If you stretch a spring to 5 ft, it has 310 ft-lb of potential energy (PE). Potential energy varies directly as the square of the stretched length (l). The potential energy can be represented by the formula PE=1/2 kl^2, , where k is the spring constant. a. What is the value of the spring constant for this spring?
b. How many ft-lbs of PE would an 8 ft length of spring have?

Step-by-step explanation:

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b. 2

step-by-step explanation:

the slope of a line =δy/δx

δy= y₂-y₁

δx= x₂-x₁

therefore, using the points (-2,-3)   and (3,7) we can calculate gradient as follows:

m=(7--3)/(3--2) (a number minus a negative number the sign changes to a plus)

m=(7+10)/(3+2)

=10/5

=2

45

step-by-step explanation:

to find a percentage, add a decimal in front and multiply it by the other number.

.25 times 180= 45