Question 1

Solve the problem.
-Three receiving stations at (8, 11), (4, 6), and (12, 9)record distances to an earthquake of 2 units, 5 units, and 4 units, respectively.Find the coordinates of the epicenter.

Accepted Answer

The answer of Solve the problem.
-Three receiving stations at (8,...

Question 2

Find the center and radius of the circle.
-$x ^ { 2 } + y ^ { 2 } - 16 x + 18 y = - 141$

Accepted Answer

The answer of Find the center and radius of the...

Question 3

Graph the circle.
-$$( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 4$$

Accepted Answer

The answer of Graph the circle.
-\[( x - 3 )...

Question 4

Graph the circle.
-$(x-4)^{2}+y^{2}=25$

Accepted Answer

The answer of Graph the circle.
-$(x-4)^{2}+y^{2}=25$

...

Question 5

Find the center and radius of the circle.
-$x ^ { 2 } + y ^ { 2 } + 18 x + 2 y = - 46$

Accepted Answer

To find the center and radius of the circle given by the equation $x^2 + y^2 + 18x + 2y = -46$, we complete the square for both $x$ and $y$. For $x$, we take the coefficient of $x$, which is 18, divide it by 2 to get 9, and square it to get 81. We add and subtract 81 inside the equation. For $y$, we take the coefficient of $y$, which is 2, divide it by 2 to get 1, and square it to get 1. We add and subtract 1 inside the equation. The equation becomes $x^2 + 18x + 81 + y^2 + 2y + 1 = -46 + 81 + 1$, which simplifies to $(x + 9)^2 + (y + 1)^2 = 36$. This shows the circle has a center at $(-9, -1)$ and a radius of $\sqrt{36} = 6$.

Question 6

Graph the circle.
-$x^{2}+y^{2}-6 x-8 y+16=0$

Accepted Answer

The answer of Graph the circle.
-$x^{2}+y^{2}-6 x-8 y+16=0$

...

Question 7

Solve the problem.
-The enrollment at State University for 2002 was 12 thousand students. The projected enrollment for 2008 is 28 thousand students. Use the midpoint formula to estimate the enrollment for
2005.

Accepted Answer

The midpoint formula is used to find the average of two numbers. The enrollment in 2002 is 12 thousand and in 2008 is 28 thousand. The midpoint for 2005 is (12 + 28) / 2 = 20 thousand students.

Question 8

Find the center and radius of the circle.
-$2 x ^ { 2 } + 2 y ^ { 2 } + 8 x + 12 y + 24 = 0$

Accepted Answer

To find the center and radius, first divide the equation by 2: $x^2 + y^2 + 4x + 6y + 12 = 0$. Complete the square: $(x+2)^2 - 4 + (y+3)^2 - 9 + 12 = 0$. Simplify to $(x+2)^2 + (y+3)^2 = 1$. The center is $(-2, -3)$ and the radius is 1.

Question 9

Solve the problem.
-If the coordinates of one endpoint of a line segment are (4, 2)and the coordinates of the midpoint of the segment are (6, 2), what are the coordinates of the other endpoint?

Accepted Answer

The midpoint formula is [(x1 + x2)/2, (y1 + y2)/2]. Given one endpoint (4, 2) and the midpoint (6, 2), we can find the other endpoint by setting up equations: (4 + x2)/2 = 6 and (2 + y2)/2 = 2. Solving these gives x2 = 8 and y2 = 2, so the other endpoint is (8, 2).

Question 10

Find the center and radius of the circle.
-$x ^ { 2 } + ( y - 7 ) ^ { 2 } = 25$

Accepted Answer

The equation of a circle in the standard form is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. Comparing the given equation $x^2 + (y - 7)^2 = 25$ with the standard form, we find $h = 0$, $k = 7$, and $r^2 = 25$, which means the center is $(0, 7)$ and the radius $r = \sqrt{25} = 5$.

Question 11

Find the center and radius of the circle.
-$x ^ { 2 } + y ^ { 2 } = 9$

Accepted Answer

The equation $x^2 + y^2 = 9$ is in the standard form for a circle centered at the origin $(0,0)$ with radius $\sqrt{9} = 3$.

Question 12

Find the center and radius of the circle.
-$( x + 9 ) ^ { 2 } + y ^ { 2 } = 49$

Accepted Answer

The equation of a circle in the form $(x - h)^2 + (y - k)^2 = r^2$ has a center at $(h, k)$ and a radius of $r$. The given equation $(x + 9)^2 + y^2 = 49$ can be rewritten to match this form as $(x - (-9))^2 + (y - 0)^2 = 7^2$, indicating a center at $(-9, 0)$ and a radius of $7$.

Question 13

Graph the circle.
-$x ^ { 2 } + y ^ { 2 } + 8 x + 6 y + 16 = 0$

Accepted Answer

The answer of Graph the circle.
-\(x ^ { 2 }...

Question 14

Decide whether the statement is true or false.
-If a point lies on the y-axis, its x-coordinate is zero.

Accepted Answer

A point on the y-axis means it is directly above or below the origin along the y-axis, which implies its distance from the y-axis (x-coordinate) is zero.

Question 15

Decide whether the statement is true or false.
-If a point lies on the x-axis, its x-coordinate is zero.

Accepted Answer

If a point lies on the x-axis, its y-coordinate is zero, not its x-coordinate.

Question 16

Find the center and radius of the circle.
-$( x - 1 ) ^ { 2 } + ( y + 4 ) ^ { 2 } = 64$

Accepted Answer

The equation of a circle in the form $(x - h)^2 + (y - k)^2 = r^2$ has a center at $(h, k)$ and a radius $r$. For the given equation, $h = 1$, $k = -4$, and $r = \sqrt{64} = 8$, making the center $(1, -4)$ and the radius $8$.

Question 17

Find the center and radius of the circle.
-$x ^ { 2 } + y ^ { 2 } + 18 x + 4 y + 85 = 36$

Accepted Answer

To find the center and radius of the circle given by the equation $x^2 + y^2 + 18x + 4y + 85 = 36$, we complete the square for both $x$ and $y$. The equation can be rewritten as $x^2 + 18x + y^2 + 4y = -49$. Completing the square for $x$ and $y$ separately, we add and subtract $(18/2)^2 = 81$ for $x$ and add and subtract $(4/2)^2 = 4$ for $y$, leading to $x^2 + 18x + 81 + y^2 + 4y + 4 = -49 + 81 + 4$. This simplifies to $(x + 9)^2 + (y + 2)^2 = 36$, indicating a circle with center $(-9, -2)$ and radius $r = \sqrt{36} = 6$.

Question 18

Graph the circle.
-$x^{2}+(y-1)^{2}=9$

Accepted Answer

The answer of Graph the circle.
-$x^{2}+(y-1)^{2}=9$

...

Question 19

Find the center and radius of the circle.
-$x ^ { 2 } - 10 x + 25 + y ^ { 2 } + 14 y + 49 = 36$

Accepted Answer

To find the center and radius, we complete the squares for x and y. The given equation is $x^2 - 10x + 25 + y^2 + 14y + 49 = 36$. Completing the square for x and y gives us $(x-5)^2 + (y+7)^2 = 36$. Thus, the center is $(5, -7)$ and the radius $r = \sqrt{36} = 6$.

Question 20

Graph the circle.
-$x^{2}+y^{2}=25$

Accepted Answer

The answer of Graph the circle.
-$x^{2}+y^{2}=25$

...

Quiz 8: Graphs, Functions, and Systems of Equations and Inequalities

Solve the problem. -Three receiving stations at (8, 11) , (4, 6) , and (12, 9) record distances to an earthquake of 2 units, 5 units, and 4 units, respectively.Find the coordinates of the epicenter.

Find the center and radius of the circle. - $x ^ { 2 } + y ^ { 2 } - 16 x + 18 y = - 141$

Graph the circle. - $( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 4$ $Graph the circle. - ( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 4 A) B) C) D)$

Graph the circle. - $(x-4) ^{2}+y^{2}=25$ $Graph the circle. - (x-4) ^{2}+y^{2}=25 A) B) C) D)$

Find the center and radius of the circle. - $x ^ { 2 } + y ^ { 2 } + 18 x + 2 y = - 46$

Graph the circle. - $x^{2}+y^{2}-6 x-8 y+16=0$ $Graph the circle. - x^{2}+y^{2}-6 x-8 y+16=0 A) B) C) D)$

Solve the problem. -The enrollment at State University for 2002 was 12 thousand students. The projected enrollment for 2008 is 28 thousand students. Use the midpoint formula to estimate the enrollment for 2005.

Find the center and radius of the circle. - $2 x ^ { 2 } + 2 y ^ { 2 } + 8 x + 12 y + 24 = 0$

Solve the problem. -If the coordinates of one endpoint of a line segment are (4, 2) and the coordinates of the midpoint of the segment are (6, 2) , what are the coordinates of the other endpoint?

Find the center and radius of the circle. - $x ^ { 2 } + ( y - 7 ) ^ { 2 } = 25$

Find the center and radius of the circle. - $x ^ { 2 } + y ^ { 2 } = 9$

Find the center and radius of the circle. - $( x + 9 ) ^ { 2 } + y ^ { 2 } = 49$

Graph the circle. - $x ^ { 2 } + y ^ { 2 } + 8 x + 6 y + 16 = 0$ $Graph the circle. - x ^ { 2 } + y ^ { 2 } + 8 x + 6 y + 16 = 0 A) B) C) D)$

Decide whether the statement is true or false. -If a point lies on the y-axis, its x-coordinate is zero.

Decide whether the statement is true or false. -If a point lies on the x-axis, its x-coordinate is zero.

Find the center and radius of the circle. - $( x - 1 ) ^ { 2 } + ( y + 4 ) ^ { 2 } = 64$

Find the center and radius of the circle. - $x ^ { 2 } + y ^ { 2 } + 18 x + 4 y + 85 = 36$

Graph the circle. - $x^{2}+(y-1) ^{2}=9$ $Graph the circle. - x^{2}+(y-1) ^{2}=9 A) B) C) D)$

Find the center and radius of the circle. - $x ^ { 2 } - 10 x + 25 + y ^ { 2 } + 14 y + 49 = 36$

Graph the circle. - $x^{2}+y^{2}=25$ $Graph the circle. - x^{2}+y^{2}=25 A) B) C) D)$

Quiz 8: Graphs, Functions, and Systems of Equations and Inequalities

Solve the problem. -Three receiving stations at (8, 11) , (4, 6) , and (12, 9) record distances to an earthquake of 2 units, 5 units, and 4 units, respectively.Find the coordinates of the epicenter.

Find the center and radius of the circle. - x2+y2−16x+18y=−141x ^ { 2 } + y ^ { 2 } - 16 x + 18 y = - 141x2+y2−16x+18y=−141

Graph the circle. - (x−3) 2+(y−1) 2=4( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 4(x−3) 2+(y−1) 2=4

Graph the circle. - (x−4) 2+y2=25(x-4) ^{2}+y^{2}=25(x−4) 2+y2=25

Find the center and radius of the circle. - x2+y2+18x+2y=−46x ^ { 2 } + y ^ { 2 } + 18 x + 2 y = - 46x2+y2+18x+2y=−46

Graph the circle. - x2+y2−6x−8y+16=0x^{2}+y^{2}-6 x-8 y+16=0x2+y2−6x−8y+16=0

Solve the problem. -The enrollment at State University for 2002 was 12 thousand students. The projected enrollment for 2008 is 28 thousand students. Use the midpoint formula to estimate the enrollment for 2005.

Find the center and radius of the circle. - 2x2+2y2+8x+12y+24=02 x ^ { 2 } + 2 y ^ { 2 } + 8 x + 12 y + 24 = 02x2+2y2+8x+12y+24=0

Solve the problem. -If the coordinates of one endpoint of a line segment are (4, 2) and the coordinates of the midpoint of the segment are (6, 2) , what are the coordinates of the other endpoint?

Find the center and radius of the circle. - x2+(y−7) 2=25x ^ { 2 } + ( y - 7 ) ^ { 2 } = 25x2+(y−7) 2=25

Find the center and radius of the circle. - x2+y2=9x ^ { 2 } + y ^ { 2 } = 9x2+y2=9

Find the center and radius of the circle. - (x+9) 2+y2=49( x + 9 ) ^ { 2 } + y ^ { 2 } = 49(x+9) 2+y2=49

Graph the circle. - x2+y2+8x+6y+16=0x ^ { 2 } + y ^ { 2 } + 8 x + 6 y + 16 = 0x2+y2+8x+6y+16=0