SOLUTION: Juan has a rectangular corral for his horses. The length of his horse corral is 10 ft longer than 3 times it’s width. Jeremiah has a rectangular corral for his cattle. The length

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Question 1179865: Juan has a rectangular corral for his horses. The length of his horse corral is 10 ft longer than 3 times it’s width. Jeremiah has a rectangular corral for his cattle. The length of his cattle corral is 8 ft longer than 4 times it’s width. Both corrals have the same width. Let x represent this width, in feet.
1. White a polynomial, in standard form, for each of the following. Show your work. Classify each polynomial by its degree and by its number of terms.
(a) the perimeter of each corral
(b) the difference between the perimeters of Jeremiah’s and Juan’s corrals
(c) the area of each corral
(d) the Sum of the areas of both corrals
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
for juan:
L1 = 3W + 10
for jeremiah:
L2 = 4W + 8
both corrals have the same width.

let x = W.

equations becomes:

L1 = 3x + 10
L2 = 4x + 8

perimeter for juan corral = 2L1 + 2x
perimeter for jeremiah corral = 2L2 + 2x

replace L1 with 3x + 10 and replace L2 with 4x + 8 to get:

perimeter for juan corral = 2 * (3x + 10 + x) = 6x + 20 + 2x = 8x + 20
perimeter for jermiah corral = 2 * (4x + 8 + x) = 8x + 16 + 2x = 10x + 16

both equations are degree 1 because the highest exponent of x is 1.

that takes care of (a).

(b).

10x + 16 minus 8x + 20 = 2x - 4.

jeremiah's perimeters is 2x - 4 more than juan's.
juan's perimeter is 2x - 4 less than jeremiah's.

(c).

the area of juan's corral is length * width = x * (3x + 10) = 3x^2 + 10x.

the area of jeremiah's corral is length * width = x * (4x + 8) = 4x^2 + 8x.

both these equations are second degree because the highest exponent in each is 2.

(d).

the sum of the areas of both corrals is:

3x^2 + 10x + 4x^2 + 8x = 7x^2 + 18x.