Vinculum

(13*5*12)/sqrt((13+5+12)*(5-13+12)*(13-5+12)*(13+5-12)) using circumradius formula:

you get 6.5

Question 2 is 4/25

As stated in:

https://web2.0calc.com/questions/attempted-twice-i-can-t-find-the-answer

Number 3:

https://web2.0calc.com/questions/help-last-answer-was-wrong-please-ignore-dollar-signs

Tell me if that is correct or wrong....

I double-checked 1 myself:

The area of a sector is x/360*pi*r^2 

60/360*pi*1

= pi/6

The area of the triangle is 1/2*s1*s2 *sin(60deg)

1/2*3*3 *sin(60deg)

= (9sqrt(3)/4)

The prob is the area of the sector / area of the triangle.

giving us:

\(\frac{2\sqrt{3}\pi }{81}\)

Try the answer above^^^^^

I will probably answer 2 and 3 shortly

This question has been answered by Alan before:

Check it out, it is a great explanation in my opinion:

https://web2.0calc.com/questions/let-triangle-abc-be-equilateral-where-the-side-length

Dang, there are a lot of CPhills here!!!

Yes pleas, I think I might have messed up somewhere

y = 11x + 14

2x - 4 = ax + 7 - 4x

y = 11x + 14

y=6x -ax- 11 

We need to find the value of \(a\) which makes 6x-ax=11x

ax = 11x-6x

a = -5

y = 11x + 14

y= 11x - 11 

They both done intercept, so the value of \(a\) must be -5

Also the check I attached the graph below:

https://www.desmos.com/calculator/jfzvbcaqth

See if this helps:

https://www.mathsisfun.com/geometry/triangle-inequality-theorem.html

\(\frac{\left(x+4\right)}{\left(x+5\right)}=\frac{\left(x-4\right)}{\left(2x\right)}+\frac{\left(x+3\right)}{\left(x+5\right)}\)

Multiply by LCM :

\(2x\left(x+4\right)=\left(x-4\right)\left(x+5\right)+2x\left(x+3\right)\)

\(2x^2+8x=3x^2+7x-20\)

Using quadratic formula to solve, 

\(x_1=\frac{-\left(-1\right)+9}{2\cdot \:1},\:x_2=\frac{-\left(-1\right)-9}{2\cdot \:1}\)

\(x = 5, -4\)