Question

Let y = safe load in pounds and x = length in feet of a horizontal beam. A constant of proportionality k exists such that Y=k/x If a beam can hold 2,000 pounds at 15 feet, what is the safe load if the length of the beam is 10 feet?

Answers: 300 pounds, 3,000 pounds, 20,000 pounds

Answer 1

3000 pounds

Explanation:

first sub in info to find k

2000=k/15 ; multiply both sides by 15 ; k=30000. if k is the constant, then to find the safe load (y) with the new beam (x), we input our new info into the equation.

y=30000/10 ; y=3000

Answer 2

Safe load of the beam is 3000 pounds.

Explanation:

If a horizontal beam of x feet length can hold y pounds safe load, the expression that represents the relation between load and length of the beam is

y =
`(k)/(x)`

If y = 2000 pounds and x = 15 feet

then 2000 =
`(k)/(15)`

k = 15×2000 = 30000

Now we will calculate the safe load when beam is 10 feet long.

From the formula,

y =
`(30000)/(10)=3000` pounds

Therefore, safe load of the beam is 3000 pounds.