Triangle abc is being dilated with the center of dilation at the origin. the image of c, point c' has coordinates of (7.2, 3.6). what is the scale factor of the dilation and what are the coordinates of point a'

General Idea:

When we are given a point P(x, y) centered at origin with a scale factor of k, then the dilated point will be given by P' (kx, ky)

Applying the concept:

In the diagram given, the coordinate of C is (6, 3). Triangle ABC is being dilated with the center of dilation at the origin. The image of C, point C' has coordinates of (7.2, 3.6).

If C (x, y) centered at origin with a scale factor of k, then the dilated point will be given by C' (kx, ky)

We know &

Substituting 6 for x in the equation , we get .

Dividing 6 on both sides

Simplifying fractions on both sides

Point A from the diagram is (-3, 3)

x-coordinate of A' =

y-coordinate of A' =

Point A' is given by (-3.6, 3.6)

Conclusion:

Scale factor of dilation is 1.2

The coordinates of point A' is (-3.6, 3.6)

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}

42

step-by-step explanation: