Determine the missing coordinate of the point (4/7,y) that lies on the graph of the unit circle in quadrant IV. Please show the steps because I need to solve more. Thank you.
x^2 + y^2 = 1
16/49 + y^2 = 49/49
y^2 = 33/49
y = +/- sqrt (33) /7
so since in Q IV where y is negative
(4/7, -sqrt(33)/7 )
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To find the missing coordinate y of the point (4/7, y) on the graph of the unit circle in quadrant IV, we can use the equation of the unit circle:
x^2 + y^2 = 1
Since the point (4/7, y) lies on the unit circle, we substitute x = 4/7 into the equation and solve for y:
(4/7)^2 + y^2 = 1
16/49 + y^2 = 1
y^2 = 1 - 16/49
y^2 = 49/49 - 16/49
y^2 = 33/49
Taking the square root of both sides, we have:
y = ±√(33/49)
Since the point (4/7, y) lies in quadrant IV, where y is negative, the missing coordinate y is:
y = -√(33/49)
And there you have it, the missing coordinate of the point (4/7, y) on the graph of the unit circle in quadrant IV is -√(33/49).
To determine the missing coordinate (y) of the point (4/7, y) that lies on the unit circle in quadrant IV, we can use the Pythagorean identity and the fact that the unit circle equation is x^2 + y^2 = 1.
1. Assign the value of x as 4/7.
x = 4/7
2. Use the equation of the unit circle to solve for y.
x^2 + y^2 = 1
(4/7)^2 + y^2 = 1
16/49 + y^2 = 1
3. Simplify the equation.
Multiply both sides of the equation by 49 to eliminate the fraction:
16 + 49y^2 = 49
4. Rearrange the equation to isolate the term with y^2 on one side.
49y^2 = 49 - 16
49y^2 = 33
5. Divide both sides of the equation by 49 to solve for y^2.
y^2 = 33/49
6. Take the square root of both sides of the equation to solve for y.
y = ±√(33/49)
7. Simplify the square root.
y = ±(√33)/7
Since the point lies on the unit circle in quadrant IV, the y-coordinate must be negative. Therefore, the missing coordinate of the point (4/7, y) is:
y = -(√33)/7
To determine the missing coordinate of the point (4/7, y) that lies on the graph of the unit circle in quadrant IV, we can use the equation of the unit circle. The equation of the unit circle in Cartesian coordinates is:
x^2 + y^2 = 1
Since the point lies on the unit circle, we can substitute the given x-coordinate (4/7) into the equation:
(4/7)^2 + y^2 = 1
To solve for y, we need to isolate y^2 on one side of the equation. First, square the x-coordinate:
(16/49) + y^2 = 1
Next, subtract (16/49) from both sides:
y^2 = 1 - (16/49)
Simplifying further:
y^2 = (49/49) - (16/49)
y^2 = 33/49
To find y, take the square root of both sides (we will consider the negative root since the point is in quadrant IV):
y = -√(33/49)
Therefore, the missing coordinate of the point (4/7, y) on the unit circle in quadrant IV is (4/7, -√(33/49)).
Remember to apply this step-by-step method to solve similar problems in the future!