Answer:
0.375 feet-lb
Step-by-step explanation:
We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.
We can represent our given information as:

We will use Hooke's Law to solve our given problem.

Substituting this value in our integral, we will get:

Using power rule, we will get:
![6=\left[ \frac{kx^2}{2} \right ]^2_0](https://tex.z-dn.net/?f=6%3D%5Cleft%5B%20%5Cfrac%7Bkx%5E2%7D%7B2%7D%20%5Cright%20%5D%5E2_0)


We know that 6 inches is equal to 0.5 feet.
Work needed to stretch it beyond 6 inches beyond its natural length would be 
Using power rule, we will get:
![\int\limits^{0.5}_0 {3x} \, dx = \left [\frac{3x^2}{2}\right]^{0.5}_0](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B0.5%7D_0%20%7B3x%7D%20%5C%2C%20dx%20%3D%20%5Cleft%20%5B%5Cfrac%7B3x%5E2%7D%7B2%7D%5Cright%5D%5E%7B0.5%7D_0)

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.
Step-by-step explanation:
59/8=7.375
So that's mean wxyz is bigger than the ABCD for 7.375
So YZ = 7.375×4 = 29.5
Answer:
3.5
Step-by-step explanation:
edgenuity2020
So I added all of the numbers of tourists together, I got 30 tourists. Then I saw that the number of tourists that visisted the museum but not the zoo was 5. I divided 5 by 30 and got that there was a 16% chance that a randomly selected tourist would of visited the museum but NOT the zoo.
Hope this helps!
~Lizzie
Answer:
The answer is 6
Step-by-step explanation: