Algebra Pre-Algebra and Basic Algebra Math Forum

May 19th, 2018, 08:25 PM   #1
Newbie

Joined: May 2018
From: Bellingham, WA

Posts: 5
Thanks: 0 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 😣
Attached Images 20180519_211524.jpg (91.9 KB, 19 views) May 19th, 2018, 09:44 PM #2 Newbie   Joined: May 2018 From: Bellingham, WA Posts: 5 Thanks: 0 Update I had an ahah moment, got all but #6 now May 19th, 2018, 10:19 PM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,463 Thanks: 1340 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. May 19th, 2018, 10:34 PM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,463 Thanks: 1340 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 10:41 PM. May 20th, 2018, 08:14 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,594 Thanks: 1038 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 Thanks from greg1313 May 20th, 2018, 12:54 PM   #6
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Joined: May 2018
From: Bellingham, WA

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Quote:
 Originally Posted by romsek 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
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.
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