The spring of a spring balance is 5.0in. long when there is no weight on the balance, and it is 8.4in. long with 4.0 lb hung from the balance. How much work is done in stretching it from 5.0in. to a length of 12.8in.?

work is done in stretching it from 5.0in. to a length of 12.8in is 36 lb.In

Step-by-step explanation:

We know that

f(x) = kx

given that;

Natural length L = 5

x = 8.4 - 5 = 3.4

force F = 4.0 lb

f(x) = kx

we substitute

4 = k × 3.4

k = 4 / 3.4

k = 1.1764 = 20/17

Now work done from natural length to 12.8

x = 12.8 - 5 = 7.8

work done = ₀∫^7.8  f(x) dx

= ₀∫^7.8  kx dx

we substitute value of k

= ₀∫^7.8  (20/17)x dx

= 20/17 [ x²/2 ]

= 20/17 [ (7.8)²/2 - 0 ]

= 1.1764 × 30.42

= 35.786 ≈ 36 lb.In

Therefore work is done in stretching it from 5.0in. to a length of 12.8in is 36 lb.In

18.8 cubic inches.

step-by-step explanation:

volume of a cone = πr²h

r = 1.5 inches

h = 8 inches

pi = 3.14

volume of our cone = × 3.14 × 1.5² × 8 = 18.84 square inches

or equals to 18.8 square inches rounded up to nearest tenth.

i believe that the answer is

b. y-1=-2(x-4)

hope this !