Math Question Need Help?

Question

I was just wondering if someone can help me with this math problem I have because I really don't understand it. The question is(A rectangular field has a perimeter of 160m and an area of 1500m^2. find its length and width). Your help would be really appreciated in this matter. Thank You very much.

Answer 1

Perimeter is the distance around the field.

2L + 2W = 160

The area is the number of square units contained within...

LW = 1500

Solve for L

L = 1500/w

Substitute into the first equation

2(1500/w) + 2W = 160

Multiply by w

3000 + 2w^2 = 160w

Move all terms to the left side

2w^2 - 160W + 3000 = 0

Divide out the 2

w^2 - 80W + 1500 = 0

Factor

(w - 50)(w - 30) = 0

Set equal to zero and solve

w - 50 = 0
w = 50

w - 30 = 0
w = 30

The width is 30 or 50 m

Since LW = 1500

The length is 50 or 30 m

Answer 2

So 2x+2y=160 and xy=1500
Use the first equatio to solve for x in terms of y to get x=80-y; next substitute this into the second equation to get
(80-y)(y)=1500 so 80y-y^2=1500
So solve y^2-80y+1500=0
factor this to get (y-50)(y-30)=0 so y=50 or y=30
If y=50 then x=30;conversely, if y=30 then x must equal 50.

Answer 3

A=wl=1500 m^2
p=2l +2w=160 m divide both sides by 2
l+w=80 subtract w from each side
l=80-w
from above wl=1500 substitute 80-w for l
w(80-w)=1500
-w^2+80w=1500 subtract 1500 from each side
-w^2+80w-1500=0 multiply by -1 for convenience.
w^2-80w+1500=0
(w-50)(w-30)=0
w-50=0
w=50
w-30=0
w=30

the field is 30 mn by 50 m

Answer 4

2L+2w=160m
LW=1500m^2 - EQ.1

2(L+W)=160
L+W=80
L=80-W - EQ.2

equate equation 2 to 1

(80-W)W=1500
8W-W^2=1500
W^2-8W+1500=0

use the quadratic equation to W in W^2-8W+1500 and after u get W, equate W in equation 1 to get L

Answer 5

it too easy u just 160 divide 4 =40........... den the length is 40m!!! the area u use ( ) x ( ) = 1500 den u just do it and can find everything!!!