Expanding the limit, we get (x^2+2x∆x+∆x^2-2x-2∆x+1-x^2+2x-1)/<span>∆x
Crossing the 1s , the 2xs, and the x^2s out, we get
(2x</span>∆x+∆x^2-2∆x)/<span>∆x
Dividing the </span><span>∆x, we get
2x+</span><span>∆x-2.
Making the limit of </span><span>∆x=0, we get 2x-2.</span>
The question is incomplete. Here is the complete question.
Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = -
x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.
Answer: L = 1; W = 9/4; A = 2.25;
Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:
A = x.y
A = x(-
)
A = -
To maximize, we have to differentiate the equation:
=
(-
)
= -3x + 3
The critical point is:
= 0
-3x + 3 = 0
x = 1
Substituing:
y = -
x + 3
y = -
.1 + 3
y = 9/4
So, the measurements are x = L = 1 and y = W = 9/4
The maximum area is:
A = 1 . 9/4
A = 9/4
A = 2.25
Answer:I don't know if you're asking for equation or not cause you didn't mention the question so I'm just gonna write down the equation.
y=4.2x+3.4
Use this formula➡️y=mx+b
m is slope, b is y-intercept , x and y are the points like (0, 3.4)
You said it's the y-intercept but it's not. it's just a point on the graph
Step-by-step explanation: