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July 30, 2014

July 30, 2014

Posted by **Amber** on Wednesday, September 5, 2012 at 2:26am.

- math -
**KokkoDeRas**, Wednesday, September 5, 2012 at 3:04ami)the area of rectangle is x*y (whit x,y side)=>x*y=3200 Ya^2

ii)the condition imposed from the questions is: x+y+y=200 Ya (i have choise two y because the text not exsplicit the misure of one the side):

x+2y=200 => x=200-2y Ya

for (i) (200-2y)*y=3200 Ya^2

200y-2y^2=3200 second-degree equation

you know resolve

- math -
**Reiny**, Wednesday, September 5, 2012 at 7:52amlet the length of the field by y yds, (parallel to river)

leth the width of the river be x yds

first restriction:

2x + y = 200 ----> y = 200-2x

Second piece of data:

xy = 3200

x(200-2x) = 3200

200x - 2x^2 - 3200 = 0

2x^2 - 200x + 10000 = 10000-3200

x^2 - 100x = 1600

x^2 - 100x + 2500 = 2500-1600

(x-50)^2 = 900

x-50 = ± 30

x = 80 or a negative, which will be rejected

the width is 80 yds and the length is 40

check:

40 + 2(80) = 200

80(40) = 3200

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