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,y – µ * 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!
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,y – µ * 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?
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.