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Mathematics 14 Online
OpenStudy (anonymous):

http://assets.openstudy.com/updates/attachments/51af3b89e4b05b167ed23c22-pyromancer-1370438863548-here.jpg Please read number 21 and help me solve :D

OpenStudy (anonymous):

@Hero

OpenStudy (anonymous):

@amistre64 @andriod09 @allopersonwhat @abbiebosworth @aditkarekatte @bhaskarbabu @BostonBlue @brookecookie @B.rad @blue_cheetah @bcheyennehaynes @countrygirl33 @caideering

OpenStudy (anonymous):

Someone flippin help me please...

OpenStudy (andriod09):

I am so sorry to say that i can't help you. I suck at those kinds of problems, sorry.

OpenStudy (anonymous):

Aw that sucks. :)P Well whtevr.

OpenStudy (andriod09):

Sorry.

OpenStudy (amistre64):

glorified fractions .. whats the question?

OpenStudy (anonymous):

It's TOTALLY OKAY

OpenStudy (amistre64):

21, word problem ....

OpenStudy (anonymous):

Read #21

OpenStudy (anonymous):

Yes

OpenStudy (amistre64):

find a round trip flying time

OpenStudy (anonymous):

How do I start?

OpenStudy (amistre64):

im gonna say you can start by calling your variables to and from .. t and f

OpenStudy (amistre64):

you know the total distance is going to be twice the distance stated right?

OpenStudy (anonymous):

I got 138,430/59r

OpenStudy (anonymous):

But all I need to do is show my work and I can get an A

OpenStudy (amistre64):

one direction is 18% faster t + f = 2(1270) t + ( t + t(18%)) = 2540 t + t + .18t = 2540 2.18t = 2540

OpenStudy (amistre64):

the "format" in the book is just confusing to read, you need to be able to parse the instructions from the clutter

OpenStudy (amistre64):

also, keep in mind the speed = distance/time

OpenStudy (anonymous):

Ok, but how would i do this.. d/t=s 1270/1165.137615

OpenStudy (amistre64):

d from At-Al = 1270mi speed Al-At = 18% speed from At-Al r = speed from At-Al in m/h find time for a round trip ----------------------------- r = 1270mi/t hours k = 1270mi/t hours, times (1.18) round trip time = r + k thats the best i got ..... solve for t?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

So was mine right?

OpenStudy (amistre64):

s = d/t t = d/s 18% of t is equal to 18% of d/s a little adjusting ...\[t=\frac{1270}{s}+\frac{1270(1.18)}{s}\] \[t=\frac{1270(2.18)}{s}\] without knowing an actual speed or time value ... this is what we would be left with

OpenStudy (anonymous):

So we can't solve?

OpenStudy (amistre64):

it asks for a simplified expression, not for a numerical solution

OpenStudy (anonymous):

But in the basck of my book it says the answer is 138,430/59r

OpenStudy (amistre64):

the "r" is how they are formating it to their own delusions :)

OpenStudy (anonymous):

ok then how do they get the actual number?

OpenStudy (amistre64):

the trip to is given as: r = 1270/t; or t = 1270/r the trip back is 18% faster: r + r(.18) adding the 2 speeds gives us a round trip speed r + (r + r(.18)) 1270 + 1270 + 1270(.18) ------------------------- t 1270 + 1270 + 1270(.18) ------------------------- 1270/r 1270(2.18) ---------- = 2.18r 1270/r not really sure how the back of the book is trying to get at it, or even if the book is correct

OpenStudy (anonymous):

hmm ok..sorry it took so long..i had to pee lol

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