Call Now to Set Up Tutoring:

(888) 888-0446

Common Core: High School - Geometry : High School: Geometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

Create An Account

All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #9 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( 6, 8\right ) \\ \text{directrix } y = 10$

Possible Answers:

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

$y = \frac{x}{16} \left(- x + 12\right)$

$\left(x - 8\right)^{2} + \left(x - 6\right)^{2}$

$y = \frac{x}{8} \left(- x + 12\right)$

$y = \frac{x}{4} \left(- x + 12\right)$

Correct answer:

$y = \frac{x}{4} \left(- x + 12\right)$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 6 for a 8 for b and 10 for y

$y_{0}^{2} - 20 y_{0} + 100 = x_{0}^{2} - 12 x_{0} + y_{0}^{2} - 16 y_{0} + 100$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}}{4} \left(- x_{0} + 12\right)$

So our answer is then

$y = \frac{x}{4} \left(- x + 12\right)$

Report an Error

Example Question #10 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -10, -3\right ) \\\text{directrix } y = -4$

Possible Answers:

$y = \frac{x^{2}}{4} + 5 x + \frac{93}{4}$

$y = \frac{x^{2}}{8} + \frac{5 x}{2} + \frac{93}{8}$

$\left(x + 3\right)^{2} + \left(x + 10\right)^{2}$

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

$y = \frac{x^{2}}{2} + 10 x + \frac{93}{2}$

Correct answer:

$y = \frac{x^{2}}{2} + 10 x + \frac{93}{2}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -10 for a -3 for b and -4 for y

$y_{0}^{2} + 8 y_{0} + 16 = x_{0}^{2} + 20 x_{0} + y_{0}^{2} + 6 y_{0} + 109$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{2} + 10 x_{0} + \frac{93}{2}$

So our answer is then

$y = \frac{x^{2}}{2} + 10 x + \frac{93}{2}$

Report an Error

Example Question #11 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( 2, 5\right ) \\\text{directrix } y = -6$

Possible Answers:

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

$y = \frac{x^{2}}{88} - \frac{x}{22} - \frac{7}{88}$

$\left(x - 5\right)^{2} + \left(x - 2\right)^{2}$

$y = \frac{x^{2}}{22} - \frac{2 x}{11} - \frac{7}{22}$

$y = \frac{x^{2}}{44} - \frac{x}{11} - \frac{7}{44}$

Correct answer:

$y = \frac{x^{2}}{22} - \frac{2 x}{11} - \frac{7}{22}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 2 for a 5 for b and -6 for y

$y_{0}^{2} + 12 y_{0} + 36 = x_{0}^{2} - 4 x_{0} + y_{0}^{2} - 10 y_{0} + 29$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{22} - \frac{2 x_{0}}{11} - \frac{7}{22}$

So our answer is then

$y = \frac{x^{2}}{22} - \frac{2 x}{11} - \frac{7}{22}$

Report an Error

Example Question #12 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -8, 9\right ) \\\text{directrix } y = 12$

Possible Answers:

$\left(x - 9\right)^{2} + \left(x + 8\right)^{2}$

$y = - \frac{x^{2}}{24} - \frac{2 x}{3} - \frac{1}{24}$

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

$y = - \frac{x^{2}}{12} - \frac{4 x}{3} - \frac{1}{12}$

$y = - \frac{x^{2}}{6} - \frac{8 x}{3} - \frac{1}{6}$

Correct answer:

$y = - \frac{x^{2}}{6} - \frac{8 x}{3} - \frac{1}{6}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -8 for a 9 for b and 12 for y

$y_{0}^{2} - 24 y_{0} + 144 = x_{0}^{2} + 16 x_{0} + y_{0}^{2} - 18 y_{0} + 145$

Now we can simplify, and solve for $y_{0}$

$y_{0} = - \frac{x_{0}^{2}}{6} - \frac{8 x_{0}}{3} - \frac{1}{6}$

So our answer is then

$y = - \frac{x^{2}}{6} - \frac{8 x}{3} - \frac{1}{6}$

Report an Error

Example Question #13 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -6, 1\right ) \\\text{directrix } y = -5$

Possible Answers:

$y = \frac{x^{2}}{12} + x + 1$

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

$\left(x - 1\right)^{2} + \left(x + 6\right)^{2}$

$y = \frac{x^{2}}{48} + \frac{x}{4} + \frac{1}{4}$

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{1}{2}$

Correct answer:

$y = \frac{x^{2}}{12} + x + 1$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -6 for a 1 for b and -5 for y

$y_{0}^{2} + 10 y_{0} + 25 = x_{0}^{2} + 12 x_{0} + y_{0}^{2} - 2 y_{0} + 37$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{12} + x_{0} + 1$

So our answer is then

$y = \frac{x^{2}}{12} + x + 1$

Report an Error

Example Question #14 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -6, 6\right ) \\\text{directrix } y = -6$

Possible Answers:

$y = \frac{x^{2}}{48} + \frac{x}{4} + \frac{3}{4}$

$y = \frac{x^{2}}{96} + \frac{x}{8} + \frac{3}{8}$

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

$\left(x - 6\right)^{2} + \left(x + 6\right)^{2}$

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{3}{2}$

Correct answer:

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{3}{2}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -6 for a 6 for b and -6 for y

$y_{0}^{2} + 12 y_{0} + 36 = x_{0}^{2} + 12 x_{0} + y_{0}^{2} - 12 y_{0} + 72$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{24} + \frac{x_{0}}{2} + \frac{3}{2}$

So our answer is then

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{3}{2}$

Report an Error

Example Question #1 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (9,3) and (-5,3) with a major axis distance of .

Possible Answers:

$10 x^{2} + 10 y^{2} = 1$

$x^{2} + y^{2} = 1$

$\frac{1}{25} \left(x - 2\right)^{2} - \frac{1}{24} \left(y - 3\right)^{2} = 1$

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

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

Correct answer:

$\frac{1}{25} \left(x - 2\right)^{2} - \frac{1}{24} \left(y - 3\right)^{2} = 1$

Explanation:

The general equation is

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

$\left ( k, h\right )$ is the coordinates of the center of the foci.

{a} is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is .

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 10

Now solve for

a = 5.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = 2 \\y = 3.0$

So the center of the foci is at $\left ( 2, 3\right )$ .

Now we need to figure out what the distance from the center to the foci is .

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

c = -7

Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 5.0 and with -7

$b^{2} = -24.0$

Now we can substitute these values into the general equation to get.

$\frac{1}{25} \left(x - 2\right)^{2} - \frac{1}{24} \left(y - 3\right)^{2} = 1$

Report an Error

Example Question #2 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (7, 1) and (-7, 1) with a major axis distance of .

Possible Answers:

$18 x^{2} + 18 y^{2} = 1$

$\frac{x^{2}}{81} + \frac{1}{32} \left(y - 1\right)^{2} = 1$

$x^{2} + y^{2} = 1$

$\frac{x^{2}}{18} + \frac{y^{2}}{18} = 1$

$\frac{1}{18} \left(x - 7\right)^{2} + \frac{1}{18} \left(y - 1\right)^{2} = 1$

Correct answer:

$\frac{x^{2}}{81} + \frac{1}{32} \left(y - 1\right)^{2} = 1$

Explanation:

The general equation is

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

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is .

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 18

Now solve for .

a = 9.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = 0 \\y = 1.0$

So the center of the foci is at $\left ( 0, \quad 1\right )$

Now we need to figure out what the distance from the center to the foci is .

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

c = -7

Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 9.0 and with -7

$b^{2} = 32.0$

Now we can substitute these values into the general equation to get.

$\frac{x^{2}}{81} + \frac{1}{32} \left(y - 1\right)^{2} = 1$

Report an Error

Example Question #3 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (3, 4) and (-5, 4) with a major axis distance of .

Possible Answers:

$x^{2} + y^{2} = 1$

$18 x^{2} + 18 y^{2} = 1$

$\frac{x^{2}}{18} + \frac{y^{2}}{18} = 1$

$\frac{1}{18} \left(x - 3\right)^{2} + \frac{1}{18} \left(y - 4\right)^{2} = 1$

$\frac{1}{81} \left(x + 1\right)^{2} + \frac{1}{65} \left(y - 4\right)^{2} = 1$

Correct answer:

$\frac{1}{81} \left(x + 1\right)^{2} + \frac{1}{65} \left(y - 4\right)^{2} = 1$

Explanation:

The general equation is

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

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is.

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 18

Now solve for

a = 9.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = -1 \\y = 4.0$

So the center of the foci is at $\left ( -1, \quad 4\right )$

Now we need to figure out what the distance from the center to the foci is.

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

c = -4

Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 9.0 and with -4

$b^{2} = 65.0$

Now we can substitute these values into the general equation to get.

$\frac{1}{81} \left(x + 1\right)^{2} + \frac{1}{65} \left(y - 4\right)^{2} = 1$

Report an Error

Example Question #4 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (-3, 5) and (-1, 5) with a major axis distance of .

Possible Answers:

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

$\frac{1}{64} \left(x + 2\right)^{2} + \frac{1}{63} \left(y - 5\right)^{2} = 1$

$x^{2} + y^{2} = 1$

$\frac{1}{16} \left(x + 3\right)^{2} + \frac{1}{16} \left(y - 5\right)^{2} = 1$

$16 x^{2} + 16 y^{2} = 1$

Correct answer:

$\frac{1}{64} \left(x + 2\right)^{2} + \frac{1}{63} \left(y - 5\right)^{2} = 1$

Explanation:

The general equation is

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

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is.

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 16

Now solve for

a = 8.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = -2 \\ y = 5.0$

So the center of the foci is at $\left ( -2, \quad 5\right )$

Now we need to figure out what the distance from the center to the foci is .

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate.

c = 1

Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 8.0 and with 1

$b^{2} = 63.0$

Now we can substitute these values into the general equation to get.

$\frac{1}{64} \left(x + 2\right)^{2} + \frac{1}{63} \left(y - 5\right)^{2} = 1$

Report an Error

View Tutors

Latoya
Certified Tutor

Kaplan University-Maine Campus, Bachelor of Science, Criminal Justice. University of Phoenix-Chicago Campus, Masters in Busin...

View Tutors

Alec
Certified Tutor

University of Kentucky, Bachelor of Science, Economics.

View Tutors

Kunal
Certified Tutor

Indiana University-Bloomington, Bachelor of Science, Finance.

All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

MCAT Tutors in Boston, English Tutors in Philadelphia, Calculus Tutors in Washington DC, GMAT Tutors in San Diego, ACT Tutors in Washington DC, MCAT Tutors in New York City, Spanish Tutors in San Diego, Algebra Tutors in Washington DC, GRE Tutors in Miami, Algebra Tutors in Chicago

Popular Courses & Classes

SSAT Courses & Classes in Philadelphia, ISEE Courses & Classes in Phoenix, ACT Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in Philadelphia, SSAT Courses & Classes in San Diego, GMAT Courses & Classes in Miami, SSAT Courses & Classes in Houston, ISEE Courses & Classes in Houston, Spanish Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Boston

Popular Test Prep

SAT Test Prep in Phoenix, LSAT Test Prep in Dallas Fort Worth, SAT Test Prep in Denver, GMAT Test Prep in New York City, SSAT Test Prep in Phoenix, SSAT Test Prep in Chicago, GMAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Boston, MCAT Test Prep in Los Angeles, LSAT Test Prep in Miami

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Common Core: High School - Geometry : High School: Geometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #9 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #10 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #11 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #12 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #13 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #14 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #1 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #2 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #3 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #4 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

All Common Core: High School - Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: