Consider a wire of length 16 ft. The wire is to be cut into two pieces of length x and 16 − x. Suppose the length x is used to form a circle of radius r and the length 16 − x is used to form a square with side of length s. What value of x will minimize the sum of their areas

x = 7.14ft

Step-by-step explanation:

First, the circle will have a perimeter P = x.

We know that the perimeter of a circle is:

P = 2*pi*r

where pi = 3.14 and r = radius of the circle.

Then we will have:

x = 2*pi*r

r = (x/(2*3.14)) = x/6.28

And the area of a circle of radius r is:

A = pi*r^2 = 3.14*(x/6.28)^2

Now, for the square, the perimeter will be:

P = 16ft - x

And we know that the perimeter of a square of sidelengt s is:

P = 4*s

then:

16ft - x = 4*s

s = (16ft - x)/4 = 4ft - x/4

And the area of a square is:

A = s^2 = (4ft - x/4)*(4ft - x/4).

Now, the total area of the circle + square will be:

Total area =  (4ft - x/4)*(4ft - x/4) + 3.14*(x/6.28)^2

We want to find the value of x that minimizes this.

First, let's rewrite the area equation as a quadratic equation:

Ta(x) = (4ft - x/4)*(4ft - x/4) + 3.14*(x/6.28)^2

        = 16ft + x^2/16  -2ft*x + 0.08*x^2

        =  (1/16 + 0.08)*x^2 - 2ft*x + 16ft^2

        = 0.14*x^2 - 2ft*x + 16ft^2

This is a quadratic equation with a positive leading coefficient, this means that the arms of the graph will open upwards, then the minimum will be at the vertex of the equation.

To find the vertex we can just take the first derivative, and find at what value of x is equal to zero.

Ta´(x) = 2*0.14*x - 2ft

Ta´(x) = 0.28*x - 2ft

Let´s find x such that this is equal to zero:

Ta´(x) = 0 = 0.28*x - 2ft

           x = 2ft/0.28 = 7.14ft

Then x = 7.14ft minimizes the area of the sum of the areas of the circle and square.

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

3x + 18 = 147°

step-by-step explanation:

3x + 18 = 147°, because the value of the central angle is equal to the value of the arc formed by it.

i hope i you.