Width: x
Length: 2x
Area=x(2x)=2x^2
2x^2=200
x^2=100
x=10 or -10(reject as x>0) [width cannot be less than 0, doesnt make sense]
Perimeter
=x+x+2x+2x
=6x
=6(10)
=60 yd
Range is just input the domain
so find f(-2), f(0) and f(2)
f(-2)=-(-2)²+1=-4+1=-3
f(0)=-(0)²+1=0+1=1
f(2)=-(2)²+1=-4+1=-3
range={-3,1}
The LCD is 1/15 and 3/15.
Her statement is possible, the first rectangle could be equal to the second because x is unknown so if x was 5in in the first rectangle and - 1 in the second rectangle both problems would look like this 2 in & 5 in and the second one 5in & (2 in = (-1 +3)).
