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Andreyy89
3 years ago
9

Dilation of 1/2 through the point (-4,2)

Mathematics
1 answer:
GREYUIT [131]3 years ago
6 0

Answer:

  See the table attached for a list of the dilated coordinates.

Step-by-step explanation:

Each image point will be the midpoint* of the segment joining the center of dilation with the pre-image point. Since there are so many, it is convenient to use a spreadsheet to calculate them.

  • A'(-1.5, 5)
  • B'(1.5, 1.5)
  • C'(-1.5, 4.5)
  • D'(1.5, 6)
  • E'(-1.5, 2.5)
  • F'(2.5, 5)

_____

* The points are the midpoint because the scale factor is 1/2. That is, the distance from the center of dilation (COD) in the image is 1/2 the distance from the COD to a point in the pre-image. Any midpoint is the average of the two endpoints.

Here, that means the image point is half the sum of the coordinates of the pre-image point and (-4, 2). That is what the spreadsheet calculates.

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Answer: Plot (0, -1) and (3, 0) and draw a line through them.

***************************************************************************************

y = 4 +  \frac{1}{3}x    ⇒   y =  \frac{1}{3}x  + 4

Same as above.  Plot the y-intercept (0, 4) and then use rise over run from the slope to plot (3, 5).

Answer: Plot (0, 4) and (3, 5) and draw a line through them.

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Answer: No Solution

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4 years ago
Given: Circle O with diameter LN and inscribed angle LMN Prove: is a right angle. What is the missing reason in step 5? Statemen
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Anuta_ua [19.1K]

Answer:

c

Step-by-step explanation:

First, we can transform this into a matrix. The x coefficients will be the first ones for each row, the y coefficients the second column, etc.

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right]

Next, we can define a reduced row echelon form matrix as follows:

With the leading entry being the first non zero number in the first row, the leading entry in each row must be 1. Next, there must only be 0s above and below the leading entry. After that, the leading entry of a row must be to the left of the leading entry of the next row. Finally, rows with all zeros should be at the bottom of the matrix.

Because there are 3 rows and we want to solve for 3 variables, making the desired matrix of form

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] for the first three rows and columns. This would make the equation translate to

x= something

y= something

z = something, making it easy to solve for x, y, and z.

Going back to our matrix,

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right] ,

we can start by removing the nonzero values from the first column for rows 2 and 3 to reach the first column of the desired matrix. We can do this by multiplying the first row by -6 and adding it to the second row, as well as multiplying the first row by -1 and adding it to the third row. This results in

\left[\begin{array}{cccc}1&-2&3&-2\\0&14&-16&-36\\0&6&0&-36\end{array}\right]

as our matrix. * Next, we can reach the second column of our desired matrix by first multiplying the second row by (2/14) and adding it to the first row as well as multiplying the second row by (-6/14) and adding it to the third row. This eliminates the nonzero values from all rows in the second column except for the second row. This results in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&14&-16&-36\\0&0&96/14&-288/14\end{array}\right]

After that, to reach the desired second column, we can divide the second row by 14, resulting in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&1&-16/14&-36/14\\0&0&96/14&-288/14\end{array}\right]

Finally, to remove the zeros from all rows in the third column outside of the third row, we can multiply the third row by (16/96) and adding it to the second row as well as multiplying the third row by (-10/96) and adding it to the first row. This results in

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&96/14&-288/14\end{array}\right]

We can then divide the third row by -96/14 to reach the desired third column, making the reduced row echelon form of the matrix

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&1&-3\end{array}\right]

Therefore,

x=-5

y=-6

z=-3

* we could also switch the second and third rows here to make the process a little simpler

3 0
3 years ago
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