Answer: Choice A) 4/5
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Work Shown:
cos^2(theta) + sin^2(theta) = 1
(-3/5)^2 + sin^2(theta) = 1
9/25 + sin^2(theta) = 1
9/25 + sin^2(theta) - 9/25 = 1 - 9/25
sin^2(theta) = 1 - 9/25
sin^2(theta) = 25/25 - 9/25
sin^2(theta) = (25 - 9)/25
sin^2(theta) = 16/25
sqrt[sin^2(theta)] = sqrt[16/25]
sin(theta) = 4/5
The fact that sine is positive in quadrant 2 means that the result is positive.
You lay out the triangle formula for area which would be A=(1/2)(b)(h)
plug in for variables
375=(1/2)(23.8)(b)
solve for b
A <span>counterclockwise rotation of 270º about the origin is equivalent to a </span><span>clockwise rotation of 90º about the origin.
Given a point (4, 5), the x-value, i.e. 4 and the y-value, i.e. 5 are positive, hence the point is in the 1st quadrant of the xy-plane.
A clockwise rotation of </span><span>90º about the origin of a point in the first quadrant of the xy-plane will have its image in the fourth quadrant of the xy-plane. Thus the x-value of the image remains positive but the y-value of the image changes to negative.
Also the x-value and the y-value of the original figure is interchanged.
For example, given a point (a, b) in the first quadrant of the xy-plane, </span><span>a counterclockwise rotation of 270º about the origin which is equivalent to a <span>clockwise rotation of 90º about the origin will result in an image with the coordinate of (b, -a)</span>
Therefore, a </span><span>counterclockwise rotation of 270º about the origin </span><span>of the point (4, 5) will result in an image with the coordinate of (5, -4)</span> (option C)
