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High School Math : High School Math

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Example Questions

Example Question #1 : How To Find The Diameter Of A Sphere

What is the diameter of a sphere with a volume of $36\pi$ ?

Possible Answers:

Correct answer:

Explanation:

The volume of a sphere is determined by the following equation:

$V=\frac{4}{3}\pi r^3$

$36\pi=\frac{4}{3}\pi r^3$

$(\frac{3}{4})36\pi=(\frac{3}{4})(\frac{4}{3})\pi r^3$

$\frac{27\pi}{\pi}=\frac{\pi r^3}{\pi}$

27=r^3

$\sqrt[3]{27}=\sqrt[3]{r^3}$

3=r

d=2r

d=2(3)=6

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Example Question #691 : Geometry

Find the radius of a sphere whose surface area is $SA=100cm^{2}$ .

Possible Answers:

35.45cm

2.82cm

Not enough information to solve

3.54cm

8.22cm

Correct answer:

2.82cm

Explanation:

We know that the surface area of the spere is $SA=100cm^{2}$ .

$SA=4\pi r^{2}$

$100cm^{2}=4\pi r^{2}$

Rearrange and solve for .

$r^{2}=\frac{100cm^{2}}{4\pi}$

$\dpi{100} r=\sqrt{\frac{100cm^{2}}{4\pi}}$

$\dpi{100}\rightarrow 2.82cm$

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Example Question #1 : How To Find The Radius Of A Sphere

What is the radius of a sphere that has a surface area of $16\pi$ ?

Possible Answers:

$2\sqrt{\pi }$

$8\sqrt{\pi }$

$8\pi$

$\sqrt{2\pi }$

Correct answer:

Explanation:

The standard equation to find the area of a sphere is $SA=4\pi r^2$ where denotes the radius. Rearrange this equation in terms of :

$r=\sqrt{\frac{SA}{4\pi }}$

To find the answer, substitute the given surface area into this equation and solve for the radius:

$r=\sqrt{\frac{16\pi }{4\pi }}=\sqrt{4}=2$

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Example Question #3 : How To Find The Radius Of A Sphere

Given that the volume of a sphere is $12\pi$ , what is the radius?

Possible Answers:

$9\pi$

$\sqrt{9\pi }$

$\sqrt[3]{12 }$

$\sqrt[3]{9}$

Correct answer:

$\sqrt[3]{9}$

Explanation:

The standard equation to find the volume of a sphere is

$V=\frac{4}{3}\pi r^3$ where denotes the radius. Rearrange this equation in terms of :

$r=\sqrt[3]{\frac{3V}{4\pi }}$

Substitute the given volume into this equation and solve for the radius:

$r=\sqrt[3]{\frac{3V}{4\pi }}=\sqrt[3]{\frac{3\cdot 12\pi }{4\pi }}=\sqrt[3]{\frac{36\pi }{4\pi }}=\sqrt[3]{9 }$

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Example Question #4 : How To Find The Radius Of A Sphere

What is the radius of a sphere with a volume of $288\pi$ ?

Possible Answers:

Correct answer:

Explanation:

$V=\frac{4}{3}\pi r^3$

$288\pi=\frac{4}{3}\pi r^3$

$\frac{288\pi}{\pi}=\frac{4\pi r^3}{3\pi}$

$288=\frac{4r^3}{3}$

$(288)(\frac{3}{4})=\frac{3}{4}(\frac{4r^3}{3})$

216=r^3

$\sqrt[3]{216}=\sqrt[3]{r^3}$

6=r

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Example Question #1 : Pre Calculus

What is the center and radius of the circle indicated by the equation?

(x-2)^2+y^2=36

Possible Answers:

$(2,0),\ r=6$

$(-2,0),\ r=6$

$(2,0),\ r=36$

$(-2,0),\ r=36$

Correct answer:

$(2,0),\ r=6$

Explanation:

A circle is defined by an equation in the format (x-h)^2+(y-k)^2=r^2 .

The center is indicated by the point (h,k) and the radius .

In the equation (x-2)^2+(y)^2=36=6^2 , the center is (2,0) and the radius is .

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Example Question #2 : Pre Calculus

What is the shape of the graph indicated by the equation?

$\frac{x^{2}}{16}+\frac{y^2}{4}=1$

Possible Answers:

Hyperbola

Ellipse

Circle

Parabola

Correct answer:

Ellipse

Explanation:

An ellipse has an equation that can be written in the format $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ . The center is indicated by (h,k) , or in this case (0,0) .

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Example Question #1 : Conic Sections

A conic section is represented by the following equation:

$\small -4x^{2} + 15y^{2} + 343 =0$

What type of conic section does this equation represent?

Possible Answers:

Parabola

Ellipse

Hyperbola

Circle

Correct answer:

Hyperbola

Explanation:

The simplest way to know what kind of conic section an equation represents is by checking the coefficients in front of each variable. The equation must be in general form while you do this check. Luckily, this equation is already in general form, so it's easy to see. The general equation for a conic section is the following:

$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

Assuming the term Bxy is 0 (which it usually is):

If A equals C, the equation is a circle.
If A and C have the same sign (but are not equal to each other), the equation is an ellipse.
If either A or C equals 0, the equation is a parabola.
If A and C are different signs (i.e. one is negative and one is positive), the equation is a hyperbola.

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Example Question #1 : Pre Calculus

A conic section is represented by the following equation:

$\frac{(x-3)^{2}}{4}-\frac{(y+2)^{2}}{9}=1$

Which of the following best describes this equation?

Possible Answers:

vertical ellipse with center (-3,2) and a major axis length of

horizontal hyperbola with center of (3,-2) and asymptotes with slopes of $\frac{3}{2}$ and $-\frac{3}{2}$

vertical parabola with vertex (-3, 2) and a vertical stretch factor of

vertical hyperbola with center (3, -2) and asymptotes with slopes of $\frac{2}{3}$ and $-\frac{3}{2}$

horizontal hyperbola with center (3,-2) and asymptotes with slopes of $\frac{2}{3}$ and $-\frac{2}{3}$

Correct answer:

horizontal hyperbola with center of (3,-2) and asymptotes with slopes of $\frac{3}{2}$ and $-\frac{3}{2}$

Explanation:

First, we need to make sure the conic section equation is in a form we recognize. Luckily, this equation is already in standard form:

$\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1$

The first step is to determine the type of conic section this equation represents. Because there are two squared variables ( $x^{2}$ and $y^{2}$ ), this equation cannot be a parabola. Because the coefficients in front of the squared variables are different signs (i.e. one is negative and the other is positive), this equation must be a hyperbola, not an ellipse.

In a hyperbola, the squared term with a positive coefficient represents the direction in which the hyperbola opens. In other words, if the $x^{2}$ term is positive, the hyperbola opens horizontally. If the $y^{2}$ term is positive, the hyperbola opens vertically. Therefore, this is a horizontal hyperbola.

The center is always found at (h, k) , which in this case is (3, -2) .

That leaves only the asymptotes. For a hyperbola, the slopes of the asymptotes can be found by dividing by (remember to always put the vertical value, , above the horizontal value, ). Remember that these slopes always come in pairs, with one being positive and the other being negative.

In this case, is 3 and is 2, so we get slopes of $\frac{3}{2}$ and $-\frac{3}{2}$ .

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Example Question #1 : Pre Calculus

Find the vertex $\left ( x,y \right )$ for a parabola with equation

y=3x^2-6x+1

Possible Answers:

$\left ( 2,-1 \right )$

$\left ( 1,2 \right )$

$\left ( 1,1 \right )$

$\left ( 1,3 \right )$

$\left ( 1,-2 \right )$

Correct answer:

$\left ( 1,-2 \right )$

Explanation:

For any parabola of the form ax^2+bx+c , the -coordinate of its vertex is

$x = \frac{-b}{2a}$

So here, we have

$x=\frac{-(-6))}{2(3)}$

We plug this back into the original equation to find :

y = 3(1)^2-6(1)+1

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : How To Find The Diameter Of A Sphere

Example Question #691 : Geometry

Example Question #1 : How To Find The Radius Of A Sphere

Example Question #3 : How To Find The Radius Of A Sphere

Example Question #4 : How To Find The Radius Of A Sphere

Example Question #1 : Pre Calculus

Example Question #2 : Pre Calculus

Example Question #1 : Conic Sections

Example Question #1 : Pre Calculus

Example Question #1 : Pre Calculus

All High School Math Resources

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