anonymous
  • anonymous
any help on #3?? attachment?
Mathematics
  • Stacey Warren - Expert brainly.com
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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.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Attachment?
anonymous
  • anonymous
?
anonymous
  • anonymous
1 Attachment

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anonymous
  • anonymous
Ohhh haha it wasnt showing on my screen sorry
anonymous
  • anonymous
its hard as heak?
amistre64
  • amistre64
\[-1000 +\frac{1500x}{\sqrt{400+x^2}}=0\]
anonymous
  • anonymous
no #3
amistre64
  • amistre64
whicih #3 then lol
anonymous
  • anonymous
the long paragraph
amistre64
  • amistre64
ooooo.... this one again
amistre64
  • amistre64
ideally when the beach pipe cost equals the sea pipe cost you reach a balance point
amistre64
  • amistre64
b(1000) + s(1500) = Equilibrium we just have to get a formula to equate b and to plug in
amistre64
  • amistre64
s = sqrt(20^2 + (b-x)^2) where b-x is the distance between the rig and the shore pipe
amistre64
  • amistre64
thats about the best i can do for equating the s and b
amistre64
  • amistre64
b(1000) + sqrt(20^2 + (b-x)^2)(1500) = E solve for b right?
anonymous
  • anonymous
yes x is the distance from the oil rigs vertical line to the shoreline..
anonymous
  • anonymous
well it says my answer will be in terms of x?...
amistre64
  • amistre64
maybe better like this tho: when b(1000) = s(1500) b(1000) = sqrt(20^2 + (b-x)^2)(1500) b(1000)/1500 = sqrt(20^2 + (b-x)^2) 2b/3 = sqrt(20^2 + (b-x)^2) ; ^2 both sides
amistre64
  • amistre64
4b^2 ---- = 400 + (b-x)^2 9 4b^2 ----- = 400 + b^2 -2bx +x^2 9 4b^2 = 9(400 + b^2 -2bx +x^2) 4b^2 = 3600 + 9b^2 -18bx +9x^2
amistre64
  • amistre64
4b^2 = 3600 + 9b^2 -18bx +9x^2 0 = (5)b^2 -(18x)b + (9x^2 + 3600) perhaps?
amistre64
  • amistre64
to solve for b we can use the quadratic formula and insert all this other stuff as parameters
amistre64
  • amistre64
18x sqrt(324x^2 - 4(5)(9x^2 + 3600)) --- + ----------------------------- and simplify lol 10 10
amistre64
  • amistre64
sqrt(144x^2 - 72000) looks possible :)
anonymous
  • anonymous
holy bageezus ... i uderstand the transitions and all ..and everything your doing ..but what are parameters?
amistre64
  • amistre64
parameters are just the values we plug into the quadratic formula to make this easier; they relate to the coeffs of the b values
anonymous
  • anonymous
oh okay!.
anonymous
  • anonymous
so we cant reduce the last step by 12 or anything?
amistre64
  • amistre64
sqrt(144x^2 - 72000) the innards here factor to what: sqrt(144(x^2 - 500)) 12 sqrt(x^2 - 500) if i did that part right lol
amistre64
  • amistre64
our options for b are then: 9x 6sqrt(x^2 - 500) -- +- -------------- right? 5 5
anonymous
  • anonymous
would the 500 then be 100?
amistre64
  • amistre64
the 500 cant reduce the way i have it
amistre64
  • amistre64
\[\frac{9x \pm 6 \sqrt{x^2 - 500}}{5}\]
anonymous
  • anonymous
okay so we just add the numerators over 5!...yay!!
amistre64
  • amistre64
if this is b we should be able to add this to x and get 30 right?
anonymous
  • anonymous
OMGEE YES!!!!!
anonymous
  • anonymous
you are amazing!!!!
anonymous
  • anonymous
are you a professor or something?/
amistre64
  • amistre64
just a something....
anonymous
  • anonymous
haha mysterious i understand you a secret agent for the mathematical race of under delinquents?!?!?
anonymous
  • anonymous
haha jk well thanks for the help!!
amistre64
  • amistre64
:) libraries closing soon, so good luck with the rest; and I hope that helped out :)

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