Igure 1 and figure 2 are two congruent parallelograms drawn on a coordinate grid as shown below: 4 quadrant coordinate grid showing two parallelograms. figure 1 has vertices at negative 5, 2 and negative 3, 4 and negative 4, 7 and negative 6, 5. figure 2 has vertices at 5, negative 2 and 7, negative 4 and 6, negative 7 and 4, negative 5. which two transformations can map figure 1 onto figure 2? reflection across the y-axis, followed by reflection across x-axis reflection across the x-axis, followed by reflection across y-axis reflection across the x-axis, followed by translation 10 units right reflection across the y-axis, followed by translation 5 units down

Reflection across the x-axis, followed by translation 10 units right.

Step-by-step explanation:

I'm sorry, I know the question asks for two transformations, but let's look a the math before tackling the figure (see attachment).

When you are asked to do a reflection on the x-axis, they are asking you to invert the sign on the y coordinate of every point, and when you are asked to do a reflection on the y-axis, just invert the sign of the x coordinate, always following the convention of (x, y).

Translation to the right means to add the amount of units given to all the x coordinates, to the left means to subtract said number of units.

Translation down is to subtract those units to the y coordinate and translation up, is to add to that y coordinate.

So in this exercise:

Fig 1 coordinates are:

(-5, 2)  (-3, 4)  (-4, 7)  (-6, 5)

Fig 2 coordinates are:

(5, -2)  (7, -4)  (6, -7)  (4, -5)

So let's test the options given:

a. Reflection across the y-axis, followed by reflection across x-axis

Reflection across the y-axis:

Fig 1.1: (5, 2)  (3, 4)  (4, 7)  (6, 5) <- Every x coordinate with inverted sign

Then reflection across x-axis:

Fig 1.2: (5, -2)  (3, -4)  (4, -7)  (6, -5) <- Every y coordinate with inverted sign

if we compare this new Fig 1.2 with Fig 2:

(5, -2)  (3, -4)  (4, -7)  (6, -5)  ≠  (5, -2)  (7, -4)  (6, -7)  (4, -5)   Wrong

b. Reflection across the x-axis, followed by reflection across y-axis

Reflection across the x-axis:

Fig 1.1: (-5, -2)  (-3, -4)  (-4, -7)  (-6, -5) <- Every y coordinate with inverted sign

Then reflection across y-axis:

Fig 1.2: (5, -2)  (3, -4)  (4, -7)  (6, -5) <- Every x coordinate with inverted sign

if we compare this new Fig 1.2 with Fig 2:

(5, -2)  (3, -4)  (4, -7)  (6, -5)  ≠  (5, -2)  (7, -4)  (6, -7)  (4, -5)   Wrong

c. Reflection across the x-axis, followed by translation 10 units right

Reflection across the x-axis:

Fig 1.1: (-5, -2)  (-3, -4)  (-4, -7)  (-6, -5) <- Every y coordinate with inverted sign

Then translation 10 units right:

Fig 1.2: (5, -2)  (7, -4)  (6, -7)  (4, -5) <- Every x coordinate +10

if we compare this new Fig 1.2 with Fig 2:

(5, -2)  (7, -4)  (6, -7)  (4, -5)  =  (5, -2)  (7, -4)  (6, -7)  (4, -5)   Correct!

d. Reflection across the y-axis, followed by translation 5 units down

Reflection across the y-axis:

Fig 1.1: (5, 2)  (3, 4)  (4, 7)  (6, 5) <- Every x coordinate with inverted sign

Then translation 5 units down:

Fig 1.2: (5, -3)  (3, -1)  (4, 2)  (6, 0) <- Every y coordinate -5

if we compare this new Fig 1.2 with Fig 2:

(5, -3)  (3, -1)  (4, 2)  (6, 0)  ≠  (5, -2)  (7, -4)  (6, -7)  (4, -5)     Wrong

So from all options only c. works

In this problem, you want the decimal point between the 2 and the 5. This moves it 3 places to the left .

 2.5×10^3

hope that helped