Contain a while loop and a predicate. In each case show that if the predicate is true before entry to the loop, then it is also true after exit from the loop. loop: while (m≥0
and
m≤100)
,
m:=m+4 n:=n−2
, end while, predicate: m+n is odd

answer:   n ≤ -3 or n < -5

step-by-step explanation:

-5 + 2n ≥ -11

+5             +5

2n ≥ -6

2/2 ≤ -6/2

n ≤ -3

-6 - 10n > 44

+6           +6

-10n > 50

-10/-10 50/-10

n < -5

i'm sorry i don't know

step-by-step explanation:

vertical=18cm horizontal=18cm hypotenuse=25.5

step-by-step explanation: