Question 1 of 5Determine which Inequality Matches the Graph
Select the Correct Answer Below: Correct! Not Correct!
A
$$y<2x^2+4x-3$$
B
$$y>-4x^2-x-7$$
C
$$y>x^2+2x+1$$
D
$$y>2x^2-4x-3$$
E
$$y<x^2-2x+1$$
Question 2 of 5Determine which Inequality Matches the Graph
Select the Correct Answer Below: Correct! Not Correct!
A
$$x^2+(y-2)^2<3$$
B
$$x^2-(y+2)^2>9$$
C
$$(x+2)^2+y^2<3$$
D
$$x^2+(y+2)^2<9$$
E
$$(x+2)^2+y^2>9$$
Question 3 of 5Determine which Inequality Matches the Graph
Select the Correct Answer Below: Correct! Not Correct!
A
$$\frac{x^2}{25}+\frac{y^2}{25}>1$$
B
$$\frac{(x-1)^2}{5}-\frac{y^2}{5}<1$$
C
$$\frac{x^2}{9}+\frac{(y-4)^2}{25}>1$$
D
$$\frac{(x+1)^2}{5}-\frac{y^2}{3}<1$$
E
$$\frac{(x+1)^2}{25}+\frac{y^2}{9}<1$$
Question 4 of 5Determine which System Matches the Graph
Select the Correct Answer Below: Correct! Not Correct!
A
$$\frac{(x-2)^2}{9}+\frac{(y-1)^2}{36}<1$$$$x-2y≤4$$ $$\frac{(x-2)^2}{9}\hspace{.25em}+$$$$\frac{(y-1)^2}{36}<1$$$$x-2y≤4$$
B
$$\frac{(x+2)^2}{3}+\frac{(y+1)^2}{6}<1$$$$x+y≥-2$$ $$\frac{(x+2)^2}{3}\hspace{.25em}+$$$$\frac{(y+1)^2}{6}<1$$$$x+y≥-2$$
C
$$\frac{x^2}{9}+\frac{y^2}{36}<1$$$$2x-y≤1$$ $$\frac{x^2}{9}+\frac{y^2}{36}<1$$$$2x-y≤1$$
D
$$\frac{(x+4)^2}{3}-\frac{(y-2)^2}{6}<1$$$$3x-2y≥2$$ $$\frac{(x+4)^2}{3}\hspace{.25em}-$$$$\frac{(y-2)^2}{6}<1$$$$3x-2y≥2$$
E
$$\frac{x^2}{9}-\frac{(y-1)^2}{6}<1$$$$x-2y≤-1$$ $$\frac{x^2}{9}-\frac{(y-1)^2}{6}<1$$$$x-2y≤-1$$
Question 5 of 5Determine which System Matches the Graph
Select the Correct Answer Below: Correct! Not Correct!
A
$$\frac{x^2}{2}+\frac{y^2}{4}≥1$$$$(x+1)^2-(y+2)^2<7$$ $$\frac{x^2}{2}+\frac{y^2}{4}≥1$$$$(x+1)^2\hspace{.25em}-$$$$(y+2)^2<7$$
B
$$\frac{(x-2)^2}{4}-\frac{y^2}{25}≥1$$$$(x-1)^2+(y+1)^2<49$$ $$\frac{(x-2)^2}{4}-\frac{y^2}{25}≥1$$$$(x-1)^2\hspace{.25em}+$$$$(y+1)^2<49$$
C
$$\frac{(x-1)^2}{4}+\frac{(y-1)^2}{25}≤1$$$$(x+1)^2-(y-1)^2>49$$ $$\frac{(x-1)^2}{4}\hspace{.25em}+$$$$\frac{(y-1)^2}{25}≤1$$$$(x+1)^2\hspace{.25em}-$$$$(y-1)^2>49$$
D
$$\frac{(x+3)^2}{4}-\frac{(y-3)^2}{4}≥1$$$$x^2+y^2>1$$ $$\frac{(x+3)^2}{4}\hspace{.25em}-$$$$\frac{(y-3)^2}{4}≥1$$$$x^2+y^2>1$$
E
$$\frac{(x+2)^2}{4}+\frac{y^2}{4}≤1$$$$(x+7)^2+(y+7)^2<1$$ $$\frac{(x+2)^2}{4}+\frac{y^2}{4}≤1$$$$(x+7)^2\hspace{.25em}+$$$$(y+7)^2<1$$

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