Do you think I could just leave this part blank and it'd be okay? We're just going to replace the whole thing with a header image anyway, right?
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How much you wanna bet Jad tries to do it?
Anyway, I have a boatload of Algebra II homework, and there's one question I can't really complete. Here it is:
Write and solve an inequality to find three consecutive whole numbers with a sum between 13 and 16.
SHOW YOUR WORK DAD GUM IT.
The use of absolute value is not necessary, but encouraged.
Last edited by 32OrtonEdge32dh (Sep 16 2011 3:34:00 pm)
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13 < x + (x+1) + (x+2) < 16
Did I win?
Hmm, seems like you guys use different terms in USA than in Canada, I don't know what an inequality is
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Per IRC chat:
13 < x + (x+1) + (x+2) < 16
13 < 3x + 3 < 16
10 < 3x < 13
10/3 < x < 13/3
There. I robbed you of your education. lol
Algebra? No thanks.
Lemme have a go,
13 < x + (x+1) + (x+2) < 16
13 < x + (x+3) < 16
13 < (x+x)=2+3 < 16
Did I do something wrong? I fail at algebra.
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I don't know what an "inequality" is. I don't live in America therefor I don't know they're standards for math.
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Per IRC chat:
13 < x + (x+1) + (x+2) < 16
13 < 3x + 3 < 16
10 < 3x < 1310/3 < x < 13/3
There. I robbed you of your education. lol
This.
But you need to make it whole numbers, so make 10/3 to 4 and 13/3 to 4... WHAT ARWOIHWIRUCHC
That means it has to be 4!?!?!
Last edited by calculusguy (Sep 16 2011 4:56:33 pm)
This is an impossible problem IMO.
Look:
13 < (x-1) + x + (x+1) < 16 -------> 13 < 3x < 16
AND:
13 < x + (x+1) + (x+2) < 16 -------> 13 < 3x + 3 < 16 -------> 10 < 3x < 16
So:
13 < 3x < 16
10 < 3x < 13
Impossible.
How much you wanna bet Jad tries to do it?
Lolyeah
Algebra II
Bro, I'm only in Geometry.
I hate tall signatures.
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This is an impossible problem IMO.
Look:13 < (x-1) + x + (x+1) < 16 -------> 13 < 3x < 16
AND:
13 < x + (x+1) + (x+2) < 16 -------> 13 < 3x + 3 < 16 -------> 10 < 3x < 16
So:
13 < 3x < 16
10 < 3x < 13Impossible.
Well, jakery did it.... How is it impossible?
13 < x + (x+1) + (x+2) < 16
13 < 3x + 3 < 16
10 < 3x < 13
10/3 < x < 13/3
You didn't isolate the variable.
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13 < 3x < 16
10 < 3x < 13Impossible.
x isn't the same thing in those two though. In the top one x is the middle number, whereas in the bottom one x is the first number. If you're consistent with calling x the first number, the top one is really 13 < 3(x+1) < 16, which is clearly the same as the second inequality.
Baticon, your first inequality will work as well, but you have to remember that you subtracted 1 from each of the three consecutive numbers, so you must subtract three from the other two sides of the inequality:
13 < x + (x+1) + (x+2) < 16
10 < (x-1) + x + (x+1) < 13
And from there, you still get lead to:
10 < 3x < 13
look it up!
x
x + 1
x + 2
Equation: 13 < x + (x + 1) + (x + 2) < 16
Simplify the middle part first:
13 < x + x + 1 + x + 2 < 16
13 < 3x + 3 < 16
Now subtract 3 from all three sections
13 - 3 < 3x + 3 - 3 < 16 - 3
10 < 3x < 13
Divde through by 3
10/3 < x < 13/3
3.3333 < x < 4.33333
The only whole number between 3.333 and 4.3333 is 4, so the first number will be 4
x = 4
x + 1 = 5
x + 2 = 6
CHECK:
4 + 5 + 6 = 15......this is between 13 and 16
Answer: 4, 5, 6
Source(s):
Math teacher for many years
found this on yahoo answers.
jakery wrote:Per IRC chat:
13 < x + (x+1) + (x+2) < 16
13 < 3x + 3 < 16
10 < 3x < 1310/3 < x < 13/3
There. I robbed you of your education. lol
This.
But you need to make it whole numbers, so make 10/3 to 4 and 13/3 to 4... WHAT ARWOIHWIRUCHC
That means it has to be 4!?!?!
Y u no read?
ok, I love math, but jeeze. I am only 11. I did'nt know half of the words in that problem XD
Ok I wish i could solve this, but the problem is I am Norwegian (dont know what the some of the words mean), 13 years, and I havent had algebra at school yet >.<
This is an impossible problem IMO.
Look:13 < (x-1) + x + (x+1) < 16 -------> 13 < 3x < 16
AND:
13 < x + (x+1) + (x+2) < 16 -------> 13 < 3x + 3 < 16 -------> 10 < 3x < 16
So:
13 < 3x < 16
10 < 3x < 13Impossible.
Uhh... not really... one of your x values is different from the other.
The first x is one more than the second x. So your math is off...
And this has already been solved here, easily.
13<x+x+1+x+2<16
13<3x+3<16
10<3x<13
10/3<x<13/3
3.33<x<4.33
x must be a whole number, so x is 4. 4+5+6=15. Yay. Easy. Not to mention it already was solved before this...
Well, that's the best a 12 year old in Calc 1 can do...
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damn. why an inequality. thats the only thing i cant do. i can tell you that the consectutive whole numbers are 4 5 and 6, but i cant do the inequality.
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oh. well. i took algebra 2 last year, and that class was torture. i know how it feels to have to use the internet to get work done.
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Locked on request of OP.
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