# Rearranging long equationWatch

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#1
Ok guys I need a little help if possible. I have left doing a few equations until the last minute and need to hand something in tomorrow that requires me to find n for two equations:

m = n * 2 * µ * sinα * FT / g (cx,yµ * cz) * fs
m = 2 * n * Fr * (µ * fu * sinα + cosα * cosB) / g * (c - µ * fu * Cz)

My gcse level maths is just not cutting it! If anyone can help or guide my I will send them my love!

Cross fingers!
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#2
(Original post by Kingofthechicken)
Ok guys I need a little help if possible. I have left doing a few equations until the last minute and need to hand something in tomorrow that requires me to find n for two equations:

m = n * 2 * µ * sinα * FT / g (cx,yµ * cz) * fs
m = 2 * n * Fr * (µ * fu * sinα + cosα * cosB) / g * (c - µ * fu * Cz)

My gcse level maths is just not cutting it! If anyone can help or guide my I will send them my love!

Cross fingers!
is it even possible to rearrange something this big?
0
4 years ago
#3
These equations both require literally 2 steps to re-arrange. Simply multiply both sides by the denominator, then divide both sides by everything except n. This will give you n.
0
#4
cheers done
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