- 24/55 divided by 39/40
use the fact that:\[\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}\]
do i cross multiply
rewrite the division as a multiplication using the rule above and then cancel common factors between the numerators and the denominators.
how do i do that
I suggest you go through this site to learn about how to manipulate fractions: http://www.aaamath.com/fra.html#topic23
well i only ask how to set this up i never ask you to give me the answer.
I have given you the information above. That site will help you understand how to do these types of questions as it also hows you lots of examples.
*shows
ok thank you!
in your case you have:\[\frac{24}{55}\div\frac{39}{40}=\frac{24}{55}\times\frac{40}{39}\]
see the thing was i already had it up to this point I just wasnt sure what you meant for me to do next do i find the GCF of both? because if so i know how to do at least that.
no - you just need to cancel out common factors between numerators and denominators. e.g. the following fraction can be simplified as follows:\[\frac{12}{28}=\frac{3}{7}\]because I divided both the numerator and the denominator by their common factor (4 in this case)
remember that:\[\frac{24}{55}\times\frac{40}{39}=\frac{24*40}{55*39}\]so, for example, I can see that the 40 in the numerator and the 55 in the denominator can both be divided by 5 to get:\[\frac{24*40}{55*39}=\frac{24*8}{11*39}\]keep dividing the numerator and denominators by any common factors you can find.
ok thank! this website you told me that would help is it free?
yes
ok thanks!
yw
one more thing when i turn the fracton into a mulitiplication what happens to the negative sign
oh - I didn't realise you had a negative sign. the negative sign just gets placed in front of the overal fraction as follows:\[-\frac{24}{55}\times\frac{40}{39}=-\frac{24*40}{55*39}\]
would it be - 64/143
perfect!
this is very had did you see how long it took me to find this out lol
don't worry - we all had to start somewhere - with practice you will build confidence in these types of questions and I'm sure you'll be doing them in your sleep very soon! :)
trust me im far from that lol
:D
well - if you need help you only need to ask :)
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