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Please explain this in 3rd grade terms

2007-05-01 18:40:11 · 4 answers · asked by WENDY M 1 in Science & Mathematics Mathematics

4 answers

First, you have to assume the room is a rectangle. Then it has four sides Length x 2 and Width x 2.

L + L + W + W = 64 yds
L = 8
8 yds + 8 yds + W + W = 64 yds
16 yds + W + W = 64 yds
16 yds - 16 yds + W + W = 64 yds - 16 yds
W + W = 48 yds
(W + W)/2 = 48 yds / 2
W = 24 yds

2007-05-01 18:47:39 · answer #1 · answered by TychaBrahe 7 · 0 0

Let x = width of the room
x + 8 = length of the room

Total distance od the room = length + length + width + width
Assuming that the room is a rectangular in shape.

D = 2l + 2w , l = length, w = width


Equation:

2x + 2( x + 8 ) = 64
2x +2 x + 16 = 64
4x + 16 = 64
4x + 16 - 16 = 64 - 16
4x = 48
4x/4= 48/4
x =12 yards ===> the width
12 + 8 = 20 yards ==> the length

2007-05-01 18:55:50 · answer #2 · answered by detektibgapo 5 · 0 0

It is 64 yd around the room. One wall is 8 yd long. The wall on the opposite side of the room has to be 8 yd long, too. What's left when we subtract these two lengths from 64?
64
-8
---
56
-8
---
48
The 2 walls that are left have to be the same length and have to add up to 48. To find them, we divide 48 by 2:
. . 24
2)48
. 4
. ----
. . . 8
. . . 8
. . .---

So these walls are 24 yd long.

2007-05-01 19:10:08 · answer #3 · answered by Helmut 7 · 0 0

Easy!
To measure distances and angles on manifolds, the manifold must be Riemannian. A Riemannian manifold is an analytic manifold in which each tangent space is equipped with an inner product 〈⋅,⋅〉 in a manner which varies smoothly from point to point. Given two tangent vectors u and v, the inner product 〈u,v〉 gives a real number. The dot (or scalar) product is a typical example of an inner product. This allows one to define various notions such as length, angles, areas (or volumes), curvature, gradients of functions and divergence of vector fields.

Most familiar curves and surfaces, including n-spheres and Euclidean space, can be given the structure of a Riemannian manifold.

For any metric space M, one can construct a complete metric space M' (which is also denoted as M with a bar over it), which contains M as a dense subspace. It has the following universal property: if N is any complete metric space and f is any uniformly continuous function from M to N, then there exists a unique uniformly continuous function f' from M' to N which extends f. The space M' is determined up to isometry by this property, and is called the completion of M.

2007-05-01 18:46:29 · answer #4 · answered by Anonymous · 0 2

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