Now for the point A), L is in the middle of M and H, and the interval will be:
[tex](M,H)[/tex]
For the second point, We need to put the value of the segments in the draw...
From the draw, we can deduce that:
[tex]ML+LH=MH[/tex]
We replace with values:
[tex]\begin{gathered} ML+LH=MH \\ 6x-4+(10x+1)=29 \end{gathered}[/tex]
We solve to x:
[tex]\begin{gathered} 6x-4+(10x+1)=29 \\ 6x\text{ -4 +10x +1=29 ; we agroup the values with x} \\ (6x+10x)-4+1=29 \\ 16x-3=29 \\ 16x=29+3 \\ 16x=32 \\ x=\frac{32}{16}=2 \\ x=2 \end{gathered}[/tex]
Finally, if the value of x = 2, then whi can replace in:
[tex]\begin{gathered} ML=6x-4 \\ ML=6(2)-4 \\ ML=12-4 \\ ML=8 \end{gathered}[/tex]
Your answer of point B) is ML=8.
