Answer:
Work done in stretching the spring = 7.56 lb-ft.
Step-by-step explanation:
Normal length of the spring = 8 in or
ft
=
ft
If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in
=
ft
Force applied to stretch the spring = 12 lb
By Hook's law,
F = kx [where k is the spring constant and x = length by which the spring is stretched]
12 = k(
)
k = 18
Work done (W) to stretch the spring by
ft will be
W = 
= 
= ![18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0](https://tex.z-dn.net/?f=18%5B%5Cfrac%7Bx%5E%7B2%7D%7D%7B2%7D%5D%5E%7B%5Cfrac%7B11%7D%7B12%7D%7D_0)
= 9(
)²
= 7.56 lb-ft
Answer:
The graph is shifted 1 unit right
The graph is shifted 2 units down
Step-by-step explanation:
We are given
Parent function as

New function as

we can also write as

We can see that
1 is subtracted to x-value
and 2 is added to y-value
So, f(x) is shifted right by 1 unit
and
f(x) is shifted up by 2 unit
Answer: -1536
Step-by-step explanation:
Answer:
12
Step-by-step explanation:
First, multiply the length by two then, subtract the total by the sum, last divide by two.
(25 * 2)
(74 - 50)
(24/2)
Answer = 12