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
Answer:
(- 2, 4 )
Step-by-step explanation:
Given endpoints (x₁, y₁ ) and (x₂, y₂ ) , then the midpoint is
( 
, 
)
Here (x₁, y₁ ) = A (4, 6 ) and (x₂, y₂ ) = B (- 8, 2 )
midpoint = ( 
, 
) = ( 
, 
) = (- 2, 4 )
Answer:
d=32!!!
Step-by-step explanation:
So first rewrite it as an equation: d/4+12=20
d/4+12=20 then subtract 12 from both sides to balance the equation
d/4=8 so multply both sides by 4 to isolate the variable, 4(d/4)=d =d and 8x4=32, so d=32
Is skipped a number and that's not good
9514 1404 393
Answer:
C 24
D 25
Step-by-step explanation:
Consider the choices offered:
5 and 7 will not form a triangle* with any of the other lengths
7 and 12 will not form a triangle with any of the other (longer) lengths
7 and 24 will form a triangle with the other possibilities, 25 and 29. (needs further consideration)
7 and 25 will not form a right triangle with 29, because a right triangle cannot have 3 odd-length sides.
__
<em>Further consideration</em>
7^2 +24^2 = 49 +576 = 625 = 25^2 . . . . checking Pythagorean theorem
The side measures 7, 24, 25 will form a right triangle.
_____
* A triangle can be formed from three segments only if the sum of the shortest two exceeds the length of the longest one.
