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Shalnov [3]
2 years ago
6

A rectangle has one side on the x-axis and two vertices on the curve, = 1/1+x^2. find the coordinates of the vertices of the rec

tangle with the maximum area
Mathematics
2 answers:
Readme [11.4K]2 years ago
8 0
I believe that you have issues with the function for the curve. Because if that function is correct, the rectangle with the maximum area will have the two coordinates off the X-axis of (-infinity, infinity) and (infinity, infinity) and the area of the rectangle will be infinite.

Assuming correct expression is
y = 1/1 + x^2 = 1 + x^2.
Then you need to find 2 x values that have the same y value. You'll quickly realize that the values X and -X will work and give you the same Y value. And as you use large absolute values of X, the Y value will also increase. And carried to the logical limit, the largest possible rectangle will happen with X values of -infinity and +infinity.

Assuming correct expression is
1 = 1 + x^2, which simplifies to 0 = x^2, which has the exact same argument. The coordinates of the 2 points are (-infinity, infinity) and (infinity, infinity). So once again, the area of the rectangle increases without limit.

Tcecarenko [31]2 years ago
5 0
<span>The dimension of the rectangle would be length=#4/sqrt3# and width =4-#4/3= 8/3# Explanation: Let one vertex on the x axis be x, then the other vertex would be -x. Length of the rectangle would be thus 2x and width would be #4-x^2# Area of the rectangle would be A=#2x(4-x^2)= 8x-2x^3# For maximum area #(dA)/dx=0=8-6x^2#, which gives x=#2/sqrt3# The dimension of the rectangle would be length=#4/sqrt3# and width =4-#4/3= 8/3# Was this helpful? Let the contributor know!</span>
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Given the functions a(x) = 3x - 12 and b(x) = x-9, solve a[b(x)].
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Answer:

Step-by-step explanation:

hello :

a(x) = 3x - 12 and b(x) = x-9, so

a[b(x)]=a(x-9) =3(x-9)-12

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2 years ago
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Answer:

y=-5/3x+20

Step-by-step explanation:

Let the equation of the required line be represented as \[y=mx+c\]

This line is perpendicular to the line \[y=\frac{3}{5}x+10\]

\[=>m*\frac{3}{5}=-1\]

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So the equation of the required line becomes \[y=\frac{-5}{3}x+c\]

This line passes through the point (15.-5)

\[-5=\frac{-5}{3}*15+c\]

\[=>c=20\]

So the equation of the required line is \[y=\frac{-5}{3}x+20\]

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