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Eleventh Grade (Grade 11) Nonlinear Equations and Functions Questions

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Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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Substitute the equations for d1,d2 found in the second and third questions into the equation in the previous question. After some rearranging, the result is the following equation:

(x+c)2+y2=±2a+(x-c)2+y2

Why is there a plus-minus sign on the 2a term?
  1. In order to remove the absolute value sign, one has to consider the case where d1>d2 and d1<d2.
  2. To remove the square root from this term, one has to look at the positive and negative case.
  3. Since it's not certain which side of the equation this term should be on, it could be either positive or negative.
  4. This is a hyperbola and the curves can be on either the positive or negative x-axis.
Grade 11 Nonlinear Equations and Functions
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1

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Grade 11 Nonlinear Equations and Functions
Grade 11 Nonlinear Equations and Functions
Solve x3=216.
  1. x=9
  2. x=6
  3. x=81
  4. x=27
Grade 11 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve. Check for extraneous solutions.

xx-1=x+3-2x+2
  1. x=-1orx=-2
  2. x=-1
  3. x=1
  4. x=1orx=-1
Grade 11 Nonlinear Equations and Functions
The equation x2+y2+2x+1=0 represents
  1. a point.
  2. a circle.
  3. a parabola.
  4. an ellipse.
Grade 11 Nonlinear Equations and Functions
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.2
What is the equation of a parabola with the focus at (1,3) and a directrix of x=-5?
  1. (x-3)2=12(y+2)
  2. (y-3)2=12(x-1)
  3. (y+3)2=3(x-2)
  4. (y-3)2=12(x+2)
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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What does |d1-d2| equal?
  1. |d1-d2|=a
  2. |d1-d2|=c
  3. |d1-d2|=|a-c|
  4. |d1-d2|=2a
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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What is the distance, d2, between the F2 and the point (x,y)?
  1. d2=(x-c)2+(y-c)2
  2. d2=x2+y2
  3. d2=(x+c)2+y2
  4. d2=(x-c)2+y2
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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It can be shown that the value b from the previous question relates to the vertices of the minor-axis. Specifically, the vertices are (0,-b) and (0,b). Looking at the positive vertex, it forms an isosceles triangle with the two foci. What is the length of the two congruent sides? How is this related to the previous question?
  1. They each have a length of 2c. Using the distance formula and looking at the difference of distances between the lengths just found and the other vertices of the major axes, one finds that b=a2-c2.
  2. They each have a length of c. Then, the right triangle formed between the vertices and the origin, and applying the Pythagorean theorem, results in a2+b2=c2.
  3. They each have a length of a. Looking at the right triangle formed by the origin, F1, and the vertex (0,b), and applying the Pythagorean theorem results in a2=b2+c2.
  4. They each have a length of a2. Therefore, the sum of their lengths, a, can be used as a value equal to the sum of the lengths of b and c.
Grade 11 Nonlinear Equations and Functions

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Using the distances between the two foci and the vertex (-a,0), what does d1+d2 equal?
  1. d1+d2=2a-c
  2. d1+d2=2a-c
  3. d1+d2=2a
  4. d1+d2=c
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1

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Grade 11 Nonlinear Equations and Functions
Factor and solve. x3-64=0
  1. -4,2±2i3
  2. 4,-2±2i3
  3. 2,-4±4i3
  4. -4,4±4i3
Grade 11 Nonlinear Equations and Functions
Grade 11 Nonlinear Equations and Functions
Given the equation of a hyperbola (y-5)216-(x+3)24=1, identify the foci.
  1. (-3±25,0)
  2. (0,5±25)
  3. (5±5,-3)
  4. (-3,5±25)
Grade 11 Nonlinear Equations and Functions
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