perimeter of the first rectangular board is 9 feet.
step-by-step explanation:
let p = perimeter of first board.
l = length of first board
w = width of first board
length of first board = 2 * width of first board - 3
l = 2w - 3
perimeter of first board = 2l + 2w
since l = 2w-3, then
p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6
length and width of the second board are reciprocals of corresponding length and width of the first board.
length of second board = 1/(2w-3)
width of second board = 1/w
perimeter of second board is 1/5th perimeter of the first board.
perimeter of second board = p/5
perimeter of the second board = (2/(2w-3) + 2/w)
since the perimeter of the first board is 5 times the perimeter of the second board, this means that:
6w-6 = 5*(2/(2w-3) + 2/w)
which is the same as:
6w-6 = (10/(2w-3) + 10/w)
if we multiply both sides of this equation by (2w-3), we get:
(6w-6)*(2w-3) = 10 + 10*(2w-3)/w
if we multiply both sides of this equation by w, we get:
(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)
we can simplify this to become:
(6w-6)*(2w-3)*w = 10w + 20w - 30
which becomes:
(6w-6)*(2w-3)*w = 30w-30
if we divide both sides of this equation by (6w-6), we get:
(2w-3)*w = (30w-30)/(6w-6)
which becomes:
(2w-3)*w = 5
this can be simplified to:
2w^2 - 3w = 5
subtract 5 from both sides of this equation to get:
2w^2 - 3w - 5 = 0
which can be factored into:
(2w-5)*(w+1) = 0
which makes:
w = 5/2
or
w = -1.
w can't be negative so the only possible answer is w = 5/2.
if w = 5/2, then
p = 2w-3 = 5-3 = 2
we have:
p = 2
w = 5/2
p = 2l + 2w gets p = 4 + 5 = 9
the perimeter of the first board is 9 feet.
since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.
to see if that's correct, we substitute known values for l and w into the second board.
the length of the second board is 1/l.
this becomes 1/2.
the width of the second board is 1/w
this bgecomes 1/(5/2) = (2/5)
let l2 = length of second board.
let w2 = width of second board.
l2 = (1/2)
w2 = (2/5)
let p2= perimeter of second board.
then p2 = 9/5
since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.
2 * l2 = 2 * (1/2) = 1
2 * w2 = 2 * (2/5) = 4/5
1 is the same as 5/5
5/5 + 4/5 = 9/5
p2 checks out good.
the dimensions as given are good.
answer to the problem is:
perimeter of the first rectangular board is 9 feet.
![\bf \sqrt[4]{\cfrac{24x^6y}{128x^4y^5}}\implies \cfrac{\sqrt[4]{4x^6y}}{\sqrt[4]{128x^4y^5}}\quad \begin{cases} 24=2\cdot 2\cdot 2\cdot 3\\ 128=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\\ x^6=x^{4+2}\\ \qquad x^4x^2\\ y^5=y^{4+1}\\ \qquad y^4y^1 \end{cases}](/tpl/images/0324/3555/c4f86.png)
![\bf \cfrac{\sqrt[4]{\underline{2\cdot 2\cdot 2}\cdot 3\cdot \underline{x^4} x^2\underline{y}}}{\sqrt[4]{\underline{2\cdot 2\cdot 2}\cdot 2\cdot 2\cdot 2\cdot 2\cdot \underline{x^4} y^4\underline{y^1}}} \implies \cfrac{\sqrt[4]{3x^2}}{\sqrt[4]{2^4y^4}}\implies \cfrac{\sqrt[4]{3x^2}}{2y}](/tpl/images/0324/3555/62a69.png)
