# Maple TA Quiz 17-2

## Re: Maple TA Quiz 17-2

Question 4: I1 + I2. The answer is negative.

Answer may differ if a different image is loaded in your quiz (for this particular question).

Answer may differ if a different image is loaded in your quiz (for this particular question).

**helping**- Guest

## #2

thanks, but could u explain how u got the answer for #2? i tried using the R = p(L/A) and P = (V/R)V formulas, but i couldnt figure it out. thanks

**at**- Guest

## Re: Maple TA Quiz 17-2

Regarding the question above:

Here's how I did it: First I wrote out the equation R=p(L/A) so that it looked like this: R=p(L/(pi*r^2)). I wrote out the equations, one of them having the value provided (mine was 1.2 something) multiplied onto the r: R=p(L/(pi*r^2)) & R(prime)=p(L/(pi*(1.2r)^2)).

At this point, you can begin calculating the proportion: R/R(prime) = (p(L/(pi*r^2)))/(p(L/(pi*(1.2r)^2)))

Once all is said and done (and you've eliminated all like terms - p, L, pi, r^2), you're left with that same value provided, only squared. And that's your answer.

I hope that helps.

Here's how I did it: First I wrote out the equation R=p(L/A) so that it looked like this: R=p(L/(pi*r^2)). I wrote out the equations, one of them having the value provided (mine was 1.2 something) multiplied onto the r: R=p(L/(pi*r^2)) & R(prime)=p(L/(pi*(1.2r)^2)).

At this point, you can begin calculating the proportion: R/R(prime) = (p(L/(pi*r^2)))/(p(L/(pi*(1.2r)^2)))

Once all is said and done (and you've eliminated all like terms - p, L, pi, r^2), you're left with that same value provided, only squared. And that's your answer.

I hope that helps.

**helping**- Guest

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