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

- anonymous

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

- schrodinger

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- anonymous

5 people means 3.6 seconds
so y people will be t seconds
we can show it this way :
|dw:1350566474324:dw|

- anonymous

now cross multiply and get the connection between y (number of people) and t (seconds)

- anonymous

and hey .. how are you ? :)

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## More answers

- anonymous

\[t=.72p\]

- anonymous

|dw:1350566721091:dw|

- anonymous

if you multiply one side by 100 you should multiply the other by 100 too

- anonymous

we need to isolate t from the equation that i gave you

- anonymous

no .. it's final answer is correct
but i want you to be able to get it by yourself

- anonymous

there is no 75 in decart's answer :|

- 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

yes sorry i was washing dishes lol

- anonymous

division of two fractions
is like multiplying the first by the reciprocal of the second

- anonymous

\[\frac{ a }{ b } \div \frac{ c }{ d } = \frac{ a }{ b } \times \frac{ d }{ c }\]

- anonymous

so you can rewrite the problem as .. ?

- anonymous

no i just gave an example

- anonymous

you have to multiply the first fraction by the reciprocal of the second..

- anonymous

|dw:1350568226013:dw|

- anonymous

yes .. and changed the division to multiplication
this his how you divide two fractions..

- anonymous

\[\frac{ (x^2-x-6)(x^2+5x+4) }{ (x^2-2x-3)(x^2+x-12) }\]

- 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

\[\frac{ a }{ b } \neq a * b\]

- anonymous

we changed the division to multiplication because we divided FRACTION BY FRACTION

- anonymous

no it's not right

- anonymous

\[\frac{ 5 }{ 3 } * \frac{ 4 }{ 2 } = ?\]

- anonymous

solve this please

- anonymous

right you just multiplied 5 with 4
and divided by the multiplication of 3 with 2

- anonymous

|dw:1350569102823:dw|

- anonymous

so same here .. after we changed the division into multiplication
it's just multiplication of two fractions

- anonymous

you tell me .. how do we multiply those two fractions ?

- 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

i dont know why you multiply numerator by denominator

- 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

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

yes as i wrote up there

- 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

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

can you factor the rest of the terms ?

- anonymous

yes.. i just showed the way

- anonymous

you see my last expression is (x+2)(x-3)

- anonymous

now factor the rest of the terms .. what do you get ?

- anonymous

so are you factoring the rest ?

- anonymous

good so how the expression looks like ?

- anonymous

after you change every term by it's factored form

- 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

very good

- anonymous

now you can eliminate some of them..

- anonymous

if you have the same thing in the numerator and in the denominator
you can eliminate it ..

- anonymous

and one more ..

- anonymous

|dw:1350571449666:dw|

- anonymous

so what is left .. \[\frac{ x+2 }{ x-3 }\]

- anonymous

yes

- anonymous

yw :)

- anonymous

i hope the next question of this kind will be easier for you
now that you know the steps

- anonymous

Im happy i could help you

- anonymous

:)

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