Pegs

f(x) = (9x^3 - 1x^2 + 8x + 4)/(-2x^2 - 4x - 2) What is the smallest value of x at which f(x) intersects its non-vertical asymptote? I took the slant asymptote which is -9/2x+19/2 and set it equal to the function and tried to solve it as a quadratic equati...

How would I scale down an experiment? For example a procedure states that 5.0 g will be made (53% yield) but we want to propose a microscale procedure that would synthesize .120 g. Do I simply divide desired amount of product by the one given in the proce...

I'm trying to find the antiderivative of (x+3)/2 How would you go about this? The answer is (x+3)^2/4

f(x) = (x+2)/(x+6) What is f^(-1)(-6)? (what is f inverse of -6) I found f' which was [(x+6)-(x+2)]/(x+6)^2 I set the function equal to -6 and found x to be -30/7. Used the formula to find inverse function: 1/f'(f^(-1)(-6)) So I replaced f^(-1)(-6) with -...

Can someone please help me find the arc length of this curve? x = 3y^(4/3) - 3/32y^(2/3) y is between -64 and 64

How would I solve for R? R^2 + (4400)^2 = (R + 1.5)^2

No, I wrote everything correctly. Yes I did that and then I did multiply both sides by the denominator. The answer I got from the quadratic equation was not correct, however, according to my online homework.

Wow, I knew it was easier than I thought. I tried splitting them up into x/2 + 3/2 and all kinds of things. Thanks so much.

I think I did wrong math and ended up doing more than I was supposed to. Thanks for the help, I get it now!

Oh wow...so I need to find it implicitly first??? How confusing. I was just trying to figure out the arc length using the first equation and I assumed it didn't matter that the x and y terms were switched Yes, the last term is (-3/32)y^(2/3), the fraction...

Thanks! I didn't even need to use the quadratic because when I added like terms, the square canceled out :)