Given that th<span>e coordinates of the vertices of △DEF are D(2, −1) , E(7, −1) , and F(2, −3) and the coordinates of the vertices of △D′E′F′ are D′(0, −1) , E′(−5, −1) , and F′(0, −3) .
Notice that the y-coordinates of the pre-image and that of the image are the same, which means that there is a reflection across the y-axis.
A refrection across the y-axis results in the change in sign of the x-coordinates of the pre-image and the image while the y-coordinate of the image remains the same as that of the pre-image.
A refrection across the y-axis of </span>△DEF with vertices D(2, −1) , E(7, −1) , and F(2, −3)
will result in and image with vertices (-2, -1), (-7, -1) and (-2, -3) respectively.
Notice that the x-coordinate of the final image △D′E′F′ with vertices <span>D′(0, −1) , E′(−5, −1) , and F′(0, −3) is 2 units greater than the vertices of the result of recting the pre-image across the y-axis.
This means that the result of refrecting the pre-image was shifted two places to the right.
Therefore, </span>the sequence of transformations that maps △DEF to △D′E′F′ are reflection across the y-axis and translation 2 units right.
Answer:
y = 6x - 43
Step-by-step explanation:
(6, -7) and (8,5)
m=(y2-y1)/(x2-x1)
m=(5 + 7)/(8 - 6)
m= 12/2
m = 6
y - y1 = m(x - x1)
y + 7 = 6(x - 6)
y + 7 = 6x - 36
y = 6x - 43
Answer:
d
Step-by-step explanation:
Answer:
6!
Step-by-step explanation:
chord x chord = chord x chord
(5) (n+8) = (7) (n+4)
i just started plugging in numbers
(5) (6+8) = (7) (6+4)
5 x 14 = 7 x 10
70=70
Solving for the missing term and the missing coefficient (6a − )5a = ( ) a^2 − 35a
Let the missing term be X
Let the missing coefficient be Y
Therefore, (6a – X)5a = Y(a^2) – 35a
6a x 5a – X.5a = Y.a^2 – 35a
30a^2 – X.5a = Y.a^2 – 35a
Equating co-efficients,
30a^2 = Y.a^2; X.5a = 35a
30 = Y; 5X = 35
Y = 30; X = 7
Therefore, (6a-7)5a = 30 a^2 – 35a
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