To get N by itself,
subtract 5 from both sides of the equal sign.
and multiply by 2 on both sides of the equal sign.
I'm not telling you the answer...someone else might, but I won't.
ps: it's not 1!!
The answer really isn't important here. What is important is knowing how to get the answer.
EDIT:
JUST CURIOUS, WHAT THE HECK IS UP WITH THE THUMBS DOWN ON MY ANSWER AND THE ONE BELOW ME?
WAS THAT THE FIRST PERSON THAT ANSWERED REALIZING WHAT AN IDIOT THEY WERE, AND DELETING THEIR QUESTION?
I JUST DON'T UNDERSTAND PEOPLE SOMETIMES.
2006-11-28 08:29:16
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answer #1
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answered by powhound 7
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4/2+5=7.
2006-11-28 16:30:30
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answer #2
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answered by Anonymous
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You don't even have to know how to do balanced equations to figure this one out. It's as easy as what plus 5 equals 7?
We know that TWO plus five equals 7. So what would N need to be?
2006-11-28 16:31:14
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answer #3
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answered by Anonymous
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You have to solve for n. You do this by having your variable ('n' in this case) by itself on one side of the equation. In this case it would be easier to move the numbers to the right side of the equation. To move the numbers to the opposite side, you have to use the opposite operation. For example, in this case, since you have +5, you would subtract 5 from both sides of the equation, giving you n/2 + 5 - 5 = 7 - 5; or n/2 = 2. Since 'n' is divided by 2 on the left side, what would you have to do to move it to the right side? That should give you your answer for n. . . Welcome to the world of algebra!
2006-11-28 16:36:12
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answer #4
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answered by ireelimeanet 1
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First, you have to subtract five from both sides.
n/2+5=7
-5 -5
n/2=2
Next, you multiply each side by two.
2*n/2=2*2
Then you get!!!
n=? Figure it out!!!
2006-11-28 16:39:50
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answer #5
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answered by Meika 1
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1. Subtract 5 from each side
(-5) + n/2 + 5 = 7 - 5
n/2 = 2
2. Multiply each side by 2.
2*(n/2) = 2*2
n = ... (you do the final step. hint: the first anser here is WRONG)
2006-11-28 16:29:49
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answer #6
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answered by lillielil 3
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The correct answer is 4
2006-11-28 16:32:09
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answer #7
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answered by Eric F 2
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