The first one it goes up in the same increments
ANSWER
and
e have
EXPLANATION
Let us make y the subject and call it equation (2)
We put equation (2) in to equation (1)
Simplify to get,
Divide both sides by 31,
We put this value in to equation (2) to get,
We collect LCM to obtain,
Problema Solution
You have 800 feet of fencing and you want to make two fenced in enclosures by splitting one enclosure in half. What are the largest dimensions of this enclosure that you could build?
Answer provided by our tutors
Make a drawing and denote:
x = half of the length of the enclosure
2x = the length of the enclosure
y = the width of the enclosure
P = 800 ft the perimeter
The perimeter of the two enclosures can be expressed P = 4x + 2y thus
4x + 3y = 800
Solving for y:
........
click here to see all the equation solution steps
........
y = 800/3 - 4x/3
The area of the two enclosure is A = 2xy.
Substituting y = 800/3 - 4x/3 in A = 2xy we get
A = 2x(800/3 - 4x/3)
A =1600x/3 - 8x^2/3
We need to find the x for which the parabolic function A = (- 8/3)x^2 + (1600/3)x has maximum:
x max = -b/2a, a = (-8/3), b = 1600/3
x max = (-1600/3)/(2*(-8/3))
x max = 100 ft
y = 800/3 - 4*100/3
y = 133.33 ft
2x = 2*100
2x = 200 ft
Based on the measurements of the pool and the path of uniform width, the area of the width and length of the path can be found to be 28.8 meters.
<h3>How to find the width and length?</h3>
First, find the area of the pool:
= Length x Width
= 10 x 26
= 260 meters ²
The area of the path can therefore be found to be:
= Total area of pool and path combined - Area of pool
= 1,092 - 260
= 832 meters²
Seeing as the path has uniform width, that means that the width is the same and the length so the width and length of the pool is:
= √area of the path
= √832
= 28.8
In conclusion, the width and length of the path are 28.8 meters.
Find out more on combined area at brainly.com/question/12966430
#SPJ1
