To determine whether Sasha's bean bag landed at any of the given points on the edge of the hole, we need to check which points satisfy the equation of the hole: [tex]\(x^2 + y^2 = 5\)[/tex].
Let's evaluate each set of points:
1. Point [tex]\((-1, -2)\)[/tex]:
[tex]\[
x = -1, \quad y = -2
\][/tex]
[tex]\[
(-1)^2 + (-2)^2 = 1 + 4 = 5
\][/tex]
So, [tex]\((-1, -2)\)[/tex] is on the edge of the hole.
2. Point [tex]\((-2, 1)\)[/tex]:
[tex]\[
x = -2, \quad y = 1
\][/tex]
[tex]\[
(-2)^2 + 1^2 = 4 + 1 = 5
\][/tex]
So, [tex]\((-2, 1)\)[/tex] is on the edge of the hole.
3. Point [tex]\((1, -2)\)[/tex]:
[tex]\[
x = 1, \quad y = -2
\][/tex]
[tex]\[
1^2 + (-2)^2 = 1 + 4 = 5
\][/tex]
So, [tex]\((1, -2)\)[/tex] is on the edge of the hole.
4. Point [tex]\((2, 1)\)[/tex]:
[tex]\[
x = 2, \quad y = 1
\][/tex]
[tex]\[
2^2 + 1^2 = 4 + 1 = 5
\][/tex]
So, [tex]\((2, 1)\)[/tex] is on the edge of the hole.
5. Point [tex]\((-1, 2)\)[/tex]:
[tex]\[
x = -1, \quad y = 2
\][/tex]
[tex]\[
(-1)^2 + 2^2 = 1 + 4 = 5
\][/tex]
So, [tex]\((-1, 2)\)[/tex] is on the edge of the hole.
6. Point [tex]\((-2, -1)\)[/tex]:
[tex]\[
x = -2, \quad y = -1
\][/tex]
[tex]\[
(-2)^2 + (-1)^2 = 4 + 1 = 5
\][/tex]
So, [tex]\((-2, -1)\)[/tex] is on the edge of the hole.
7. Point [tex]\((1, 2)\)[/tex]:
[tex]\[
x = 1, \quad y = 2
\][/tex]
[tex]\[
1^2 + 2^2 = 1 + 4 = 5
\][/tex]
So, [tex]\((1, 2)\)[/tex] is on the edge of the hole.
8. Point [tex]\((2, -1)\)[/tex]:
[tex]\[
x = 2, \quad y = -1
\][/tex]
[tex]\[
2^2 + (-1)^2 = 4 + 1 = 5
\][/tex]
So, [tex]\((2, -1)\)[/tex] is on the edge of the hole.
Since all the given points:
[tex]\[ (-1, -2), (-2, 1), (1, -2), (2, 1), (-1, 2), (-2, -1), (1, 2), (2, -1) \][/tex]
all satisfy the equation [tex]\(x^2 + y^2 = 5\)[/tex], they all lie on the edge of the hole. Therefore, any of these points could represent where Sasha’s bean bag landed.
