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3 years ago

Identify a pair of similar triangles

Mathematics

1 answer:

Rainbow [258]3 years ago

7 0

Answer:

its ADC and ADB

Step-by-step explanation:

they both have the triangle at the top so they are similar, they just arnt the same size

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Enter the ordered pair for the vertices for (Ry-axis ○ T(2, 0))(QRST).

zhannawk [14.2K]

Answer:

The answer is:

$Q' = (-3, 3)\\\\R' = (-5, -3)\\\\S' = (-2, -2)\\\\T' = (0, 1)\\\\$

Step-by-step explanation:

All vertices of its quadrilateral were co-ordinated;

$T(-2, 1), \ Q(1, 3), \ S(0, -2), \ R(3, -3)$ Therefore, we first undertake the operation correct: $T(2, 0) \ QRST$ , there;

$T(2, 0) \ QRST$ $\to T(-2 + 2, 1), \ Q(1 + 2, 3), \ S(0 + 2, -2), \ R(3 + 2, -3)$

$T(2, 0) \ QRST$ $\to T(0, 1), \ Q(3, 3), \ S(2, -2), \ R(5, -3)$

A preimage (x, y) will be the image in the next phase R-y-axis reflecting mostly on the y-axis (-x, y) and we've got it;

R y-axis $(T(0, 1), \ Q(3, 3), \ S(2, -2), \ R(5, -3)) = \ T'(0, 1), \ Q'(-3, 3), \ S'(-2, -2), \ R'(-5, -3)$ From which we get $(Ry-axis \ T(2, 0)) \ (QRST = T'(0, 1), \ Q'(-3, 3), \ S'(-2, -2),\ R'(-5,-3).$

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