The walkway is 1.5 m wide.
The area of the pool is 12(6) = 72 m².
Adding a walkway of unknown width, x, around all 4 sides of the pool increases the width by 2x and the length by 2x; thus the area of the entire pool and walkway together would be given by
(12+2x)(6+2x)
We know that the area of just the walkway is 9 m² less than the area of the pool. This means that:
(12+2x)(6+2x)-72 = 72-9
Multiplying through we have:
12*6+12*2x+2x*6+2x*2x - 72 = 63
72 + 24x + 12x + 4x² - 72 = 63
24x + 12x + 4x² = 63
36x + 4x² = 63
Writing in standard form we have:
4x² + 36x = 63
We want to set it equal to 0 to solve, so subtract 63 from both sides:
4x² + 36x - 63 = 63 - 63
4x² + 36x - 63 = 0
Using the quadratic formula,

Since a negative width makes no sense, the walkway is 1.5 m wide.
Answer:
The answer is neither.
Step-by-step explanation:
The slope of AB is -5/0, and the slope of CD is 4/0.
Answer:
minus 9, and 1
Step-by-step explanation:
25,688 sonygfkdjfjdjfjgj needed 20 lol
Answer:
(0,5)
Step-by-step explanation:
we have
----> inequality A
----> inequality B
we know that
If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities of the system (makes true both inequalities)
Verify each ordered pair
Substitute the value of x and the value of y of each ordered pair in the inequality A and in the inequality B and then compare the results
case 1) (0,5)
For x=0,y=5
<em>Inequality A</em>

----> is true
so
The ordered pair satisfy inequality A
<em>Inequality B</em>

---> is true
so
The ordered pair satisfy inequality B
therefore
The ordered pair would be a solution of the system
case 2) (5,2)
For x=5,y=2
<em>Inequality A</em>

----> is true
so
The ordered pair satisfy inequality A
<em>Inequality B</em>

---> is not true
so
The ordered pair not satisfy inequality B
therefore
The ordered pair is not a solution of the system
case 3) (0,4)
For x=0,y=4
<em>Inequality A</em>

----> is not true
so
The ordered pair not satisfy inequality A
therefore
The ordered pair is not a solution of the system
case 4) (6,0)
For x=6,y=0
<em>Inequality A</em>

----> is not true
so
The ordered pair not satisfy inequality A
therefore
The ordered pair is not a solution of the system