Amplitude:4
Equation of Midline: 2
Period of function:3
Function shifted left:0.5
Function shifted up: 2

but anyway, the numerator will give the angles, and θ is just half of each

ok... that's a negative tiny angle, is in the 4th quadrant, if we stick to the range given, from 0 to 360, so we have to use the positive version of it, 360-4.025
so the angle is 355.975°
now, the 3rd quadrant has another angle whose sine is negative, so... if we move from the 180° line down by 4.025, we end up at 184.025°
and those are the only two angles, because, on the 2nd and 1st quadrants, the sine is positive, so it wouldn't have an angle there
First one:
The circumference of the whole circle is (2π x the radius),
and 60° is (60/360) = 1/6 of the whole circumference.
Second one:
The area of the whole circle is (π x the radius²), and
the 40° sector is (40/360) = 1/9 of the whole area.
I hope this is enough help.
3+4i - 1 + 5i= 2 + 9i
the answer to the question is d
Here, we want to check for the relationship between the image and its pre-image
The pre-image is (x,y)
The image is (3x,3y+5)
As we can see, the pre-image is not similar
This is because the transformation applied to the two values are not same
Thus, we have that;
No, the image is not similar to the pre-image as the translations applied to both coordinates are not same
The pre-image was transformed by dilating the x-coordinate of the pre-image by 3 units while the y-coordinate was transformed by dilating the y-coordinate of the pre-image by 3 and translating it upward by 5 units