am i correct?
b is the only one that has 2 hours
because problem 2 is asking how long it would take for one of them to clean it...
yeah, I think you're correct :)
lol now im not sure how to check it on the second problem with the equation, im not sure how to write it with a decimal
i was trying 1.5 whitch i think is 15/10 so im not sure how to write that other than 1/15/10
unless i get the same denominator : t(3t) and distribute it
i got nothing
brb I'm trying to figure it out... :/ this sort of stuff isn't my strong point...
ok im looking for a way to do it to :)
so the amount of time it takes henry: 2/h ammont of time for john: 2/3h if im right 3h stands for the amount of work Henry can do on one pool
so d
\[\frac{ 2 }{ h } \times 3h + \frac{ 2 }{ 3t } \times 3h \] so as a result we would have \[2 x 3 + 2 = ?\] so now to solve im still looking
hmm, idk how to solve it but logically it cannot be less than 2hours because no matter how slow john was, he had still contributed and then 8hours sound a bit extreme so if it was me, I would go with A...
wait i messed up
that should have been: 2 x 3 + 2 = 2h 6 + 2 = 2h 8 = 2h 8/2 = h 4 = h
so going back to the original i would have: 2 x 3 + 2 = 1 x 3h 6 + 2 = 3h 8 = 3h 8/3 = h 2.6 = h so ill round it up
looks like I learned something new today ;) Good job!
there is one other one and its set up as 1/2 + 3/ x = 1 for an example :)
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