What is the image of (-12,3)(−12,3) after a dilation by a scale factor of \frac{1}{3} 3 1 centered at the origin?

The image of A(-12, 3) after dilation by a scale factor of 1/3 will be: A'(-4, 1)

Step-by-step explanation:

Given the coordinates of the point

Let suppose the given point is A(-12, 3)

We know that when an object is dilated by a scale factor, it gets reduced, stretched, or remains the same, depending upon the value of the scale factor.

If the scale factor > 1, the image is enlarged

If the scale factor is between 0 and 1, it gets shrunk

If the scale factor = 1, the object and the image are congruent

Rule to calculate the dilation by a scale factor 1/3 centered at the origin

P(x, y) → P'(1/3x, 1/3y)

Here, P'(1/3, 1/3y) is the image of P(x, y).

Thus,

A(-12, 3) → A'(1/3(-12), 1/3(3)) = A'(-4, 1)

Therefore, the image of A(-12, 3) after dilation by a scale factor of 1/3 will be: A'(-4, 1)

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}