rhiannon Posted May 15, 2006 Posted May 15, 2006 please can someone solve 2e^5t to 3 significant figures.
RyanJ Posted May 15, 2006 Posted May 15, 2006 Sorry but people will not do your work for you, people will however help you if you get stuck. Here is a hint: [math]\frac{d}{dx} \, ne^{x} \, = \, ne^{x}[/math] Cheers. Ryan Jones
matt grime Posted May 15, 2006 Posted May 15, 2006 The answer is also 'no' because 2e^5t is not something that can be 'solved' in the first place, never mind to some number of significant digits.
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