## Elementary Technical Mathematics

$l = 5.75 in$ $A = 6.25 in$
To find the length "l", we just need to add both dimensions we already know (the 2.375 in-dimension and the 3.375 in-dimension). $l = 2.375 in + 3.375 in$ $5 + 5 = 10$ so we carry the one. $7 + 7 = 14$, plus the one we carried gives $15$. Again we carry a one. $3 + 3 = 6$; remember to add the one we carried $=7$. Copy the decimal point. $2 + 3 = 5$ So, $l = 5.750 in$. To find the diameter A, again we just need to add all the dimensions we know from that segment (3.750in and 1.250 in twice). So, $A = 3.750 in + 1.250 in + 1.250 in$ $0 + 0 + 0 = 0$; so we move to the next place. $5 + 5 + 5 = 15$; we copy the 5 and carry the $1$. $7 + 2 + 2 = 11$; remember to add the one we carried $= 12$. Copy the $2$ and carry the one. Copy the decimal point. $3 + 1 + 1 = 5$, plus the $1$ we carried $= 6$. So, $A = 6.25 in$