Answer:
3 3/11 im sure its correct.
1. Divide wire b in parts x and b-x.
2. Bend the b-x piece to form a triangle with side (b-x)/3
There are many ways to find the area of the equilateral triangle. One is by the formula A=

A=

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6
3. Let x be the circumference of the circle.

so

Area of circle =

4. Let f(x)=

be the function of the sum of the areas of the triangle and circle.
5. f(x) is a minimum means f'(x)=0
f'(x)=

=0



6. So one part is

and the other part is b-
Answer:
Has two unkown variables
Step-by-step explanation:
When there is only one equation provided with two unknown variables, it is difficult to solve for the unknown variables
The most simplified version of this equation would be VX = -14
More information would be needed to solve
Circle: x^2+y^2=121=11^2 => circle with radius 11 and centred on origin.
g(x)=-2x+12 (from given table, find slope and y-intercept)
We can see from the graphics that g(x) will be almost tangent to the circle at (0,11), and that both intersection points will be at x>=11.
To show that this is the case,
substitute g(x) into the circle
x^2+(-2x+12)^2=121
x^2+4x^2-2*2*12x+144-121=0
5x^2-48x+23=0
Solve using the quadratic formula,
x=(48 ± √ (48^2-4*5*23) )/10
=0.5058 or 9.0942
So both solutions are real and both have positive x-values.
Answer: The second circle
Step-by-step explanation: I got a 100