Question

Please do you know: The length of a rectangle is 3yd more than twice its width and the area of the rectangle is 77y2 find the dimension of the rectangle

Answer 1

let width b=x
length l=2x+3
area=l*b
x(2x+3)=77
solve and get the solution.

Answer 2

What is the length and width for the quadratic equations

Answer 3

\[2 x ^{2}+3x=77, or2x ^{2}+3x-77=0\]
2*-77=-154
14-11=3
14*-11=-154
\[x ^{2}+\left( 14-11 \right)x-77=0\]
\[x ^{2}+14x-11x-77=0\]
make factors and find x i.,b and l

Answer 4

\[i wrote wrong please 2 with x ^{2}\]
\[2x ^{2}+14x-11x-77=0\]
\[2x \left( x+7 \right)-11\left( x+7 \right)=0\]
\[\left( x+7 \right)\left( 2x-11 \right)=0,either x=-7 or2x=11,x=\frac{ 11 }{2 }\]
width =11/2 yards
length=2*11/2+3=14yards

Answer 5

because width or length can not be negative

Answer 6

Ok. Thanks

Answer 7

yw