Prove that the sum of the lengths of the diagonals of a quadrilateral is less than the perimeter but greater than the half of the perimeter of this quadrilateral. Given: ABCD Prove: AC + BD > 1/2 P (of ABCD)

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

after 1 year: 221 deer

step-by-step explanation:

30% of 170 is 51. 170+51=221

multiply then divide

step-by-step explanation: