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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dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier

step-by-step explanation:

given expression is .

now we need to simplify the given expression .

we know that square root is equivalent to the power 1/2. so that means we can take half of the exponents to get the square root of given expression.

expanding that will give a very very large number so let's keep answer in exponent form.