HELP Graph the image of the figure after a dilation with a scale factor of 1/4 centered at (5, −5).

Scale factor(k) = 1/4, and centered at (5, -5).

Labelled the given figure as A, B and C.

The coordinates of the given triangle ABC are;

A = (-3, 7)

B = (-7, -5)

C = (9, 3)

To find the image of the figure after a dilation with scale factor 1/4 centered at (5, -5).

The rule of dilation with scale factor 1/4 and centered at (5, -5) is given by;

(x, y) \rightarrow (\frac{1}{4}(x-5)+5, \frac{1}{4}(y+5)-5)(x,y)→(

4

1

(x−5)+5,

4

1

(y+5)−5)

or

(x, y) \rightarrow (\frac{1}{4}x+\frac{15}{4}, \frac{1}{4}y-\frac{15}{4} )(x,y)→(

4

1

x+

4

15

,

4

1

y−

4

15

)

The coordinates of the image of the figure after dilation are;

A(-3, 7) \rightarrow A'(\frac{1}{4}(-3)+\frac{15}{4}, \frac{1}{4}(7)-\frac{15}{4} )A(−3,7)→A

′

(

4

1

(−3)+

4

15

,

4

1

(7)−

4

15

)

A(-3, 7) \rightarrow A'(3, -2)A(−3,7)→A

′

(3,−2)

B(-7, -5) \rightarrow B'(\frac{1}{4}(-7)+\frac{15}{4}, \frac{1}{4}(-5)-\frac{15}{4} )B(−7,−5)→B

′

(

4

1

(−7)+

4

15

,

4

1

(−5)−

4

15

)

B(-7, -5) \rightarrow B'(2, -5)B(−7,−5)→B

′

(2,−5)

and

C(9, 3) \rightarrow C'(\frac{1}{4}(9)+\frac{15}{4}, \frac{1}{4}(3)-\frac{15}{4} )C(9,3)→C

′

(

4

1

(9)+

4

15

,

4

1

(3)−

4

15

)

C(9, 3) \rightarrow C'(6, -3)C(9,3)→C

′

(6,−3)

As, you can see the graph as shown below in the attachment.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

answer: c. is your answer

step-by-step explanation: get me a black leather coat