Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more than 132 in. find the dimensions of a rectangular package that has a square cross section and largest volume that may be sent by priority mail. hint: the length plus the girth is 4x + l.

v = 11264 in^3

Step-by-step explanation:

4 x + l =132

l=132- 4 x

v= h x^ 2

 = x ^ 2 (132 - 4 x )

= 132 x ^ 2 - 4 x^3

 = 4 ( 33 x^2 - x ^3)

d v/d x = 4 (66 x - 3 x^2)

         = 12 x ( 22 - x)

d v/d x = 0

when x = 0  or 22

So

h = 132 - 4 *22 = 44

So if the package is  44*16*16

v = 11264 in^3

So  4 x+l >= 132 But it should be minimum if 4 x+l contain smaller value

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2

answer: the angel abc is 30°,and the abc is a triangle where the angel bac is 90°

so ac is the half of 24=12ft,and from the pythagorean theorem we find ba

ba=√bc^2-ac^2=√576-144=√432=12√3ft

so bac is 12+12√3=12(1+√3)ft