2007-02-05 12:05:04 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
2(n + 1)/7 ≥ (n + 4)/5
10(n + 1) ≥ 7(n + 4)
10n + 10 ≥ 7n + 28
3n ≥ 18
n ≥ 6
2007-02-05 12:08:49 · answer #1 · answered by Tom :: Athier than Thou 6 · 1⤊ 0⤋
ok here's what to do....
2(n+1) over 7....
1. distribute the 2 to the (n+1)
2. then muliply the 1 and the 7
you should be left with 2n over 7
3. then multiply the 2 and the 7.
you should have n over 14 is > or equal to....
n+4 over 5
1. multiply the 4 and the 5
and you should be left with n over 20 is < or equal to n over 14 then see wich one is true....
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2007-02-05 12:13:03 · answer #2 · answered by soccerqueen155 2 · 0⤊ 1⤋
2(n+1)/7≥(n+4)/5
Multiply both sides by 35:
10(n+1)≥7(n+4)
10n+10≥7n+28
Subtract 7n from both sides:
3n+10≥28
Subtract 10 from both sides:
3n≥18
Divide 3 from both sides:
n≥6
Check:
(2(6+1))/7=(6+4)/5
2(7)/7=10/5
14/7=10/5
2=2
2007-02-05 12:10:43 · answer #3 · answered by Anonymous · 0⤊ 0⤋
mutiply both sides by to get:
2n+1 = (n+4)/5 * 7
2n+1= (7n + 28)/5
no mult both sides by 5 to get
10n+5= 7n+28
take 5 from each side to get
10n = 7n + 23
take 7n from both sides
3n = 23
divide ea. side by 3
n = 7.67
so 7.67+1 = 8.67 * 2 = 17.34, divide by 7 to get 2.47
and
7.67+4 = 11.67 and divide by 5 = 2.344
2(n+1)/7 is greater than n+4/5
2007-02-05 12:13:52 · answer #4 · answered by Laura 2 · 0⤊ 1⤋
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2016-12-13 09:48:00 · answer #5 · answered by Anonymous · 0⤊ 0⤋
2n+2 /7 n+4/5
take 35 as common factor =you eliminate
the denominators
5(n+2)>or= 7(n+4)=
5n+10 > or= 7n+28
5n+10-10> or = 7n+28-10=
5n>or=7n+18
5n-7n>or=7n-7n+18
-2n= or >18
-2n/-2> or=18/-2
n > or = -9
2007-02-05 12:13:56 · answer #6 · answered by Dave aka Spider Monkey 7 · 0⤊ 1⤋
It depends on what n is equal to.
2007-02-05 12:09:50 · answer #7 · answered by woodysbeard 2 · 1⤊ 0⤋