Looks basic for some, but I keep coming up with the wrong answer, I am missing a step. I love being slow at math, so much fun.
2007-03-10 17:44:06 · 6 answers · asked by Cassie C 1 in Science & Mathematics ➔ Mathematics
1/1t + 1/7t + 1/9t = 7
63(1/1t) + 63(1/7t) + 63(1/9t) = 63(7)
63t + 9 t + 7t = 441
79t = 441
79t/79 = 441 / 79
t = 441 / 79
t = 5.582278481
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2007-03-11 05:17:21 · answer #1 · answered by SAMUEL D 7 · 0⤊ 0⤋
1/t + 1/(7t) + 1/(9t) = 7
Multiply through by 63t to get rid of fractions.
63 is the least common multiple of 1, 7 and 9.
63 + 9 + 7 = 7 * 63t
79 = 441t
t = 79/441
2007-03-11 01:51:30 · answer #2 · answered by falzoon 7 · 0⤊ 0⤋
1/t+1/7t+1/9t=7
first, multiply both sides by 63t:
63 + 9 + 7 = 441t
79 = 441t
t = 79/441
2007-03-11 01:48:33 · answer #3 · answered by datz 2 · 1⤊ 0⤋
1/t+1/7t+1/9t=7
63/63t+9/63t+7/63t = 7
79/63t = 7
79/63*7 = t
79/441 = t
Check:
1/(79/441) + 1/7(79/441) + 1/9(79/441) = 7
1 + 1/7 + 1/9 = 7(79/441)
63/63 + 9/63 + 7/63 = 553/441
441*63/63 + 441*9/63 + 441*7/63 = 553
27783/63+3969/63+3087/63 = 553
27783+3969+3087 = 553*63
34839 = 34839
2007-03-11 04:18:59 · answer #4 · answered by robertonereo 4 · 0⤊ 0⤋
If you mean: 1/t + 1/(7t) + 1/(9t) = 7, then
1/t + 1/(7t) + 1/(9t) = (1/t)*(1+1/7+1/9)
= (1/t)*(63+9+7)/63
= (1/t)*(79/63)
so 7t = 79/63 and now you just divide by 7.
so t = 79/(63*7) = 0.179138...
2007-03-11 01:51:36 · answer #5 · answered by Peter 2 · 0⤊ 0⤋
1/t(1+1/7+1/9)=1/t[(63+7+9)/63]so
79=63t(7)and
t=0.179
2007-03-11 05:38:47 · answer #6 · answered by zary n 1 · 0⤊ 0⤋