Triangle PQR has vertices P(0, 1), Q(0, -4), and R(2, 5). Find the coordinates of R' to the nearest hundredth after rotating triangle PQR counterclockwise about the origin 45º.

the radius to (2,5) is at an angle θ such that tanθ = 5/2

So
cosθ = 2/√29
sinθ = 5/√29

R' is at an angle (θ+45), so
x = √29 cos(θ+45)
y = √29 sin(θ+45)

x = √29 (cosθ cos45 - sinθ sin45)
= √29 (2/√29 * 1/√2 - 5/√29 * 1/√2)
= 2/√2 - 5/√2
= -3/√2

y = √29 (sinθ cos45 + cosθ sin45)
= √29 (5/√29 * 1/√2 + 2/√29 * 1/√2)
= 7/√2

So, R' = (-3/√2, 7/√2)

wow

To find the coordinates of R' after rotating triangle PQR counterclockwise about the origin 45º, we can use the following steps:

1. Find the direction cosines of the angle of rotation:
- The angle of rotation is 45º.
- The direction cosines of a counterclockwise rotation about the origin can be computed as:
- cos(45º) = √2/2
- sin(45º) = √2/2

2. Apply the rotation formula to each vertex of the triangle:
- For a counterclockwise rotation about the origin, the rotation formula is:
- x' = x * cos(θ) - y * sin(θ)
- y' = x * sin(θ) + y * cos(θ)

Let's apply these steps to find the coordinates of R':

1. Find the direction cosines of the angle of rotation:
- cos(45º) = √2/2
- sin(45º) = √2/2

2. Apply the rotation formula to vertex R(2, 5):
- x' = 2 * (√2/2) - 5 * (√2/2) = (√2 - 5√2)/2 = -3√2/2
- y' = 2 * (√2/2) + 5 * (√2/2) = (√2 + 5√2)/2 = 7√2/2

Therefore, the coordinates of R' after rotating triangle PQR counterclockwise about the origin 45º are approximately (-3√2/2, 7√2/2) to the nearest hundredth.