Recent content by Nawz

ok sweet i think i got it.. this is what i did. 13=5.85wd+16.1-16.1wd -3.1=-10.25wd wd=0.302439024 Since both must equal 100 in this problem. that is why they do 1-wd. What i originally did on the quiz was i just plugged in the answers wd and solved and then w/e was left must...

Homework Statement A. 53% B. 40% C. 30% D. 70% E. 36% Solution: WACC = wd kd (1 - T) + wsks ; 13.00% = wd (9% (1 - 0.35)) + (1 - wd) ($2(1.075)/$25.00 + 7.5%) ; Solve for wd = 30%. Homework Equations The Attempt at a Solution...

Homework Statement I think it was: 3ln5x Homework Equations The Attempt at a Solution I just took it on a test. My professor said it was 3/x but I don't see how you get that? When you get the integral of 3/x how do you get the 3ln5x? Where does the 5 go? I don't understand :(

What is n?

Yea that's what I meant. I don't understand where the Du goes. 8\int\frac{((u+4)/5)^3 + 10}{u^(1/3)} So that is right? Then does du takes the spot of the 5?

So do you get something like: 8\int\frac{(u+4/5)^3 + 10}{u^(1/3)} That's U raised to the 1/3 on the denominator ?

Where do you get the 5x=u+4?

It would be 1/x What is the point of the dx after it too: is it just there for the added C? I got this right now and it is close: \int\frac{(u)^3-4(u)^2+5}{x}dx \frac{(u)^4/4-(4/3)u^3+5x}{x} + C I guess that 5x is suppose to be a U but I don't know why. I still have it divided...

Homework Statement Evaluate using any method: \int\frac{8x^3+10} {\sqrt[3]{5x-4}}dx Homework Equations The Attempt at a Solution I'm lost on this one.

Homework Statement \int\frac{ [(lnx)^3 - 4(lnx)^2 + 5]}{x}dx Homework Equations The Attempt at a Solution xlnx^4 - (4/3x)lnx^3 +5x^2 my attempt at a soultion was all wrong. The divided by x is confusing me on this one. The right answer is: ((lnx)^4/4) -(4/3)(lnx)^3+5lnx+C

If you set x^3=0 You get x=0? If you set e^-x=0 You multiply both by (ln) so -x =ln(0) divide both by -1. That is what I thought? ln(0) / -1

Wow. Yes that makes it much easier. When can you do this? Can you do it For all constants in front of an x? Is it recommended to do that?

Yep, okay i got it. Maximum is 4.6888 at x=4. and Min is 0 at x=0 I don't know if (x3)(e^-x)=0 gives you ln0/-1 but either way 4 is the value I was looking for. Thank You

okay i see that now. so you get something like: (x3)(e-x)[4-x] so then (x3)(e-x)=0 and (4-x)=0 so you get x=4 and then i think the other one is ln0/-1 which is undefined if you set it to 0.

Homework Statement Evaluate: Integral sign: Square root of 3x times dx Homework Equations The Attempt at a Solution integral of 3x^1/2 dx (3x^(3/2)) / (3/2) so I multiplied it by 2/3 and got: (2/3x^(3/2)) +C but the answer is 2 times the square root of 3...