Verify that the point ( square root(31)/7, -3square root(2)/7 ) lies on the graph of the unit circle. If it does, use symmetry to find three other points that also lie on the graph of the unit circle.
(direction - Type ordered pairs).
PS: The -3 is outside the square root ( but it's in the numerator). 2 in inside the square root. 7 is not included in the square root(7 is the denominator). 31 is inside the square root and 7 is not included in the square root (7 is the denominator).
Please show the steps & answer because I have to do so many. Thank You.
All you need to do is show that x^2+y^2=1. Don't forget your Algebra I now that you're in pre-cal!
(√31/7)^2 = 31/49
(-3√2/7)^2 = 18/49
Looks like they add up to 1, so the point is on the unit circle.
This is very, very similar to one I already helped you with. Please try them yourself. It does not do you any good to have us do all the problems for you. The idea is that we can help, but not do virtually the same problem over and over for you.
http://www.jiskha.com/display.cgi?id=1469345829
To verify if the point (sqrt(31)/7, -3sqrt(2)/7) lies on the graph of the unit circle, we need to check if the coordinates satisfy the equation of the unit circle, which is x^2 + y^2 = 1.
Substituting the given coordinates into the equation, we have:
(sqrt(31)/7)^2 + (-3sqrt(2)/7)^2 = 1
Expanding and simplifying, we get:
(31/49) + (18/49) = 1
49/49 = 1
Since the equation is true, the point (sqrt(31)/7, -3sqrt(2)/7) lies on the graph of the unit circle.
To find three other points on the graph using symmetry, we use the trigonometric properties of the unit circle. The three other quadrants on the unit circle will have the same x-coordinate but different y-coordinates due to symmetry.
1) First Quadrant:
The point (sqrt(31)/7, -3sqrt(2)/7) lies in the second quadrant, so its corresponding point in the first quadrant will have the same x-coordinate but a positive y-coordinate:
Point 1: (sqrt(31)/7, 3sqrt(2)/7)
2) Third Quadrant:
The point (sqrt(31)/7, -3sqrt(2)/7) lies in the second quadrant, so its corresponding point in the third quadrant will have a negative x-coordinate and a negative y-coordinate:
Point 2: (-sqrt(31)/7, -3sqrt(2)/7)
3) Fourth Quadrant:
The point (sqrt(31)/7, -3sqrt(2)/7) lies in the second quadrant, so its corresponding point in the fourth quadrant will have a negative x-coordinate and a positive y-coordinate:
Point 3: (-sqrt(31)/7, 3sqrt(2)/7)
So, the four points on the graph of the unit circle are:
(sqrt(31)/7, -3sqrt(2)/7)
(sqrt(31)/7, 3sqrt(2)/7)
(-sqrt(31)/7, -3sqrt(2)/7)
(-sqrt(31)/7, 3sqrt(2)/7)
To verify if the point ( sqrt(31)/7, -3sqrt(2)/7 ) lies on the graph of the unit circle, we need to check if the x-coordinate squared plus the y-coordinate squared equals 1.
Let's calculate:
(x-coordinate)^2 + (y-coordinate)^2 = (sqrt(31)/7)^2 + (-3sqrt(2)/7)^2
= (31/49) + (18/49)
= 49/49
= 1
Since the sum of the squares of the x-coordinate and y-coordinate is equal to 1, the point lies on the graph of the unit circle.
Now let's use symmetry to find three other points on the unit circle:
1. To find a point in the second quadrant, we change the sign of the y-coordinate and keep the sign of the x-coordinate the same. Therefore, the point becomes ( sqrt(31)/7, 3sqrt(2)/7 ).
2. To find a point in the third quadrant, we change the sign of both the x-coordinate and the y-coordinate. Therefore, the point becomes (-sqrt(31)/7, -3sqrt(2)/7 ).
3. To find a point in the fourth quadrant, we change the sign of the x-coordinate and keep the sign of the y-coordinate the same. Therefore, the point becomes (-sqrt(31)/7, 3sqrt(2)/7 ).
The four points on the graph of the unit circle are:
1. ( sqrt(31)/7, -3sqrt(2)/7 )
2. ( sqrt(31)/7, 3sqrt(2)/7 )
3. (-sqrt(31)/7, -3sqrt(2)/7 )
4. (-sqrt(31)/7, 3sqrt(2)/7 )
I hope this helps! Let me know if you have any further questions.