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=
![\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7Dsin60%5E%7Bo%7Dside%2Aside%3D%20%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B2%7D%20%20%28%5Cfrac%7Bb-x%7D%7B3%7D%29%20%5E%7B2%7D%3D%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B36%7D%28b-x%29%5E%7B2%7D%20%20%20%20)
A=
![\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B36%7D%28b-x%29%5E%7B2%7D%3D%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B36%7D%28%20b%5E%7B2%7D-2bx%2B%20x%5E%7B2%7D%20%20%29%3D%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B36%7Db%5E%7B2%7D-%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7Dbx%2B%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B36%7Dx%5E%7B2%7D)
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.
![2 \pi r=x](https://tex.z-dn.net/?f=2%20%5Cpi%20r%3Dx)
so
![r= \frac{x}{2 \pi }](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7Bx%7D%7B2%20%5Cpi%20%7D%20)
Area of circle =
![\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}](https://tex.z-dn.net/?f=%20%5Cpi%20%20r%5E%7B2%7D%3D%20%5Cpi%20%20%28%20%5Cfrac%7Bx%7D%7B2%20%5Cpi%20%7D%20%29%5E%7B2%7D%20%3D%20%5Cfrac%7B%20%5Cpi%20%7D%7B%204%20%5Cpi%20%5E%7B2%7D%20%20%7D%2A%20x%5E%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B4%20%5Cpi%20%7D%20x%5E%7B2%7D%20%20)
4. Let f(x)=
![\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B36%7Db%5E%7B2%7D-%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7Dbx%2B%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B36%7Dx%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B4%20%5Cpi%20%7D%20x%5E%7B2%7D)
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)=
![\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x](https://tex.z-dn.net/?f=%5Cfrac%7B%20-%5Csqrt%7B3%7D%20%7D%7B18%7Db%2B%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7Dx%2B%5Cfrac%7B1%7D%7B2%20%5Cpi%20%7D%20x)
=0
![\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0](https://tex.z-dn.net/?f=%5Cfrac%7B%20-%5Csqrt%7B3%7D%20%7D%7B18%7Db%2B%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7Dx%2B%5Cfrac%7B1%7D%7B2%20%5Cpi%20%7D%20x%3D0)
![(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b](https://tex.z-dn.net/?f=%28%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7D%2B%5Cfrac%7B1%7D%7B2%20%5Cpi%20%7D%29%20x%3D%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7Db)
![x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7Db%7D%7B%28%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7D%2B%5Cfrac%7B1%7D%7B2%20%5Cpi%20%7D%29%20%7D%20)
6. So one part is
![\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7Db%7D%7B%28%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B18%7D%2B%5Cfrac%7B1%7D%7B2%20%5Cpi%20%7D%29%20%7D%20)
and the other part is b-
Answer:
True, all integers are rational numbers.
Step-by-step explanation:
Because each integer can be written as n/1 and integers can be positive and negative. For example, 3= 3/1 , 3 is the rational number. But all rational numbers like 1/2 =0.5 are not an integer. (fractions, decimals are not integers.)
Answer:
d = 9.4
Step-by-step explanation:
d=![\sqrt{(3-(-5))^2 +(-3-2)^2\\](https://tex.z-dn.net/?f=%5Csqrt%7B%283-%28-5%29%29%5E2%20%2B%28-3-2%29%5E2%5C%5C)
d=![\sqrt{(8)^2+(-5)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%288%29%5E2%2B%28-5%29%5E2%7D)
d=![\sqrt{64+25}](https://tex.z-dn.net/?f=%5Csqrt%7B64%2B25%7D)
d=![\sqrt{89](https://tex.z-dn.net/?f=%5Csqrt%7B89)
d=9.4