Suppose that 6 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 55 cm. (a) How much work is needed to stretch the spring from 42 cm to 47 cm? (Round your answer to two decimal places.) (b) How far beyond its natural length will a force of 25 N keep the spring stretched? (Round your answer one decimal place.)

answer is x<

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

idk

step-by-step explanation:

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40