Respuesta :
Hello!
Let's solve this using a system of equations. We will have w be the width and L be the length.
w=0.5L
wL=50
Now we will substitute the value for y in the second equation.
0.5L(L)=50
0.5L²=50
L²=100
L=10
Now we plug our L value into the first equation.
w=0.5(10)
w=5
The width is 5 and the length is 10.
I hope this helps!
Let's solve this using a system of equations. We will have w be the width and L be the length.
w=0.5L
wL=50
Now we will substitute the value for y in the second equation.
0.5L(L)=50
0.5L²=50
L²=100
L=10
Now we plug our L value into the first equation.
w=0.5(10)
w=5
The width is 5 and the length is 10.
I hope this helps!
"The width of a rectangle is half as long as the length"
So use this equation W=(1/2)*L
"The rectangle has an area of 50 square feet"
A=L*W=50
Substitute,
L*W=50
L*(1/2)*L=50
(L^2)/2=50
L^2=100
Length=10 feet
Then,
W=(1/2)*L
W=(1/2)*10
Width=5 feet
So use this equation W=(1/2)*L
"The rectangle has an area of 50 square feet"
A=L*W=50
Substitute,
L*W=50
L*(1/2)*L=50
(L^2)/2=50
L^2=100
Length=10 feet
Then,
W=(1/2)*L
W=(1/2)*10
Width=5 feet