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The answer is 7s + 49.
The equivalent of 7s plus 2s is 9s.
5.28751 7s equals 37.
8 / 7 in long division is however many 7s go into 8. so there's 1x 7 in 8 with 1 remainder. For this example, assume every number beyond is a 10, multiplied by the remainder. so, it'd be 7s into 10, which is 1 again. Then 7s into 30, which is 4. Then 7s into 20, which is 2. Then 7s into 60, which 8. Then 7s into 40, which is 5. Then 7s into 50, which is 7. And this is a reoccuring number, making it 1.142857142857 and so on.
You said that 4(2s - 1) = 7s + 12Eliminate parentheses: 8s - 4 = 7s + 12Add 4 to each side: 8s = 7s + 16Subtract 7s from each side: s = 16
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The answer is 7s + 49.
There is one 7s orbital with two sub-orbitals: 7s(+1/2) and 7s(-1/2) . A picture of this 7s orbital is in 'Related links'
The equivalent of 7s plus 2s is 9s.
5.28751 7s equals 37.
The Arsenal SAM-7S is currently valued at 1500$.
Expressed algebraically, this would equal 7s - 5.
nothing
8 / 7 in long division is however many 7s go into 8. so there's 1x 7 in 8 with 1 remainder. For this example, assume every number beyond is a 10, multiplied by the remainder. so, it'd be 7s into 10, which is 1 again. Then 7s into 30, which is 4. Then 7s into 20, which is 2. Then 7s into 60, which 8. Then 7s into 40, which is 5. Then 7s into 50, which is 7. And this is a reoccuring number, making it 1.142857142857 and so on.
You said that 4(2s - 1) = 7s + 12Eliminate parentheses: 8s - 4 = 7s + 12Add 4 to each side: 8s = 7s + 16Subtract 7s from each side: s = 16
7s + 12 = 131 Therefore, 7s = 119 s = 119/7 Therefore, s = 17
You can't go through the Lucky 7s. The Lucky 7s broke up after Mick's crash. Remember? Your not going to race the Lucky 7s at all. The only person you are going to race from the team is Sara in the end. She still has the whole Lucky 7s logo, so she is the only one still on the team.