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Please help! I'll mark brainiest!

Match the x-coordinates with their corresponding pairs of y-coordinates on the unit circle.

Please help Ill mark brainiest Match the xcoordinates with their corresponding pairs of ycoordinates on the unit circle class=

Respuesta :

Lynnsanity

Answer:

Step-by-step explanation:

2  on  top goes  to  last  on  the  bottom  or  b  goes  to  d

1st one one top goes to the 2nd one on bottom or a goes to b

last one on top goes to the third one on bottom or d goes to c

The last two witch are 3rd on top and first one together

Hope this helped it took me a long time :)

The x and y coordinates on the circle will be such that they satisfy the equation of the unit circle.

  • [tex]y = \pm \dfrac{\sqrt{5}}{{3}} \rightarrow \left(\dfrac{2}{3}, y\right)[/tex]
  • [tex]y = \pm \dfrac{\sqrt{7}}{{3}} \rightarrow \left(\dfrac{\sqrt{2}}{3}, y\right)[/tex]
  • [tex]y = \pm \dfrac{3}{5} \rightarrow \left(\dfrac{4}{5}, y\right)[/tex]
  • [tex]y = \pm \dfrac{2\sqrt{2}}{{3}} \rightarrow \left(\dfrac{1}{3}, y\right)[/tex]

What is the equation of the circle with radius r units, centered at (x,y) ?

If a circle O has radius of r units length and that it has got its center positioned at (h, k) point of the coordinate plane, then, its equation is given as:

[tex](x-h)^2 + (y-k)^2 = r^2[/tex]


A unit circle refers to a circle with unit radius (r = 1 unit) and positioned at center ( coordinates of origin = (h,k) = (0,0))

Thus, the equation of unit circle would be:

[tex]x^2 + y^2 =1[/tex]

Getting expression for y in terms of x,

[tex]x^2 + y^2 =1\\\\y = \pm \sqrt{1 - x^2}[/tex]

Using this equation to evaluate x for all given y:

  • Case 1: y = ±√5/3

[tex]\pm \dfrac{\sqrt{5}}{3} = \pm \sqrt{1-x^2}\\\\\text{Squaring both the sides}\\\\\dfrac{5}{9} = 1 - x^2\\\\x^2 = \dfrac{4}{9}\\\\x = \pm \dfrac{2}{3}[/tex]

From the options available, the fourth block seems valid.

Thus, we get:

[tex]y = \pm \dfrac{\sqrt{5}}{{3}} \rightarrow \left(\dfrac{2}{3}, y\right)[/tex]

  • Case 2: y = ±√7/3

[tex]\pm \dfrac{\sqrt{7}}{3} = \pm \sqrt{1-x^2}\\\\\text{Squaring both the sides}\\\\\dfrac{7}{9} = 1 - x^2\\\\x^2 = \dfrac{2}{9}\\\\x = \pm \dfrac{\sqrt{2}}{3}[/tex]

From the options available, the fourth block seems valid.

Thus, we get: [tex]y = \pm \dfrac{\sqrt{7}}{{3}} \rightarrow \left(\dfrac{\sqrt{2}}{3}, y\right)[/tex]

  • Case 3: y = ±3/5

[tex]\pm \dfrac{3}{5} = \pm \sqrt{1-x^2}\\\\\text{Squaring both the sides}\\\\\dfrac{9}{25} = 1 - x^2\\\\x^2 = \dfrac{16}{25}\\\\x = \pm \dfrac{4}{5}[/tex]

From the options available, the fourth block seems valid.

Thus, we get: [tex]y = \pm \dfrac{3}{5} \rightarrow \left(\dfrac{4}{5}, y\right)[/tex]

  • Case 4: y = ±2√2/3

[tex]\pm \dfrac{2\sqrt{2}}{3} = \pm \sqrt{1-x^2}\\\\\text{Squaring both the sides}\\\\\dfrac{8}{9} = 1 - x^2\\\\x^2 = \dfrac{1}{9}\\\\x = \pm \dfrac{1}{3}[/tex]

From the options available, the fourth block seems valid.

Thus, we get: [tex]y = \pm \dfrac{2\sqrt{2}}{{3}} \rightarrow \left(\dfrac{1}{3}, y\right)[/tex]

Thus, the x and y coordinates on the circle will be such that they satisfy the equation of the unit circle.

  • [tex]y = \pm \dfrac{\sqrt{5}}{{3}} \rightarrow \left(\dfrac{2}{3}, y\right)[/tex]
  • [tex]y = \pm \dfrac{\sqrt{7}}{{3}} \rightarrow \left(\dfrac{\sqrt{2}}{3}, y\right)[/tex]
  • [tex]y = \pm \dfrac{3}{5} \rightarrow \left(\dfrac{4}{5}, y\right)[/tex]
  • [tex]y = \pm \dfrac{2\sqrt{2}}{{3}} \rightarrow \left(\dfrac{1}{3}, y\right)[/tex]

Learn more about equation of a circle here:

https://brainly.com/question/10165274

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