Question

Please solve the attachment below

Answer 1

Answer:

Part 1) $x=(+/-)\frac{1}{\sqrt{11}}$ -----> $(x,\frac{\sqrt{110}}{11})$

Part 2) $x=(+/-)\frac{6\sqrt{2}}{11}$ ----> $(x,\frac{7}{11})$

Part 3) $x=(+/-)\frac{4\sqrt{6}}{11}$  ----> $(x,\frac{5}{11})$

Part 4) $x=(+/-)\frac{2\sqrt{10}}{11}$   ---> $(x,\frac{9}{11})$

Step-by-step explanation:

we know that

In the unit circle

The coordinates of a point have the following rule

$x^{2} +y^{2} =r^{2}$

where

(x,y) are the coordinates of the point a r is the radius

but remember that in a unit circle the radius is equal to 1

so

$x^{2} +y^{2} =1$

$x^{2}=1-y^{2}$

$x=(+/-)\sqrt{1-y^{2}}$

Find the x-coordinate of each case

Part 1) we have the point

$(x,\frac{\sqrt{110}}{11})$

so

The y-coordinate is

$y=\frac{\sqrt{110}}{11}$  

Find the value of the x-coordinate

substitute

$x=(+/-)\sqrt{1-y^{2}}$  

$x=(+/-)\sqrt{1-(\frac{\sqrt{110}}{11})^{2}$

$x=(+/-)\sqrt{1-(\frac{110}{121})$

$x=(+/-)\sqrt{\frac{11}{121})$

$x=(+/-)\frac{\sqrt{11}}{11}$

$x=(+/-)\frac{1}{\sqrt{11}}$

Part 2) we have the point

$(x,\frac{7}{11})$

so

The y-coordinate is

$y=\frac{7}{11}$  

Find the value of the x-coordinate

substitute

$x=(+/-)\sqrt{1-y^{2}}$  

$x=(+/-)\sqrt{1-(\frac{7}{11})^{2}$

$x=(+/-)\sqrt{1-(\frac{49}{121})$

$x=(+/-)\sqrt{\frac{72}{121})$

$x=(+/-)\frac{6\sqrt{2}}{11}$  

Part 3) we have the point

$(x,\frac{5}{11})$

so

The y-coordinate is

$y=\frac{5}{11}$  

Find the value of the x-coordinate

substitute

$x=(+/-)\sqrt{1-y^{2}}$  

$x=(+/-)\sqrt{1-(\frac{5}{11})^{2}$

$x=(+/-)\sqrt{1-(\frac{25}{121})$

$x=(+/-)\sqrt{\frac{96}{121})$

$x=(+/-)\frac{4\sqrt{6}}{11}$  

Part 4) we have the point

$(x,\frac{9}{11})$

so

The y-coordinate is

$y=\frac{9}{11}$  

Find the value of the x-coordinate

substitute

$x=(+/-)\sqrt{1-y^{2}}$  

$x=(+/-)\sqrt{1-(\frac{9}{11})^{2}$

$x=(+/-)\sqrt{1-(\frac{81}{121})$

$x=(+/-)\sqrt{\frac{40}{121})$

$x=(+/-)\frac{2\sqrt{10}}{11}$