Set up the given triangle on x-y coordinates with right angle at (0,0). So the two vertices are at (5,0) and (0,2
![sqrt{x} n]{3}](https://tex.z-dn.net/?f=sqrt%7Bx%7D%20n%5D%7B3%7D%20)
)
let (a,0) and (0,b) be two vertices of the<span> equilateral triangle. So the third vertex must be at </span>

for a pt (x,y) on line sx+ty=1, the minimum of

equals to

smallest value happens at

so area is

hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145
Use the formula of an area of a tringle:

We have b = 11 yd, c = 24 yd and m∠A = 67°


Answer:
All of the above
Step-by-step explanation:
Also, this is a history, not a maths problem
$78 for each tire
4,992 divided by 64 is 78
<span>-34-r=-7(8r-3)
- 34 - r = - 56r + 21
- r + 56r = 21 + 34
55r = 55
r = 55/55
r = 1</span>