Answer: 
Step-by-step explanation:
Given : Two opposite sides of a rectangle are each of length x.
Let the other adjacent side be y.
The perimeter of the rectangle is 12 units.
Perimeter of rectangle is given by :-

The area of rectangle is given by :-

Hence, the area as a function x = 
Answer:
b I'm sure byyyyyyyyyyyyy
Answer:
The equation of the line is y - 3 = -2(x + 4)
Step-by-step explanation:
* Lets explain how to solve the problem
- The slope of the line which passes through the points (x1 , y1) and
(x2 , y2) is 
- The product of the slopes of the perpendicular lines = -1
- That means if the slope of a line is m then the slope of the
perpendicular line to this line is -1/m
- The point-slope of the equation is 
* lets solve the problem
∵ A given line passes through points (-4 , -3) and (4 , 1)
∴ x1 = -4 , x2 = 4 and y1 = -3 , y2 = 1
∴ The slope of the line 
- The slope of the line perpendicular to this line is -1/m
∵ m = 1/2
∴ The slope of the perpendicular line is -2
- Lets find the equation of the line whose slope is -2 and passes
through point (-4 , 3)
∵ x1 = -4 , y1 = 3
∵ m = -2
∵ y - y1 = m(x - x1)
∴ y - 3 = -2(x - (-4))
∴ y - 3 = -2(x + 4)
* The equation of the line is y - 3 = -2(x + 4)
The cost of a utility pole with a diameter of 1.5 ft and a height of 40 ft is $782.1
<h3>How to calculate the cost of a utility pole with a diameter of 1.5 ft and a height of 40 ft?</h3>
The given parameters are:
Cost per cubic foot = $11.07 per cubic foot
Height = 40 feet
Diameter = 1.5 feet
Shape = Cylinder
The volume of the cylinder is calculated as:
V = πr^2h
Where
r = Diameter/2
So, we have
r = 1.5/2
Evaluate
r = 0.75
The volume of the cylinder becomes
V = 3.14 * 0.75^2 * 40
Evaluate
V = 70.65
The cost per cubic foot is $11.07 per cubic foot.
So, the cost of a utility pole is
Cost = $11.07 * 70.65
Evaluate
Cost = $782.1
Hence, the cost of a utility pole with a diameter of 1.5 ft and a height of 40 ft is $782.1
Read more about volume at:
brainly.com/question/1972490
#SPJ1
Answer:
11
Step-by-step explanation:

<h2>
<em>OladipoSeun</em><em>♡˖꒰ᵕ༚ᵕ⑅꒱</em></h2>