The intensity, I, of light received at a source varies inversely as the square of the distance, d, from the source. If the light intensity is 33 foot-candles at 20 feet, find the light intensity at 5fee

The light intensity is 528 foot-candles at 5 feet

Step-by-step explanation:

If x and y are in inverse variation, then x y = k, where k is the constant of variation

In the given question

∵ The intensity I, of light received at a source, varies inversely as the

   square of the distance d, from the source

→ That means I × d² = k, where k is the constant of variation

∴ I × d² = k

∵ I = 33 foot-candle at d = 20 feet

→ Substitute them in the rule above to find k

∴ 33 × (20)² = k

∴ 33 × 400 = k

∴ 13200 = k

→ Substitute the value of k in the equation

∴ I × d² = 13200 ⇒ equation of variation

∵ d = 5

→ Substitute it in the equation to find I

∵ I × (5)² = 13200

∴ I × 25 = 13200

→ Divide both sides by 15

∴ I = 528

∴ The light intensity is 528 at 5 feet

24

step-by-step explanation:

if it takes 8 pounds to make 12 divide 12 by 8 and you get 1.5, 1.5 x 16 = 24

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: