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

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

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eudora eudora · Answer 2 · 2019-08-18T08:23:48+02:00

Answer:

Safe load of the beam is 3000 pounds.

Step-by-step 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 = [tex]\frac{k}{x}[/tex]

If y = 2000 pounds and x = 15 feet

then 2000 = [tex]\frac{k}{15}[/tex]

k = 15×2000 = 30000

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

From the formula,

y = [tex]\frac{30000}{10}=3000[/tex] pounds

Therefore, safe load of the beam is 3000 pounds.

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

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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