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.

4 0

3 years ago

Hello, can anyone help please? :D

Lubov Fominskaja [6]

Answer:

just guess

Step-by-step explanation:

i got it right

6 0

3 years ago