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

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

answer: 5 17/33 % (to the tenth)

step-by-step explanation:

he guessed: 34.5 pounds

actual weight is 32.7

there for he is off:

34.5 -32.7 = 1.8

32.7 is the 100%

1.8 represents %? of 32.7

therefore:

32.7   - 100%

1.8     -     x%

x= 1.8*100%/32.7% = 18*100%/327%

x=1800%/327%   = (5 165/327)%

to the tenth:

5 17/33 %