anonymous
  • anonymous
Help!? PLEASE! Screenshot attached! 2 questions! 1 question answered! ONE TO GO!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
5 people means 3.6 seconds so y people will be t seconds we can show it this way : |dw:1350566474324:dw|
anonymous
  • anonymous
now cross multiply and get the connection between y (number of people) and t (seconds)
anonymous
  • anonymous
and hey .. how are you ? :)

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

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