Question about Educational & Reference Software

It varies by the shape. Anything sort of circular usually involves equations with Pi, but objects with corners are different. Since Fixya will not allow me to type equations very well here, I'll refer you to

http://www.cliffsnotes.com/math/geometry/perimeter-and-area/formulas-perimeter-circumference-area

http://www.coolmath.com/reference/perimeters.html

http://www.mathsisfun.com/geometry/ellipse-perimeter.html

http://math2.org/math/geometry/areasvols.htm

http://www.cliffsnotes.com/math/geometry/perimeter-and-area/formulas-perimeter-circumference-area

http://www.coolmath.com/reference/perimeters.html

http://www.mathsisfun.com/geometry/ellipse-perimeter.html

http://math2.org/math/geometry/areasvols.htm

Oct 12, 2014 | Mathsoft StudyWorks! Middle School Deluxe...

A set is a collection (list, basket, bag, ...) of objects (things, letters, numbers, ...) In a set the order in which objects are enumerated (listed) is unimportant. In a set there is no duplication: each element of the set is listed ONLY once. That is about it, as far as a simple definition is concerned.

Sep 17, 2014 | Educational & Reference Software

3 (6c + 5) + 2 (c + 4) now multiply into the brackets

= 18c + 15 + 2c + 8 now gather like terms

= 20c + 23 and that's it

= 18c + 15 + 2c + 8 now gather like terms

= 20c + 23 and that's it

Apr 10, 2014 | Educational & Reference Software

Is this a right circular cone or a right circular cylinder?

Volume of cylinder= Pi*(r^2)*h

Volume of cone is 1/3 of volume of cylinder with same height and radius.

Solve for h

**h=(Volume)/(PI*r^2) for cylinder**

Volume of cylinder= Pi*(r^2)*h

Volume of cone is 1/3 of volume of cylinder with same height and radius.

Solve for h

Dec 05, 2013 | The Learning Company Achieve! Math &...

You first need to know how to calculate volume of a cube. Take the length of the side of the cube and multiply it by itself, and then multiply by itself one final time. For example, if you have a cube that has a side length of 2 cm, you would multiply 2 x 2 to get 4, and then 4 x 2 to get 8 cubic cm.

You are asking about a 1 cubic meter cube. I know that a side of that cube is 1 m (1 x 1 x 1 = 1). Convert one meter to cm. Therre are 100 cm in one meter. The new math will be 100 x 100 x 100 = 1,000,000 cubic cm.

You are asking about a 1 cubic meter cube. I know that a side of that cube is 1 m (1 x 1 x 1 = 1). Convert one meter to cm. Therre are 100 cm in one meter. The new math will be 100 x 100 x 100 = 1,000,000 cubic cm.

Jan 18, 2010 | Mathsoft StudyWorks! Middle School Deluxe...

Hi there,

Say you have an equation x = 2y.

On a graph, you plot the points representing the values that fit into the equation.

X Y

2 1

4 2

6 3

etc.

And to throw a curve in there:

X Y

-2 -1

-4 -2

-6 -3

If have further questions repost, I'll be around for a while.

Hope it helps

Mike

Say you have an equation x = 2y.

On a graph, you plot the points representing the values that fit into the equation.

X Y

2 1

4 2

6 3

etc.

And to throw a curve in there:

X Y

-2 -1

-4 -2

-6 -3

If have further questions repost, I'll be around for a while.

Hope it helps

Mike

Nov 24, 2009 | The Learning Company Achieve! Math &...

A two by two grid forms four quadrants. In the upper left quadrant, place a 4. In the upper right quadrant place a 2. In the lower left quadrant, place an eight. In the lower right quadrant place a 5. Horizontally, the numbers 42 and 85 are formed. Vertically, you form the numbers 48 and 25. The sum total of these numbers is 200. 42 + 85 + 48 + 25 = 200.

Nov 11, 2009 | Mathsoft StudyWorks! Middle School Deluxe...

Hello,

I will give you some hints, and indications. You do the Math.

You have a certain volume of dirt that has to be moved. For that you use a pale the volume of which you do not know right now, but can calculate.

The final question is: How many time do you have to scoop with the pale to move the volume of dirt.

Answer: The number of times I have to fill the pale is equal to the Total volume of dirt divided by the as yet unknown volume of the pale.

As you can see we cannot proceed further until we find the volume of the pale.

**Volume of pale **

Solid Shape? Cylinder (its base is circular)

Volume of a cylinder: Area of the circular base times the height (10inc.)

To calculate the volume I need the area of the circular base.

Area of the base =Pi* square of radius = Pi *(6)^2 in^2 (do not forget the units)

Volume of pale = Pi* (6^2)*10 in^2 * in =Pi*10*6^2 cubic inches.

Number of times one has to scoop.

**Number = Volume of dirt / volume of pale**

You cannot proceed further until you correct the error in the data. As you can see, when you calculate the volume of the pale the unit is cubic inches. However, your question states 5600 cubic of dirt.

Do you see what is missing?

Now you are on your own. Good sailing.

I will give you some hints, and indications. You do the Math.

You have a certain volume of dirt that has to be moved. For that you use a pale the volume of which you do not know right now, but can calculate.

The final question is: How many time do you have to scoop with the pale to move the volume of dirt.

Answer: The number of times I have to fill the pale is equal to the Total volume of dirt divided by the as yet unknown volume of the pale.

As you can see we cannot proceed further until we find the volume of the pale.

Solid Shape? Cylinder (its base is circular)

Volume of a cylinder: Area of the circular base times the height (10inc.)

To calculate the volume I need the area of the circular base.

Area of the base =Pi* square of radius = Pi *(6)^2 in^2 (do not forget the units)

Volume of pale = Pi* (6^2)*10 in^2 * in =Pi*10*6^2 cubic inches.

Number of times one has to scoop.

You cannot proceed further until you correct the error in the data. As you can see, when you calculate the volume of the pale the unit is cubic inches. However, your question states 5600 cubic of dirt.

Do you see what is missing?

Now you are on your own. Good sailing.

Nov 01, 2009 | The Learning Company Achieve! Math &...

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