The length of a new rectangular playing field is 8 yards longer than double the width. if the perimeter of the rectangular playing field is 280 yards, what are its dimensions.

Let

width,w

length, l = 2w +8...eqn 1

perimeter, p = 2 (l + w)...eqn 2

subst for l in eqn 2...

p = 2(2w +8 + w) = 2 (3w + 8) = 280

280 = 6w + 16

w = 264 /6 = 44

subst for w in eqn 1

l = 2× 44 + 8 = 96

length = 96yds

width = 44 yds

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.