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May 19th, 2018, 09:25 PM   #1
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Question In need of some guidance!

I've been racking my brain trying to figure out how to do these problems (except #4, and I JUST figured out #2,) but i'm not even sure how to set up equations for the rest.
I don't want the answers, please just nudge me in the right direction 😣
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May 19th, 2018, 10:44 PM   #2
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Update
I had an ahah moment, got all but #6 now
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May 19th, 2018, 11:19 PM   #3
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the key to these sort of rate problems is that rates add, not durations.

so for example (1)

$r_{riding} = \dfrac{1~lawn}{4~hr}$

$r_{push} = \dfrac{1~lawn}{9~hr}$

we can add these two rates

$r_{combined} = \left(\dfrac{1}{4} + \dfrac{1}{9}\right)~\dfrac{lawn}{hr} = \dfrac{13}{36}~\dfrac{lawn}{hr}$

I leave it to you to determine how long it takes the two of them to cut a lawn.

I suspect the rest of the problems use a similar technique.
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May 19th, 2018, 11:34 PM   #4
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6) well let's see

$t_s = \dfrac{390}{2(v_l - 5)}$

$t_f = \dfrac{390}{2(v_l + 5)}$

$t_s = t_f + \dfrac 2 5$

A fairly straightforward system to solve.

You should get

$t_s =3~hr,~t_f = \dfrac{13}{5}~hr$

and the speed limit is

$v_l = 70~mph$

Last edited by romsek; May 19th, 2018 at 11:41 PM.
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May 20th, 2018, 09:14 AM   #5
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u = speed limit

@(u-5).......195.......>@(u+5).......195.......>WHOA!

195/(u-5) - 195/(u+5) = 2/5 (2/5 hour = 24 min)

Simplifies to u^2 = 4900 : so u = 70
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May 20th, 2018, 01:54 PM   #6
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Quote:
Originally Posted by romsek View Post
6) well let's see

$t_s = \dfrac{390}{2(v_l - 5)}$

$t_f = \dfrac{390}{2(v_l + 5)}$

$t_s = t_f + \dfrac 2 5$

A fairly straightforward system to solve.

You should get

$t_s =3~hr,~t_f = \dfrac{13}{5}~hr$

and the speed limit is

$v_l = 70~mph$
I did wind up getting 70mph, but not like that haha
Your way looks much simpler.
Mine involved the quadratic formula to solve for T, then finding the speed for each portion of the drive. 65mph & 75mph, making the limit 70.
Thanks guys!
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