Chapter 11

$20\text{inches}$

${\left(x-9\right)}^2$

$x=6\pm 4\sqrt{3}$

$x=3\pm \sqrt{3}$

$y=3{\left(x-1\right)}^2+2$

${\left(x-4\right)}^2=8$

$y=2{\left(x-3\right)}^2-4$

$x=\pm 2\sqrt{3}$

$x^2-8x+16$

$\left(3,2\right)$

$\left(2,\frac{3}{2}\right)$

$\left(1,3\right)$

$d=5$

$d=10$

$d=15$

$d=13$

$d=\sqrt{130},d\approx 11.4$

$d=\sqrt{2},d\approx 1.4$

$x^2+y^2=36$

$x^2+y^2=64$

${\left(x-2\right)}^2+{\left(y+4\right)}^2=49$

${\left(x+3\right)}^2+{\left(y+5\right)}^2=81$

${\left(x-2\right)}^2+{\left(y-1\right)}^2=25$

${\left(x-7\right)}^2+{\left(y-1\right)}^2=100$

ⓐ The circle is centered at $\left(3,\mathrm{-4}\right)$ with a radius of 2.
ⓑ

ⓐ The circle is centered at $\left(3,1\right)$ with a radius of 4.
ⓑ

ⓐ The circle is centered at $\left(0,0\right)$ with a radius of 3.
ⓑ

ⓐ The circle is centered at $\left(0,0\right)$ with a radius of 5.
ⓑ

ⓐ The circle is centered at $\left(\text{3},\text{4}\right)$ with a radius of 4.
ⓑ

ⓐ The circle is centered at $\left(\text{−}\text{3},\text{1}\right)$ with a radius of 3.
ⓑ

ⓐ The circle is centered at $\left(1,0\right)$ with a radius of 2.
ⓑ

ⓐ The circle is centered at $\left(0,6\right)$ with a radius of 5.
ⓑ

ⓐ $y=2{\left(x+1\right)}^2+3$
ⓑ

ⓐ $y=\mathrm{-2}{\left(x-2\right)}^2+1$
ⓑ

ⓐ $x=3{(y+1)}^2+4$
ⓑ

ⓐ $x=\mathrm{-4}{(y+2)}^2+4$
ⓑ

$y=-\frac{1}{20}{\left(x-20\right)}^2+20$

$y=-\frac{1}{5}{\left(x-5\right)}^2+5$

$\frac{x^2}{4}+\frac{y^2}{25}=1$

$\frac{x^2}{9}+\frac{y^2}{4}=1$

ⓐ $\frac{{\left(x+1\right)}^2}{6}+\frac{{\left(y-4\right)}^2}{9}=1$
ⓑ

ⓐ $\frac{{\left(x-2\right)}^2}{4}+\frac{{\left(y-3\right)}^2}{16}=1$
ⓑ

$\frac{x^2}{625}+\frac{y^2}{600}=1$

$\frac{x^2}{1225}+\frac{y^2}{1000}=1$

ⓐ $\frac{{\left(x+1\right)}^2}{16}-\frac{{\left(y-2\right)}^2}{9}=1$
ⓑ

ⓐ $\frac{{\left(x+3\right)}^2}{25}-\frac{{\left(y+1\right)}^2}{16}=1$
ⓑ

ⓐ circle ⓑ ellipse ⓒ parabola ⓓ hyperbola

ⓐ ellipse ⓑ parabola ⓒ circle ⓓ hyperbola

No solution

$\left(-\frac{4}{5},\frac{6}{5}\right),\left(0,2\right)$

No solution

$\left(\frac{4}{9},-\frac{2}{3}\right),\left(1,1\right)$

$\left(\mathrm{-3},0\right),\left(3,0\right),\left(\mathrm{-2}\sqrt{2},\mathrm{-1}\right),\left(2\sqrt{2},\mathrm{-1}\right)$

$\left(\mathrm{-1},0\right),\left(0,1\right),\left(0,\mathrm{-1}\right)$

$\left(\mathrm{-3},\mathrm{-4}\right),\left(\mathrm{-3},4\right),\left(3,\mathrm{-4}\right),\left(3,4\right)$

$\left(\mathrm{-2},0\right),\left(2,0\right)$

4 and 6

$\mathrm{-18}$ and 17

If the length is 12 inches, the width is 16 inches. If the length is 16 inches, the width is 12 inches.

If the length is 12 inches, the width is 9 inches. If the length is 9 inches, the width is 12 inches.

$d=5$

13

5

13

$d=3\sqrt{5},d\approx 6.7$

$d=2\sqrt{17},d\approx 8.2$

ⓐ Midpoint: $\left(2,\mathrm{-4}\right)$
ⓑ

ⓐ Midpoint: $\left(3\frac{1}{2},\mathrm{-1}\frac{1}{2}\right)$
ⓑ

$x^2+y^2=49$

$x^2+y^2=2$

${\left(x-3\right)}^2+{\left(y-5\right)}^2=1$

${\left(x-1.5\right)}^2+{\left(y+3.5\right)}^2=6.25$

${\left(x-3\right)}^2+{\left(y+2\right)}^2=64$

${\left(x-4\right)}^2+{\left(y-4\right)}^2=8$

ⓐ The circle is centered at $\left(\mathrm{-5},\mathrm{-3}\right)$ with a radius of 1.
ⓑ

ⓐ The circle is centered at $\left(4,\mathrm{-2}\right)$ with a radius of 4.
ⓑ

ⓐ The circle is centered at $\left(0,\mathrm{-2}\right)$ with a radius of 5.
ⓑ

ⓐ The circle is centered at $\left(1.5,\mathrm{-2.5}\right)$ with a radius of $0.5.$
ⓑ

ⓐ The circle is centered at $\left(0,0\right)$ with a radius of 8.
ⓑ

ⓐ The circle is centered at $\left(0,0\right)$ with a radius of 2.
ⓑ

ⓐ Center: $\left(\mathrm{-1},\mathrm{-3}\right),$ radius: 1
ⓑ

ⓐ Center: $\left(2,\mathrm{-5}\right),$ radius: 6
ⓑ

ⓐ Center: $\left(0,\mathrm{-3}\right),$ radius: 2
ⓑ

ⓐ Center: $\left(\mathrm{-2},0\right),$ radius: 2
ⓑ

Answers will vary.

Answers will vary.

ⓐ $y=\text{−}{(x-1)}^2-3$
ⓑ

ⓐ $y=\mathrm{-2}{(x+1)}^2-3$
ⓑ

ⓐ $x={\left(y+2\right)}^2-9$
ⓑ

ⓐ $x=\mathrm{-2}{\left(y+3\right)}^2+2$
ⓑ

ⓐ

ⓑ

ⓓ

$y=-\frac{1}{15}{\left(x-15\right)}^2+15$

$y=-\frac{1}{10}{\left(x-30\right)}^2+90$

Answers will vary.

Answers will vary.

$\frac{x^2}{9}+\frac{y^2}{25}=1$

$\frac{x^2}{9}+\frac{y^2}{16}=1$

ⓐ $\frac{{\left(x-2\right)}^2}{9}+\frac{{\left(y-3\right)}^2}{25}=1$
ⓑ

ⓐ $\frac{y^2}{4}+\frac{{(x-3)}^2}{25}=1$
ⓑ

$\frac{x^2}{400}+\frac{y^2}{300}=1$

$\frac{x^2}{2500}+\frac{y^2}{1275}=1$

Answers will vary.

Answers will vary.

ⓐ $\frac{{\left(x-1\right)}^2}{4}-\frac{{\left(y-1\right)}^2}{9}=1$
ⓑ

ⓐ $\frac{{\left(y-2\right)}^2}{9}-\frac{{\left(x-1\right)}^2}{9}=1$
ⓑ

ⓐ $\frac{{\left(y+1\right)}^2}{1}-\frac{{\left(x+2\right)}^2}{9}=1$
ⓑ

ⓐ parabola ⓑ circle ⓒ hyperbola ⓓ ellipse

Answers will vary.

Answers will vary.

$\left(\mathrm{-1},0\right),\left(0,3\right)$

$\left(2,0\right)$

$\left(12,\mathrm{-5}\right),\left(12,5\right)$

No solution

$\left(0,\mathrm{-4}\right),\left(1,\mathrm{-3}\right)$

$\left(3,4\right),\left(5,0\right)$

$\left(0,\mathrm{-4}\right),\left(\text{−}\sqrt{7},3\right),\left(\sqrt{7},3\right)$

$\left(0,\mathrm{-2}\right),\left(\text{−}\sqrt{3},1\right),\left(\sqrt{3},1\right)$

$\left(\mathrm{-2},0\right),\left(1,\text{−}\sqrt{3}\right),\left(1,\sqrt{3}\right)$

$\left(\mathrm{-2},\mathrm{-4}\right),\left(\mathrm{-2},4\right),\left(2,\mathrm{-4}\right),\left(2,4\right)$

$\left(\mathrm{-4},0\right),\left(4,0\right)$

$\left(\text{−}\sqrt{3},0\right),\left(\sqrt{3},0\right)$

$\left(\mathrm{-2},\mathrm{-3}\right),\left(\mathrm{-2},3\right),\left(2,\mathrm{-3}\right),\left(2,3\right)$

$\left(\mathrm{-1},\mathrm{-3}\right),\left(\mathrm{-1},3\right),\left(1,\mathrm{-3}\right),\left(1,3\right)$

$\mathrm{-3}$ and 14

$\mathrm{-7}$ and $\mathrm{-8}$ or 8 and 7

$\mathrm{-6}$ and $\mathrm{-4}$ or $\mathrm{-6}$ and 4 or 6 and $\mathrm{-4}$ or 6 and 4

If the length is 11 cm, the width is 15 cm. If the length is 15 cm, the width is 11 cm.

If the length is 10 inches, the width is 24 inches. If the length is 24 inches, the width is 10 inches.

The length is 40 inches and the width is 30 inches. The TV will not fit Donnette’s entertainment center.

Answers will vary.

Answers will vary.

$d=3$

$d=\sqrt{17},d\approx 4.1$

$\left(4,4\right)$

$\left(\frac{3}{2},-\frac{7}{2}\right)$

$x^2+y^2=7$

${\left(x+2\right)}^2+{\left(y+5\right)}^2=49$

${\left(x-2\right)}^2+{\left(y-2\right)}^2=8$

ⓐ radius: 12, center: $\left(0,0\right)$
ⓑ

ⓐ radius: 7, center: $\left(\mathrm{-2},\mathrm{-5}\right)$
ⓑ

ⓐ radius: 8, center: $\left(0,2\right)$
ⓑ

ⓐ $y=2{\left(x-1\right)}^2-4$
ⓑ

ⓐ $y=\text{−}{\left(x-6\right)}^2+1$
ⓑ

ⓐ $x={\left(y+2\right)}^2+1$
ⓑ

ⓐ $x=\mathrm{-2}{\left(y-1\right)}^2+2$
ⓑ

$y=-\frac{1}{9}{x}^2+\frac{10}{3}x$

$\frac{x^2}{36}+\frac{y^2}{64}=1$

ⓐ $\frac{{\left(x-3\right)}^2}{4}+\frac{{\left(y-7\right)}^2}{25}=1$
ⓑ

ⓐ $\frac{x^2}{9}+\frac{{\left(y-7\right)}^2}{4}=1$
ⓑ

ⓐ $\frac{{\left(x+1\right)}^2}{16}-\frac{{\left(y-3\right)}^2}{4}=1$
ⓑ

ⓐ $\frac{{\left(y-1\right)}^2}{16}-\frac{{\left(x-1\right)}^2}{4}=1$
ⓑ

ⓐ hyperbola ⓑ circle ⓒ parabola ⓓ ellipse

$\left(\mathrm{-1},4\right)$

No solution

$\left(\text{−}\sqrt{7},3\right),\left(\sqrt{7},3\right)$

$\left(\mathrm{-3},0\right),\left(0,\mathrm{-2}\right),\left(0,2\right)$

$\mathrm{-3}$ and $\mathrm{-4}$ or 4 and 3

If the length is 14 inches, the width is 15 inches. If the length is 15 inches, the width is 14 inches.

distance: 10, midpoint: $\left(\mathrm{-7},\mathrm{-7}\right)$

$x^2+y^2=121$

${\left(x+2\right)}^2+{\left(y-3\right)}^2=52$

ⓐ ellipse
ⓑ

ⓐ circle
ⓑ

ⓐ ellipse
ⓑ

ⓐ hyperbola
ⓑ

ⓐ circle
ⓑ ${\left(x+5\right)}^2+{\left(y+3\right)}^2=4$
ⓒ

ⓐ hyperbola
ⓑ $\frac{{\left(x-2\right)}^2}{25}-\frac{{\left(y+1\right)}^2}{9}=1$
ⓒ

No solution

$\left(3,0\right),\left(\mathrm{-3},0\right)$

$\frac{x^2}{2025}+\frac{y^2}{1400}=1$

The length is 44 inches and the width is 33 inches. The TV will fit Olive’s entertainment center.