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(z^2)/(2z+14) multiply (2z+14)/(2z^6)
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\[\frac{z^2}{2z + 14} \times \frac{2z + 14}{2z^6}\] note (2z + 14) cancels, and z^2 will cancel partially with 2z^6, leaving: \[\frac{1}{2z^4} \]
thanks again I'm a bit hopeless at this, right now anyway
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yes I understand it very well
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