Which triangle defined by the given points on the coordinate plane is similar to the triangle illustrated?

The correct option in the attached image is:

D. (-1, 1), (-1, 5), (-7, 1)

Step-by-step explanation:

Please refer to the attached image.

The length of the horizontal segment of the graphed triangle is 3.  The length of the vertical segment of the triangle graphed is 2. We will use the Pythagorean theorem to find the length of the missing side thus:

x² = 2²+3²

x² = 4+9

x² = 13

x = √13 or 3.6

This makes our side lengths 2, 3, and 3.6

For option A, our vertices are (-1, 1), (-1, 4), (-6, 1). This means the length of the horizontal segment 5 and the length of the vertical segment 3.  This missing side length is calculated thus:

x² = 5²+3²

x² = 25+9

x² = 34

x = √34 or 5.83

This is not proportional to the triangle drawn, so it is not similar.

For option B, our vertices are (-1, 1), (-1, 5), and (-6, 1).  This makes the horizontal segment 5 units long and the vertical segment 4 units long.  This makes the missing side length

x² = 4²+5²

x² = 16+25

x² = 41

x = √41 or 6.4

These are not proportional to the triangle drawn, so the triangles are not similar.

For option C, our vertices are (-1, 1), (-1, 4) and (-7, 1).  This makes the length of the horizontal segment 6 and the length of the vertical segment 3.  This makes the missing side length

x² = 3²+6²

x² = 9+36

x² = 45

x = √45 or 3√5 or 6.7

These sides are not proportional to the sides of the triangle drawn, so the triangles are not similar.

For option D, our vertices are (-1, 1), (-1, 5) and (-7, 1).  This makes the length of the horizontal segment 6 and the length of the vertical segment 4.  This makes the missing side length

x² = 6²+4²

x² = 36+16

x² = 52

x = √52 or 2√13 or 7.21

These sides are all twice as long as the sides of the triangle drawn, this indicates that the triangles are similar. Therefore our answer is D.

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}

answer: the four apples you took away.

step-by-step explanation: