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May 19th, 2018, 09:25 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 ðŸ˜£ 
May 19th, 2018, 10:44 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, 11:19 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,317 Thanks: 1230 
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, 11:34 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,317 Thanks: 1230 
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. 
May 20th, 2018, 09:14 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,981 Thanks: 994 
u = speed limit @(u5).......195.......>@(u+5).......195.......>WHOA! 195/(u5)  195/(u+5) = 2/5 (2/5 hour = 24 min) Simplifies to u^2 = 4900 : so u = 70 
May 20th, 2018, 01:54 PM  #6  
Newbie Joined: May 2018 From: Bellingham, WA Posts: 5 Thanks: 0  Quote:
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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