Question 393740
1) A helicopter can travel 25 miles in one hour. What is the speed of this helicopter in miles per minute?
<pre><font face = "batangche" size = 2><b>
"Miles per hour" means the same as "miles in ONE hour" which also means 
"miles OVER hours" or {{{mi/h}}} where {{{mi}}} means "miles" and {{{h}}} means hours.

So to convert 

{{{25mi/h}}}

to {{{mi/m}}}  (where {{{m}}} means "minutes")

We use the fact that {{{60m}}} = {{{1h}}} to build a unit fraction with 
one of those in the numerator and the other in the denominator, either
{{{(60m)/(1h)}}} or {{{(1h)/(60m)}}}, whichever serves the purpose of getting rid of the hours {{{h}}} and bringing in the minutes {{{m}}}.

To decide which one we want in the numerator and which in the denominator,
we observe that  

{{{(25mi)/h}}}

has {{{h}}} in the denominator.  So to cancel away the {{{h}}} in the denominator we need 
an {{{h}}} in the numerator,  So we make the unit fraction {{{(1h)/(60m)}}} 
and multiply it by {{{2.5mi/h}}} like this
 
{{{expr((25mi)/h)*expr((1h)/(60m))}}}
 
Then we can cancel the hours ({{{h}}}'s) like this:

{{{expr((25mi)/cross(h))*expr((1cross(h))/(60m))}}}

and all that's left is

{{{(25*1mi)/(60m)}}}

which is the same as

{{{25/60}}}{{{mi/m}}}

which can be reduced to

{{{5/12}}}{{{mi/m}}}

or as a decimal .4166666667{{{mi/m}}}

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2) An ant can travel 25 feet per minute. What is the speed of the ant in miles per hour?


So to convert 

{{{25f/m}}}

to {{{mi/h}}}  (where {{{f}}} means "feet")

We again use the fact that {{{60m}}} = {{{1h}}} to build a unit fraction with
 one of those in the numerator and the other in the denominator, either
{{{(60m)/(1h)}}} or {{{(1h)/(60m)}}}, whichever serves the purpose of getting rid of the minutes {{{m}}} and bringing in the hours {{{h}}}.

To decide which one we want in the numerator and which in the denominator,
we observe that  

{{{(25f)/m}}}

has {{{m}}} in the denominator.  So to cancel away the {{{m}}} in the denominator we need 
an {{{m}}} in the numerator,  So we make the unit fraction {{{(60m)/(1h)}}} 
and multiply it by {{{25f/m}}} like this
 
{{{expr((25f)/m)*expr((60m)/(1h))}}}
 
Then we can cancel the hours ({{{m}}}'s) like this:

{{{expr((25f)/cross(m))*expr((60cross(m))/(1h))}}}

But now we must also get rid of the feet and bring in the miles:

So now we use the fact that {{{5280f}}} = {{{1mi}}} to build a unit fraction with
 one of those in the numerator and the other in the denominator, either
{{{(5280f)/(1mi)}}} or {{{(1mi)/(5280f)}}}, whichever serves the purpose of getting rid of the feet {{{f}}} and bringing in the miles {{{mi}}}.

To decide which one we want in the numerator and which in the denominator,
we observe that  

{{{expr((25f)/cross(m))*expr((60cross(m))/(1h))}}}

has {{{f}}} in the numerator.  So to cancel away the {{{f}}} in the numerator we need 
an {{{f}}} in the denominator,  So we make the unit fraction {{{(1mi)/(5280f)}}} 
and multiply it by the above like this

{{{expr((25f)/cross(m))*expr((60cross(m))/(1h))*expr((1mi)/(5280f))}}}

Then we can cancel the feet {{{f}}}

{{{expr((25cross(f))/cross(m))*expr((60cross(m))/(1h))*expr((1mi)/(5280cross(f)))}}}


and all that's left is

{{{(25*60mi)/(1*5280h)}}}

which is the same as

{{{25*60/(1*5280)}}}{{{mi/h}}}

which reduces to fraction form

{{{25/88))){{{mi/h}}}

or decimal form 0.2840909091{{{mi/h}}}

Edwin</pre>