Question 241374
{{{(186000mi)/(sec)}}}
<pre><font size = 4 color = "indigo"><b>
We will take the miles to feet to inches,

{{{(186000mi)/(sec)}}}{{{"×"}}}{{{(5280ft)/(1mi)}}}{{{"*"}}}{{{(12in)/(1ft)}}}


and we will take the seconds to minutes to hours to days to years

{{{(186000mi)/(sec)}}}{{{"×"}}}{{{(5280ft)/(1mi)}}}{{{"*"}}}{{{(12in)/(1ft)}}}{{{"×"}}}{{{(60sec)/(1min)}}}{{{"×"}}}{{{(60min)/(1hr)}}}{{{"×"}}}{{{(24hr)/(1da)}}}{{{"×"}}}{{{(365.25da)/(1yr)}}}

Cancelling units which are in a numerator and a denominator:

{{{(186000cross(mi))/(sec)}}}{{{"×"}}}{{{(5280cross(ft))/(1cross(mi))}}}{{{"*"}}}{{{(12in)/(1cross(ft))}}}{{{"×"}}}{{{(60cross(sec))/(1cross(min))}}}{{{"×"}}}{{{(60cross(min))/(1cross(hr))}}}{{{"×"}}}{{{(24cross(hr))/(1cross(da))}}}{{{"×"}}}{{{(365.25cross(da))/(1yr)}}}

{{{186000*5280*12*60*60*24*365.25}}}{{{(in)/(yr)}}}

Change everything to scientific notation:

{{{(1.86*10^5)(5.28*10^3)(6*10^1)(6*10^1)(2*10^1)(3.6525*10^2)}}}{{{(in)/(yr)}}}

Group the numbers for multiplication and the exponents of 10
for addition:


{{{(1.86)(5.28)(6)(6)(2)(3.6525)*10^(5+3+1+1+1+2)}}}{{{(in)/(yr)}}}

Now your calculator can make the multiplication of those numbers:

{{{2582.673984*10^13}}}{{{(in)/(yr)}}}

Next we change {{{2582.673984}}} to scientific notation {{{2.582673984*10^3}}}

and we have:

{{{2.582673984*10^3*10^13}}}{{{(in)/(yr)}}}

And we add those powers of 10

{{{2.582673984*10^16}}}{{{(in)/(yr)}}}

With scientific notation your calculator can handle
all numbers regardless of how large.

Edwin</pre>