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Nonlinear Equations and Functions Questions - All Grades

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Grade 10 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve for [math]y[/math].
[math](20y-12)/(4y)=4[/math]
  1. [math]3[/math]
  2. [math]5[/math]
  3. [math]2[/math]
  4. [math]-1[/math]
Grade 10 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve for [math]t[/math].
[math]sqrt(5t+11)=6[/math]
  1. [math]11[/math]
  2. [math]46/55[/math]
  3. [math]10[/math]
  4. [math]5[/math]
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3
Letting [math]b^2 = c^2 - a^2[/math], the equation no becomes:
[math]b^2 x^2 - y^2 = b^2 a^2[/math]
[math]x^2 / a^2 - y^2 / b^2 = 1[/math]
This is the equation of a hyperbola, centered at the origin with foci [math](-c,0)[/math] and [math](c,0)[/math]. Where does the substitution [math]b^2 = c^2 - a^2[/math] come from?
  1. Since [math]b[/math] usually appears in the equation for a hyperbola, it must be included. Using the Pythagorean theorem, [math]a^2 +b^2 = c^2[/math], simply rearrange the equation.
  2. It has to be done, to ensure the asymptotes are related to the equation. The equations of the asymptotes are [math]y= pm b/a[/math], and knowing that [math]|c| > |a|[/math], squaring and rearranging results in [math]b^2 = c^2 - a^2[/math].
  3. It's done to simply the equation. [math]b[/math] is not defined yet, and since [math]|c| > |a|[/math], [math]c^2 > a^2[/math], and so there must be a positive number, [math]b^2[/math] such that [math]b^2 = c^2 - a^2[/math].
  4. Knowing that [math]b[/math] is the length of the semi-minor axis, a right triangle can be formed with the center of the hyperbola and either foci, with [math]b[/math] as the length of one leg of this triangle. Applying the Pythagorean theorem results in [math]b^2 + c^2 = a^2[/math], and simply rearrange.
Grade 11 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Grade 10 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3
Let [math](x,y)[/math] be a point on the ellipse. Which equation represents the distance, [math]d_1[/math], between [math]F_1[/math] and the point [math](x,y) ?[/math]
  1. [math]d_1 = sqrt( (x+c)^2 + y^2)[/math]
  2. [math]d_1 = sqrt( (x-c) + y^2)[/math]
  3. [math]d_1 = sqrt( (-x + c)^2 + y^2[/math]
  4. [math]d_1 = sqrt( (x-c)^2 + (y-x)^2)[/math]
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1
Having found the equation of the circle centered at [math](3,4)[/math] with radius of length 2 in the previous question, which of the following best describes the meaning of this equation?
  1. There are always two points inside a circle with which to form a right triangle using the radius of the circle as the hypotenuse.
  2. Any point (x,y) that is 2 units away from the center of this circle lies on the circle.
  3. There exists one point, (x,y), which solves this equation.
  4. All points that lie on this circle also must be part of a right triangle.
Grade 11 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve for all possible values of [math]x[/math]. [math]sqrt(x^2+4x-16)=4[/math]
  1. No Solution
  2. [math]-8, 4[/math]
  3. All real numbers
  4. [math]-4, 16[/math]
Grade 11 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Grade 9 Nonlinear Equations and Functions CCSS: HSF-BF.A.1, HSF-BF.A.1a, HSF-LE.B.5
In the year 2000, the population of Virginia was about 7,400,000. Between the years 2000 and 2004, the population in Virginia grew at a rate of 5.4%. At this growth rate, what function gives the population x years after 2000?
  1. [math] f(x) = 7,400,000 (1 + .054)^x [/math]
  2. [math] f(x) = 7,400,000 (1 - .054)^x [/math]
  3. [math] f(x) = 7,400,000 (.054)^x [/math]
  4. [math] f(x) = 7,400,000x^(.054) [/math]
Grade 10 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve for [math]x[/math].
[math](2x-3)/(x-5)=3[/math]
  1. [math]12[/math]
  2. [math]21/7[/math]
  3. [math]2[/math]
  4. [math]2/3[/math]
Grade 10 Nonlinear Equations and Functions
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3
After some more algebra, the equation becomes:
[math](c^2-a^2) x^2 - y^2 = (c^2 - a^2) y^2[/math]
What happened to the plus-minus sign that was in the original equation in the seventh question?
  1. Since there are no more square root terms, it is unnecessary.
  2. Each term with it was eventually squared, and a squared term must be positive it.
  3. Because the terms were moved from one side to another multiple times, it is no longer necessary.
  4. Any terms with it canceled out.
Grade 11 Nonlinear Equations and Functions
Squaring both sides, expanding, and combing like terms, dividing by any common constants, and putting all terms without a square root on the left side of the equation results in which of the following equations?
  1. [math]2cx - a = 0[/math]
  2. [math]cx pm a^2 = -2 sqrt((x-c)^2 + y^2) [/math]
  3. [math]cx - a^2 = 0[/math]
  4. [math]cx - a^2 = pm a sqrt( (x-c)^2 + y^2) [/math]
Grade 12 Nonlinear Equations and Functions
Grade 9 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Grade 10 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve for [math]x[/math].
[math]sqrt(7x)/x=x/7[/math]
  1. [math]1[/math]
  2. [math]1/7[/math]
  3. [math]7[/math]
  4. [math]-1[/math]
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3
Substitute the equations for [math]d_1, d_2[/math] found in the second and third questions into the equation in the previous question. After some rearranging, the result is the following equation:

[math]sqrt( (x+c)^2 + y^2) = pm 2a + sqrt( (x-c)^2 + y^2)[/math]

Why is there a plus-minus sign on the [math]2a[/math] term?
  1. In order to remove the absolute value sign, one has to consider the case where [math]d_1>d_2[/math] and [math]d_1 < d_2[/math].
  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 CCSS: HSG-GPE.A.3
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3
Let [math](x,y)[/math] be a point on the hyperbola. What is the distance, [math]d_1[/math], between [math]F_1[/math] and the point [math](x,y) ?[/math]
  1. [math] d_1 = sqrt( (x+c)^2 + y^2) [/math]
  2. [math] d_1 = sqrt( (x-c)^2 + y^2)[/math]
  3. [math] d_1 = sqrt( (x^2 + y^2) [/math]
  4. [math]d_1 = sqrt( (x-c)^2 + (y-c)^2)[/math]
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