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breadofknowledgePosted - 12 April 2011 19:24  Show Profile
My name is chris and I have noticed through Kaplan how many people out there need help! well I plan to be a student at kaplan university one day and
I am taking my GED as of now I have learned many ways to study and I learned it is more easy to studie when a group of people study together and work together -a sudy group with people who actually want to learn that are professional and -LEAVE SEX OFF TOPIC

I ask you guys to bare with me
I am not a teacher
I do not Tollerate sexual harrassment in
the website I am creating
and this website will be STRICTLY FOR STUDYING AND SOCIALIZING

Reason why? I am 20 years old I have a wanderfull girlfriend whom I plan to marrie and I do not need her parinoid...
and I do not want to have minors publishing sexaul things like WHATS UP HOES! they will be deleted for one it is disrespectfull and to it is unrealated to focusing on the GED.

If you need help and want help on the GED -my email is jesuschosetoloveyou@yahoo.com
the domain on this website has not yet been completed and I am looking to hire ... this is non-profit so I need some admins NOTE: THEY WILL NOT GET PAID
I need some admins to help me veiw the website and focus on helping people with learning impaired issues and also just people in general I have 4 teachers from my old school going to help me
and questions are under policy that no
abusive language is allowed.
other than that talk about any thing and
be safe I wish you all luck
oh yeah i can get fined for having
abusive language as well so erm
yeah any ways if ya need some help
I have 4 teachers that I can call
my aunt is a professer and I have been
having some small issues obtaining 100% on the GED i can only get 80-90% :(
so i am creating a website its text based
and forum so there is a chat sight and music there if you wander what kind I have a Youtube widget linked up you can search for your own music this is non-profit so its free and I DO NOT ACCEPT DONATIONS.... thank you for your time and have a blessed day
good luck on the testing-and stay out of trouble lol

breadofknowledge Posted - 13 April 2011 1:41  Show Profile
ok so here is some basic tips you will need...this website rocks I was stuck on a problem now I am teaching people pre-algerbra ok sstab me if you want dont go and do the GED book it teaches algerbra you need PRE-ALGERBRA first <BR>this is why people have most issues understanding the ged books you get :)<BR>trust me dude ... <BR>This page is designed to help you better understand, work with, and solve equations. Click any of the links below to go to that section and begin understanding equations. </P><P>Addition and/or subtraction in equations <BR>Multiplication and/or division in equations <BR>Combinations of the basic operations in equations <BR>Quiz on Basic Equations </P><P><BR>--------------------------------------------------------------------------------</P><P> <BR>Equations are something that you will constantly be using throughout your math career. Learning and understanding the basics is an integral part of &quot;getting off on the right foot&quot; when dealing with math. </P><P>This section will help you better understand, work with, and solve equations when they have addition and/or subtraction in them. </P><P> </P><P>Changing the order of the addends (numbers you're adding) doesn't change their sum (what they equal when added together). Example: <BR> a + (b + c) = (a + b) + c<BR>Any number plus 0 (zero) equals itself. Example: <BR> a + 0 = a<BR>If two sides of an equation are equal, you can add or subtract the same amount to both sides, and they will still be equal. Example: <BR> a = b<BR> a + c = b + c<BR> a - c = b - c<BR> </P><P> <BR>When solving equations, remember that addition and subtraction are inverse operations - they undo each other (i.e., 10 + 9 - 9 = 10). To solve equations using addition and subtraction, first decide which operation has been applied, then use the inverse operation to undo this (remember to add or subtract from both sides of the equation). </P><P> 1. Solve: x + 79 = 194</P><P> Solution: <BR> x + 79 = 194<BR> x + 79 - 79 = 194 - 79<BR> x = 115<BR> <BR>You need to get the variable by itself (isolate the variable). <BR>To undo adding 79, subtract 79 from both sides. <BR>2. Solve: x - 56 = 604</P><P> Solution: <BR> x - 56 = 604<BR> x - 56 + 56 = 604 + 56<BR> x = 660<BR> <BR>You need to isolate the variable. <BR>To undo subtracting 56, add 56 to both sides. </P><P><BR>Back to Top </P><P></P><P>--------------------------------------------------------------------------------</P><P> <BR>This section will help you understand, work with, and solve equations of a slightly more complex nature - equations involving the use of multiplication and/or division. </P><P> </P><P>Order of operations: <BR>The operations inside parentheses () and brackets [] are done first. <BR>Then any operations involving exponents (which you will learn about later). <BR>Then do all multiplying and dividing from left to right. <BR>Finally, do all addition and subtraction from left to right. </P><P><BR>Multiplication can be written three different ways: </P><P> 9 * x <BR> 9x <BR> 9(x) </P><P><BR>A fraction bar is also a division symbol. </P><P><BR>Changing the order of multipliers (numbers you're multiplying together) doesn't change their product (total when the numbers are multiplied together). Example: </P><P>ab = ba </P><P><BR>Zero times any number is zero and 1 times any number is the number. Examples: </P><P>x(0) = 0 <BR>(0)x = 0 <BR>x(1)= x <BR>1 * x = x </P><P><BR>If two sides of an equation are equal, you can multiply or divide each side by the same quantity (number or equation) and it will still be equal. Examples: </P><P>a = b, c &lt;&gt; 0 <BR>ac = bc <BR>(a / c) = (b / c) </P><P> <BR>When solving equations, remember that multiplication and division are inverse operations, therefore they undo each other (i.e., (4 * 8)/8 = 4). To solve equations using multiplication or division, first decide which operation has been applied, then use the inverse operation to undo this (remember to multiply or divide on both sides of the equation). </P><P> 1. Solve: 6x = 36</P><P> Solution: <BR> 6x = 36<BR> (6x) / 6 = 36 / 6<BR> x = 6<BR> <BR>You need to get the variable by itself (isolate the variable). <BR>To undo multiplying by 6, divide by 6 on both sides. <BR>2. Solve: x / 5 = 10</P><P> Solution: <BR> x / 5 = 10<BR> 5(x / 5) = 10(5)<BR> x = 50<BR> <BR>You need to isolate the variable. <BR>To undo dividing by 5, multiply both sides by 5 </P><P><BR>Back to top </P><P></P><P>--------------------------------------------------------------------------------</P><P> <BR>This section will help you understand, work with, and solve complex equations that involve different combinations of multiplication, division, addition, and subtraction. </P><P> </P><P>Order of operations: <BR>The operations inside parentheses () and brackets [] are done first. <BR>Then any operations involving exponents (which you will learn about later). <BR>Then do all multiplying and dividing from left to right. <BR>Finally, do all addition and subtraction from left to right. </P><P><BR>Multiplication can be written three different ways: </P><P>7 * x <BR>7x <BR>7(x) </P><P><BR>A fraction bar is also a division symbol. </P><P><BR>Also, be sure to refer to the above sections if you have forgotten or need to review any of the other material covered. </P><P><BR>When solving complex equations, like the ones used in the examples below, be sure to remember that multiplication and division are inverse operations along with addition and subtraction. Therefore, they undo each other (i.e., (5 * 2)/2 = 5 or 10 + 4 - 4 = 10). To solve these equations, first decide which operation has been applied and then use the inverse operation to undo this (remember to apply the operation to both sides of the equation). </P><P> 1. Solve: 7x - 7 = 42</P><P> Solution: <BR> 7x - 7 = 42<BR> 7x - 7 + 7 = 42 + 7<BR> 7x = 49<BR> (7x) / 7 = 49 / 7<BR> x = 7<BR> The variable needs to be isolated. <BR>To undo subtracting 7, add 7 to both sides. <BR>Adding 7 hasn't isolated the variable, so we need to continue. <BR>To undo multiplying by 7, divide both sides by 7. <BR>2. Solve: 5(x + 2) = 25</P><P> Solution: <BR> 5(x + 2) = 25<BR> [5(x + 2]/5 = 25/5<BR> x + 2 = 5<BR> x + 2 -2 = 5 -2<BR> x = 3<BR> <BR>The variable needs to be isolated. To undo <BR>multiplying by 5, divide by 5 on both sides. <BR>Dividing by 5 hasn't isolated the variable, so we need to continue. <BR>To undo adding 2, subtract 2, from both sides </P><P>

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