We can show that ∆ABC is congruent to ∆A′B′C′ by a translation of

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We can show that ΔABC is congruent to ΔA'B'C' by a translation of 2 unit(s) Left and a Reflection across the x axis.

Step-by-step explanation:

We were given triangles ABC and A'B'C' of which were told are congruents,

Now we can provide the coordinates of A and A' from the given triangles ΔABC and ΔA'B'C' ,if we choose a point of A from ΔABC and A' from ΔA'B'C' we have these coordinates;

A as (8,8) and A' (6,-8) from the two triangles.

If we shift A to A' , we have (8_6) = 2 unit for that of x- axis

If we try the shift on the y-coordinates we will see that there is no translation.

Hence, the only translation that take place is of 2 units left.

It can also be deducted that there is a reflection

by x-axis to form A'B'C' by the ΔABC.

BEST OF LUCK

step-by-step explanation:

the answer is d

step-by-step explanation: