briana.img
  • briana.img
Need help on work problem.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
briana.img
  • briana.img
1 Attachment
Arihangdu
  • Arihangdu
I'm sorry @briana it beyond my knowledges
briana.img
  • briana.img
@satellite73 hey can you help me out please

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

sleepyjess
  • sleepyjess
Hmmm... let me get all of my thoughts out, then see what hapens from there :)
sleepyjess
  • sleepyjess
Nevermind... this is just making me go around in circles...
sleepyjess
  • sleepyjess
@Astrophysics
caozeyuan
  • caozeyuan
lets go step by step
caozeyuan
  • caozeyuan
from statement 1 we know B+C+R+F=360
briana.img
  • briana.img
mmm yeah!
caozeyuan
  • caozeyuan
2F=B, 2R=C, ok?
briana.img
  • briana.img
yeah that makes sense
caozeyuan
  • caozeyuan
1.5R=F, last one
caozeyuan
  • caozeyuan
4 unknows, 4 eqas. all of them are simple
briana.img
  • briana.img
so you put that all together in an equation?
caozeyuan
  • caozeyuan
yep, there you go
whpalmer4
  • whpalmer4
This is pretty easy. \[B + C + F + R = 360\] We know we have 2x as many B as F, so \[B = 2F\] We have 2x as many C as R, so \[C = 2R\]and we have 1 1/2 times as many F as R, so \[F = \frac{3}{2}R\] Now just take the original equation and do all of those substitutions. For example, the first one might be \[B+C+F+R=360\]\[2F+C+F+R=360\]etc. Eventually you will have an equation where R is the only variable, and you can solve that. Then work backwards.
briana.img
  • briana.img
okay so i would do 2F+C+F+R=360 find r? and then do B+2R+F+R=360 and then keep going, right?
caozeyuan
  • caozeyuan
no, you throw the last three into the first one, solve it, then move on
whpalmer4
  • whpalmer4
No, don't substitute variables that you've eliminated back in: \[2F + C + F + R = 360\]\[3F + C + R = 360\]Okay, now pick one of your equations that has either F, C, or R on the left side and use it to get rid of another variable.
whpalmer4
  • whpalmer4
I would just go straight down the list that I gave you substitute \(2R\) everywhere you see \(C\) next then substitute \(\frac{3}{2}R\) every place you see \(F\)
briana.img
  • briana.img
2R+2R+1.5R=360 ??
whpalmer4
  • whpalmer4
\[3F+C+R = 360\]\[3F + (2R) + R = 360\]\[3(\frac{3}{2}F) + 2R + R = 360\]
briana.img
  • briana.img
where did the three come from??
briana.img
  • briana.img
the final equation is the only equation right?
whpalmer4
  • whpalmer4
Sorry, I mistyped something! \[3F + (2R) + R = 360\]Now we substitute \(F = \frac{3}{2}R\) in there, but we don't have 1 F, we have 3 F, so it is \[3(\frac{3}{2}R) + (2R) + R = 360\]which is\[\frac{9}{2}R + 2R + R = 360\]\[\frac{15}{2}R = 360\]can you solve that?

Looking for something else?

Not the answer you are looking for? Search for more explanations.