OKay help me out guys and girls, my little cousin needs help and I don’t know how to solve any of these. 
supppose u have 320 feet of fencing and want to completely enclose a rectangular space what is the maximum area you can enclose and wat are the dimensions?
write an equation that defines a rational function with a vertical asymptote as x=2 and a horizontal asymptote at y=0
Help:blush:
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eventhough i am good in math.. but rite now i can’t think of anythin… u gotta apply some kinda formula to dat rectangular space in order to find maximan area and dimesions. i think u would use area of a rectangle..:halo:
and then take a slope of dat rectangular space and write an equation: y=something+b 
and after dat u r gonna google ver. and hori. asymptote.. okay
good luck.
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awww I tried google....I guess i'll just let the poor kid down.
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That's okay I know no one wants to help poor old me, I heard cafe members were nice so I thought I would check.
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dimension = length 20
wi= 16
max= 320
sorry homiei am still thinking about the function
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chalo jee.. insan (da math teach is hea) :D
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i guess function will be y=5/4x
coz if leng is =y, 20
and wid = x , 16 20/16 give slope =5/4
not sure i don’t remeber this math anymore 
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Sounds good to me now on to the second prob.
Lots of cookies, cakes and dua's for you guys.
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well i know that i have space of 320 feet
so i just take two number which i multiply and give 320 or close to it .
coz are of rectangular is = l*w
so i choose two number which gave me 16 * 20 = 320
max is obvious coz all the space i have alredy filled up so there is no more space :) so again max will be 320 :)
still i a, not sure coz i did this long time ago and my mind is not working right now coz iam doing my hw
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so i was right, to some extent :D
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thanks for duas and cookies now what is ur second problem :halo:
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i guess
only thing is still itching me is assytope
coz if assy. is x=2 and y=0 then there is something should be there which i ma missing coz ic an’t think anything right now
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write an equation that defines a rational function with a vertical asymptote as x=2 and a horizontal asymptote at y=0
How did you find the slope insane uncle?
Shoaib...no cookies for you.
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check it out on ur graphing calculator yo :D
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it won’t make any different coz i now how the graphs look of tthis function and thats why i have dobts
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Now we’re having doubts…guys guys what happend? 
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ia m uncle for god sake i am younger than belle baji
well if y = 20 and x = 16 the y/x give u slope
ok
so
we know when x = 2 then y = 0
we know when x = 16 then y = 20
what is difference in x and y
x went from 2 to 16 so change is = 14
y went from 0 to 20 so the change is =20
y/x give u a slope so 20/14=1.43
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i think is correct coz i am trying to rechecking different ways and ia m getting it but still can excatly remember but is much better u have nothing on the paper at least u try