Juan has a rectangular corral for his horses. The length of his horse corral is 10 ft longer than 3 times its width. Jeremiah has a rectangular corral for his cattle. The length of his cattle corral is 8 ft longer than 4 times its width. Both corrals have the same width. Let x represent this width, in feet.
1. Write 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
To solve this problem, we need to carefully read the given information and use it to formulate the expressions for each item requested.
(a) The perimeter of each corral:
The perimeter of a rectangle is given by the formula: P = 2(L + W), where P represents the perimeter, L represents the length, and W represents the width.
For Juan's corral:
The length of Juan's corral is 10 ft longer than 3 times its width.
So, the length of Juan's corral can be represented as: L = 3x + 10 ft.
The width of both corrals is given as x.
Therefore, the perimeter of Juan's corral is: P1 = 2(3x + 10 + x) = 2(4x + 10) = 8x + 20 ft.
For Jeremiah's corral:
The length of Jeremiah's corral is 8 ft longer than 4 times its width.
So, the length of Jeremiah's corral can be represented as: L = 4x + 8 ft.
Therefore, the perimeter of Jeremiah's corral is: P2 = 2(4x + 8 + x) = 2(5x + 8) = 10x + 16 ft.
(b) The difference between the perimeters of Jeremiah’s and Juan’s corrals:
To find the difference, we subtract the perimeter of Juan's corral (P1) from the perimeter of Jeremiah's corral (P2).
Therefore, the difference in perimeters is: P2 - P1 = (10x + 16) - (8x + 20) = 2x - 4 ft.
(c) The area of each corral:
The area of a rectangle is given by the formula: A = L × W, where A represents the area, L represents the length, and W represents the width.
For Juan's corral:
The length of Juan's corral is 3x + 10 ft and the width is x ft.
Therefore, the area of Juan's corral is: A1 = (3x + 10) × x = 3x^2 + 10x ft^2.
For Jeremiah's corral:
The length of Jeremiah's corral is 4x + 8 ft and the width is x ft.
Therefore, the area of Jeremiah's corral is: A2 = (4x + 8) × x = 4x^2 + 8x ft^2.
(d) The sum of the areas of both corrals:
To find the sum, we add the area of Juan's corral (A1) and the area of Jeremiah's corral (A2).
Therefore, the sum of the areas is: A1 + A2 = (3x^2 + 10x) + (4x^2 + 8x) = 7x^2 + 18x ft^2.
Now, let's classify each polynomial by its degree and number of terms.
(a) The perimeter of each corral: 8x + 20 (1 term, degree 1)
(b) The difference between the perimeters of Jeremiah’s and Juan’s corrals: 2x - 4 (1 term, degree 1)
(c) The area of each corral:
- Juan's corral: 3x^2 + 10x (2 terms, degree 2)
- Jeremiah's corral: 4x^2 + 8x (2 terms, degree 2)
(d) The sum of the areas of both corrals: 7x^2 + 18x (2 terms, degree 2)
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
However I will start you out.
L = 3x + 10
L = 4x + 8
1a. 2x + 2(3x+10) = perimeter of Juan