anonymous
  • anonymous
Use the general formula to write the area A of the enclosure as a function of w (that is, only the variable w should appear in the function). it should be a quadratic function
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
What enclosure?
myininaya
  • myininaya
what is the enclosure? are we enclosing a rectangle or something? or can we enclose whatever we want?
anonymous
  • anonymous
hold on im trying to get the file of the actuall question its part of a project

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anonymous
  • anonymous
Its to question number 5
anonymous
  • anonymous
I am completely lost
myininaya
  • myininaya
maybe this will help
1 Attachment
anonymous
  • anonymous
Okay, so you have 682 feet of fencing to make up two "width sides" and one "length side" this translates to the equation: 682 = 2W + L You know from basic geometry that the area of a rectangle is length times width: A = L*W What you need to do is solve the first equation for L in terms of W: 682 = 2W + L L = 682 -2W Then substitute that L value into the second function so that you have A in terms of just W: A = L*W A = (682 -2W)W
anonymous
  • anonymous
anyone?
myininaya
  • myininaya
?
myininaya
  • myininaya
anyone what?
anonymous
  • anonymous
the site froze on me im taking a look at it now
myininaya
  • myininaya
ok lol
anonymous
  • anonymous
now if i can get the window correct on my ti 84 :P
anonymous
  • anonymous
any help with that?
myininaya
  • myininaya
go to Y=
myininaya
  • myininaya
type 682x-2x^2
myininaya
  • myininaya
then go to window
anonymous
  • anonymous
yeah i got the equation in its just getting a clear shot of the parabola
myininaya
  • myininaya
go to where it says x min and type 100 go to where it says x max and type 200 go to where it says y min and type 80 go to where it says y max and type 60000 and see if it looks viewable
myininaya
  • myininaya
try going to the highest point on the curve and approximate what it is
anonymous
  • anonymous
it works but im going to need both sides of the parabola can i just extend the xmin
anonymous
  • anonymous
holy crap I got it
myininaya
  • myininaya
no you are fine but you can if you want to
anonymous
  • anonymous
you are amazing!
myininaya
  • myininaya
did you use the maximum feature on your calculator?
anonymous
  • anonymous
ximn -100 xmax 400 ymin 80 ymax 60000
myininaya
  • myininaya
no thats not what i meant
myininaya
  • myininaya
do you see calc?
anonymous
  • anonymous
i havent taken advantage of that yet
myininaya
  • myininaya
lets do it
anonymous
  • anonymous
i got the max
myininaya
  • myininaya
do you see maximum
anonymous
  • anonymous
yeah did the whole left and right bound
anonymous
  • anonymous
thats going to give me the best area correct?
myininaya
  • myininaya
oh nice i got (170.50001,58140.5) this is just an appoximation I arleady found the exact on that pdf file
myininaya
  • myininaya
your approximation doesn't have to be mine but both of ours should be pretty close to the same thing is yours close to my?
anonymous
  • anonymous
its exact
myininaya
  • myininaya
k now you remember the file i gave you right? that vertex of the parabola i found by putting the parabola in vertex form gave me the exact (max width,max area)
anonymous
  • anonymous
what file
myininaya
  • myininaya
scroll up if you haven't seen the file and click on it
anonymous
  • anonymous
looking at it now
myininaya
  • myininaya
you remember when you uploaded a file i uploaded a file like a few minutes afterwards
anonymous
  • anonymous
yeah i see it
anonymous
  • anonymous
if i have any more questions ill post them
myininaya
  • myininaya
ok make a new post though because i'm about to leave or do it so you can get someone's attention and bring them back here to help k?
myininaya
  • myininaya
or 3wood might come back
myininaya
  • myininaya
gn guys and peace
anonymous
  • anonymous
thanks for all the help both of you are life savers

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