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Help!? PLEASE! Screenshot attached! 2 questions! 1 question answered! ONE TO GO!

Mathematics
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5 people means 3.6 seconds so y people will be t seconds we can show it this way : |dw:1350566474324:dw|
now cross multiply and get the connection between y (number of people) and t (seconds)
and hey .. how are you ? :)

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Other answers:

\[t=.72p\]
|dw:1350566721091:dw|
if you multiply one side by 100 you should multiply the other by 100 too
we need to isolate t from the equation that i gave you
no .. it's final answer is correct but i want you to be able to get it by yourself
there is no 75 in decart's answer :|
\[\frac{ 5 }{ y } = \frac{ 3.6 }{ t }\] multiply both sides by \[y \times t\] we get: \[5t = 3.6y\] now divide both sides of the equation by 5 we get: \[t = \frac{ 3.6 }{ 5 } \times y\] which is the same as \[t=0.72y\]
yes sorry i was washing dishes lol
division of two fractions is like multiplying the first by the reciprocal of the second
\[\frac{ a }{ b } \div \frac{ c }{ d } = \frac{ a }{ b } \times \frac{ d }{ c }\]
so you can rewrite the problem as .. ?
no i just gave an example
you have to multiply the first fraction by the reciprocal of the second..
|dw:1350568226013:dw|
yes .. and changed the division to multiplication this his how you divide two fractions..
\[\frac{ (x^2-x-6)(x^2+5x+4) }{ (x^2-2x-3)(x^2+x-12) }\]
this is now what you have to simplify "so, do I multiply x^2 + x - 12 by x^2 + 5x + 4 now? " no.. why you say that ?
\[\frac{ a }{ b } \neq a * b\]
we changed the division to multiplication because we divided FRACTION BY FRACTION
no it's not right
\[\frac{ 5 }{ 3 } * \frac{ 4 }{ 2 } = ?\]
solve this please
right you just multiplied 5 with 4 and divided by the multiplication of 3 with 2
|dw:1350569102823:dw|
so same here .. after we changed the division into multiplication it's just multiplication of two fractions
you tell me .. how do we multiply those two fractions ?
i wont get mad why would i :| look : \[\frac{ x^2-x-6 }{ x^2-2x-3 } \times \frac{ x^2+5x+4 }{ x^2+x-12 } = \frac{ (x^2-x-6)(x^2+5x+4) }{ (x^2-2x-3)(x^2+x-12) }\]
i dont know why you multiply numerator by denominator
this is what you get after the multiplication of the two fractions now we better factor every term than open the brackes like x^2-x-6 = x^2-3x+2x-6 = x(x-3)+2(x-3) = (x+2)(x-3)
yes look : whenever there is multiplication of two fractions you multiply numerator by numerator and denom by denom.. whenever there is division of two fractions you keep the first fraction as it is you change the division into multiplication but then for the second fraction you flip numerator and denominator and then you have multiplication of two fractions..
yes as i wrote up there
I'll just to butt in for a bit, the general flow of fraction division goes like this: 1. Turn the division into a multiplication by inverting the second fraction 2. Factorise as much as possible and eliminate common terms. 3. Solve by multiplying the numerator by the numerator and denominator by denominator. Yes I wrote the same thing as cool sorry :/.
now it's better to factor every term than open the brackets .. so factoring x^2-x-6 gives = x^2-3x+2x-6 = x(x-3)+2(x-3) = (x+2)(x-3)
can you factor the rest of the terms ?
yes.. i just showed the way
you see my last expression is (x+2)(x-3)
now factor the rest of the terms .. what do you get ?
so are you factoring the rest ?
good so how the expression looks like ?
after you change every term by it's factored form
you have to replace every term by it's factored from like replaceing " x^2 - x- 6" by " (x+2)(x-3) " in the last expression
very good
now you can eliminate some of them..
if you have the same thing in the numerator and in the denominator you can eliminate it ..
and one more ..
|dw:1350571449666:dw|
so what is left .. \[\frac{ x+2 }{ x-3 }\]
yes
yw :)
i hope the next question of this kind will be easier for you now that you know the steps
Im happy i could help you
:)

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