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Tom [10]
4 years ago
6

A rectangular field with an area of 5000m^2 is enclosed by 300m of fencing. Find the dimensions of the field? How?

Mathematics
1 answer:
garri49 [273]4 years ago
6 0
Area of field=xy=5000
perimeter of field=2(x+y)300
x+y=150
x=150-y
y(150-y)=5000
y^2-150y+5000=0
solve it
u get y= 50 or 100
then  x=100 or 50
therefore the dimensions are 100m and 50m 
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Answer:

Width=37in

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Step-by-step explanation:

Let x in be the width dimension, the length dimension will be x+18 in. Area of a rectangle is calculated as A=lw. Given  the area as2143 sq in, x can be calculated as:

Area= x(18+x)\\\\=18x+x^2\\\\

#Substitute for every x in our function to find the true value of x:

x=37\\\\A=x^2+18x\\\\=37^2+18\times37\\\\=2035 \ in^2#l=37+18=55in

for x=38:

x=38\\\\A=x^2+18x\\\\=38^2+18\times38\\\\=2128 \ in^2#l=38+18=56in

For x=39:

x=39\\\\A=x^2+18x\\\\=39^2+18\times39\\\\=2223 \ in^2#l=39+1854in

From the calculations above, A television with a width of 37 in and length 55in has its area closest to 2143 sq in.

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