Triangle ABC has vertices A(–2, –3), B(–5, –3), and C(–4, –2). The triangle is rotated 90° counterclockwise around the origin. What is the correct set of image points for triangle A’B’C’? A’(3, 2), B’(3, 5), C’(2, 4) A’(3, –2), B’(3, –5), C’(2, –4) A’(–3, –2), B’(–3, –5), C’(–2, –4) A’(–3, 2), B’(–3, 5), C’(–2, 4)

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step-by-step explanation:

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the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

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