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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 $y$.
$(20y-12)/(4y)=4$
1. $3$
2. $5$
3. $2$
4. $-1$
Grade 10 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve for $t$.
$sqrt(5t+11)=6$
1. $11$
2. $46/55$
3. $10$
4. $5$
Grade 10 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Grade 11 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Grade 11 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve for all possible values of $x$. $sqrt(x^2+4x-16)=4$
1. No Solution
2. $-8, 4$
3. All real numbers
4. $-4, 16$
Grade 11 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Grade 10 Nonlinear Equations and Functions
Grade 9 Nonlinear Equations and Functions
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. $f(x) = 7,400,000 (1 + .054)^x$
2. $f(x) = 7,400,000 (1 - .054)^x$
3. $f(x) = 7,400,000 (.054)^x$
4. $f(x) = 7,400,000x^(.054)$
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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After some more algebra, the equation becomes:
$(c^2-a^2) x^2 - y^2 = (c^2 - a^2) y^2$
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 CCSS: HSG-GPE.A.3

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After some more algebra, and then moving all terms with $x$ or $y$ to the left side of the equation (and all others to the right), the resulting equation is:

 $a^2 x^2 - c^2 x^2 + a^2 y^2$ $=$ $a^4 -a^2 c^2$ $(a^2 - c^2)x^2 + a^2 y^2$ $=$ $(a^2 - c^2) a^2$ $b^2 x^2 + a^2 y^2$ $=$ $a^2 b^2$ $x^2 / a^2 + y^2 / b^2$ $=$ $1$

This is the formula for the equation of an ellipse centered at $(0,0)$. Why can the substitution $b^2 = a^2-c^2$ be made?
1. Since $b$ is undefined so far, it can be defined as any value. Then, using the Pythagorean theorem, $a^2 + b^2 = c^2$, simply rearrange to solve for $b^2$.
2. Knowing that the semi-minor axis is $b$ units long, one can substitute the square of this value for $a^2-c^2$.
3. Since $b$ is not yet defined, it can be used to simplify the equation by defining $b^2 = a^2 - c^2$. A positive value for $a^2-c^2$ exists since $|a| > |c|$.
4. Projecting $b$, the length of the semi-minor axis, onto the semi-major axis it is seen that $b=a-c$. Then, simply square both sides of the equation.
Grade 10 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve for $x$.
$(2x-3)/(x-5)=3$
1. $12$
2. $21/7$
3. $2$
4. $2/3$
Grade 9 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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Substituting the equations for $d_1, d_2$ found in the second and third questions into the above equation, and then moving the second square root term to the right side of the equation results in the following:
$sqrt( (x+c)^2 + y^2) = 2a - sqrt( (x-c)^2 + y^2)$
The next step will be to square both sides. Why was the second square root term moved to the right side of the equation before squaring, instead of just squaring both sides immediately?
1. It gets rid of all square root terms immediately.
2. It will make the algebra easier later on, since the two square root terms are not multiplied together.
3. This ensures that there won't be any multiple answers that usually result from square roots (taking the positive and negative).
4. This makes that equation more balanced, which looks nicer.
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1

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Having found the equation of the circle centered at $(3,4)$ 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 10 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve for $x$.
$sqrt(7x)/x=x/7$
1. $1$
2. $1/7$
3. $7$
4. $-1$
Grade 9 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
Solve the equation $5/(2x-3)=3/(x+5)$.
1. $x=16$
2. $x=4$
3. $x=19$
4. $x=34$
Grade 12 Nonlinear Equations and Functions
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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Letting $b^2 = c^2 - a^2$, the equation no becomes:
$b^2 x^2 - y^2 = b^2 a^2$
$x^2 / a^2 - y^2 / b^2 = 1$
This is the equation of a hyperbola, centered at the origin with foci $(-c,0)$ and $(c,0)$. Where does the substitution $b^2 = c^2 - a^2$ come from?
1. Since $b$ usually appears in the equation for a hyperbola, it must be included. Using the Pythagorean theorem, $a^2 +b^2 = c^2$, 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 $y= pm b/a$, and knowing that $|c| > |a|$, squaring and rearranging results in $b^2 = c^2 - a^2$.
3. It's done to simply the equation. $b$ is not defined yet, and since $|c| > |a|$, $c^2 > a^2$, and so there must be a positive number, $b^2$ such that $b^2 = c^2 - a^2$.
4. Knowing that $b$ is the length of the semi-minor axis, a right triangle can be formed with the center of the hyperbola and either foci, with $b$ as the length of one leg of this triangle. Applying the Pythagorean theorem results in $b^2 + c^2 = a^2$, and simply rearrange.
Grade 9 Nonlinear Equations and Functions
$4x^3 = - 1/54$
1. $- 1/3$
2. $- 1/6$
3. $- 1/9$
4. $- 1/12$
Grade 9 Nonlinear Equations and Functions CCSS: HSA-REI.A.2
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