 Please Help me - XP Math - Forums Sign Up FREE! | Sign In | Classroom Setup | Common Core Alignment  XP Math - Forums Please Help me 07-06-2009 #1 Last   Join Date: Jul 2009 Posts: 5 Please Help me hello i am new at this site, so i am hoping i can get help b4 this evenings class. i have a problem i am not able to correctly solve. it goes like this Find 3 consecutive even integers such that 15 times the largest is 136 more than 4 times the sum of the smaller ones. can you please show me the work so that i can compare it to what i did thank you Theresa  07-08-2009   #2
MAS1   Join Date: Dec 2008
Posts: 249 Quote:
 Originally Posted by Last hello i am new at this site, so i am hoping i can get help b4 this evenings class. i have a problem i am not able to correctly solve. it goes like this Find 3 consecutive even integers such that 15 times the largest is 136 more than 4 times the sum of the smaller ones. can you please show me the work so that i can compare it to what i did thank you Theresa
HI Theresa:

Three consecutive even integers can be represented by n, n + 2, and n + 4.

15 x (n + 4) = 136 + 4 x (n + n + 2)
15n + 60 = 136 + 4 x (2n + 2)
15n + 60 = 136 + 8n + 8
15n + 60 = 8n + 144
15n - 8n = 144 - 60
7n = 84
n = 12

So the three consecutive even integers are 12, 14, and 16.

To check: 15 x 16 = 136 + 4 x (12 + 14)
240 = 136 + 4 x 26
240 = 136 + 104
240 = 240  07-13-2009   #3
Last

Join Date: Jul 2009
Posts: 5 thank you, I finally figured it out, it took all day to do so, but i did it,

Quote:
 Originally Posted by MAS1 HI Theresa: Hope this does not come too late to help you. Three consecutive even integers can be represented by n, n + 2, and n + 4. 15 x (n + 4) = 136 + 4 x (n + n + 2) 15n + 60 = 136 + 4 x (2n + 2) 15n + 60 = 136 + 8n + 8 15n + 60 = 8n + 144 15n - 8n = 144 - 60 7n = 84 n = 12 So the three consecutive even integers are 12, 14, and 16. To check: 15 x 16 = 136 + 4 x (12 + 14) 240 = 136 + 4 x 26 240 = 136 + 104 240 = 240
thankyou for your help, i had finally figured it out , it taken all day to do it, and i got it correct. i am having trouble with graphing, so i will post later on to get help with that , Thank you again
Last  07-13-2009 #4 Last   Join Date: Jul 2009 Posts: 5 Thank you I hope you get this post, the problem i had done taken me a very long time to do but i did it and i got it correct, i am hoping to come back on later and get help with graphing, i am struggling with that one too much too fast, Thanks again Last   09-09-2009   #5
allie_poo123

Join Date: Sep 2009
Posts: 2 sorry

Quote:
 Originally Posted by Last hello i am new at this site, so i am hoping i can get help b4 this evenings class. i have a problem i am not able to correctly solve. it goes like this Find 3 consecutive even integers such that 15 times the largest is 136 more than 4 times the sum of the smaller ones. can you please show me the work so that i can compare it to what i did thank you Theresa
im sorry i cant help you. Id like to but i cant im not good at math and i dont really like it my mom made me sighn up hoping that i would get good at it i am really really sorry   10-06-2009   #6
Last

Join Date: Jul 2009
Posts: 5 fractinal equations HELP !!!!!!!!!!!!!!!!!

mr Hui,,,,, Are you available to help me understand some algebraic factoring please,, Example 3 over 2x + 2 + 4 over x + 1 = 11/12

Quote:
 Originally Posted by MAS1 HI Theresa: Hope this does not come too late to help you. Three consecutive even integers can be represented by n, n + 2, and n + 4. 15 x (n + 4) = 136 + 4 x (n + n + 2) 15n + 60 = 136 + 4 x (2n + 2) 15n + 60 = 136 + 8n + 8 15n + 60 = 8n + 144 15n - 8n = 144 - 60 7n = 84 n = 12 So the three consecutive even integers are 12, 14, and 16. To check: 15 x 16 = 136 + 4 x (12 + 14) 240 = 136 + 4 x 26 240 = 136 + 104 240 = 240  10-07-2009   #7
MAS1   Join Date: Dec 2008
Posts: 249 Quote:
 Originally Posted by Last mr Hui,,,,, Are you available to help me understand some algebraic factoring please,, Example 3 over 2x + 2 + 4 over x + 1 = 11/12

3/(2x + 2) + 4/(x + 1) = 11/12

Then we can say
3/(2(x + 1)) + 4/(x + 1) = 11/12
3/(2(x + 1)) + 8/(2(x + 1)) = 11/12
(3 + 8)/(2(x + 1)) = 11/12
11/(2(x + 1) = 11/12
2(x + 1) = 12
x + 1 = 6
x = 5  01-28-2010 #8 jessie27   Join Date: Jan 2010 Posts: 1 hello iim jessie27 can you help me with probability of independent events please   01-31-2010 #9 Mr. Hui       Join Date: Mar 2005 Posts: 10,609 Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. For instance, the event of obtaining a head on the up face when a fair coin is tossed and the event of obtaining a five on the up face when a fair die is rolled are independent. __________________ Do Math and you can do Anything!  02-16-2010   #10
Math Tyrant Join Date: Sep 2008
Posts: 37 for jessie27

Quote:
 Originally Posted by jessie27 hello iim jessie27 can you help me with probability of independent events please Also If the outcome of those effects do not effect eachother in any shape or form, for example the outcome of flipping heads or rolling a six have nothing to do with eachother depending on the problem therefore making them independent of eachother. Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
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