are:

(5,2) and (12,2).

The distance between them is |12-5|=7.

Since the opposite sides of a parrlallelogram are congruent, the other horizontal vertex for (1,5) is (1+7,5), which is (1,5)

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}

e

step-by-step explanation:

an equation by definition has to have two or more variables (x and 2 in this case)