Answer:
The sequence of transformations that maps ΔABC to ΔA'B'C' is the reflection across the <u>line y = x</u> and a translation <u>10 units right and 4 units up</u>, equivalent to T₍₁₀, ₄₎
Step-by-step explanation:
For a reflection across the line y = -x, we have, (x, y) → (y, x)
Therefore, the point of the preimage A(-6, 2) before the reflection, becomes the point A''(2, -6) after the reflection across the line y = -x
The translation from the point A''(2, -6) to the point A'(12, -2) is T(10, 4)
Given that rotation and translation transformations are rigid transformations, the transformations that maps point A to A' will also map points B and C to points B' and C'
Therefore, a sequence of transformation maps ΔABC to ΔA'B'C'. The sequence of transformations that maps ΔABC to ΔA'B'C' is the reflection across the line y = x and a translation 10 units right and 4 units up, which is T₍₁₀, ₄₎
9 numbers of nickels
13 numbers of quarters
Divide both sides by -3, and replace 
with 
. Then
Factorize the quadratic in to get
which in turn means
But 
for all real 
, so we can ignore the first solution. This leaves us with
If we allow for any complex solution, then we can continue with the solution we ignored:
Answer:
Step-by-step explanation:
Distance between two points 
and 
is given by the formula,
Distance = 
a). A(-6, -4), B(-3, -1)
Distance between A and B = 
= 
= 3√2
= 4.2 units
b). C(3.5, 1), D(-4, 2.5)
Distance between C and D = 
= 
= 7.65
≈ 7.7 units
c). Distance between X(5, -5) and Y(-5, 5)
Distance = 
= 
= 10√2
= 14.14
≈ 14.1 units
