Here you're being asked to find the "perimeter" of the space, even tho' the problem doesn't specifically ask for it.
The formula for P is P = 2W + 2L.
Here the width, W, is 3 1/2 yds, and the length, L, is 4 2/3 yds. Subbing these two values into the formula for P (above) results in:
P = 2(3 1/2 yds) + 2(4 2/3 yds)
= 7 yds + 9 1/3 yds = 16 1/3 yds, total.
Answer:
Step-by-step explanation:
C.
Answer: (A) CDD
if D=C^-1
then
C D D = (C C^-1) D = I D = D
(where I is the identity matrix)
Answer:
Hi
Step-by-step explanation:
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