Question

College Algebra Help please!

Answer 1

I am actually clueless... anyway @agent0smith could save the day again or even someone else????

Answer 2

Well if we set up an equation to represent this... we would have

\[\large I = \frac{k}{d^2}\]

Notice the 'd^2' term in the denominator since this varies INVERSELY *Not directly* and it is squared because it is stated in the question

K is your constant of variation *Which we need to solve for to begin*

So what are we given? I = 250 [w/m^2] when d = 5 [m]
So lets plug that in...

\[\large 250 = \frac{k}{(5)^2}\]
\[\large 250 = \frac{k}{25}\]
\[\large k = 6250\]

This is our constant that we will use in the final equation we now set up

\[\large I = \frac{6250}{d^2}\]

Now we just need to solve for 'd' when I = 40

Answer 3

I was totally leaving out the d^2 when I plugged the I back in at the end. So much stress for one silly mistake. Thank you. I got 12.5 for that minus the 5 that I started with to get the correct answer of 7.5. I appreciate it.

Answer 4

Exactly! And not a problem!

Answer 5

@johnweldon1993 sighting.