Recent content by Schaus

1) No solution 2) Infinitely many solutions 3) One unique solution I got the a=15 by making 2a-16 = (4/3a-6)

2a-16 = ((4/3)a-6) a=15 No solution = Never Unique solution ---- a≠15 Infinitely many solutions--- a=15?

So I should have multiplied by -(a-6)? -(a-6)R2+R3 1 0 -2 ----------- 5/3 0 1 -2 ------------ 4/3 0 0 (2a-16) -------- ((4/3)a-6)

So I take (a+6)R2-R3? 1 0 -2 ----------- 5/3 0 1 -2 ------------ 4/3 0 0 (-2a-16) -------- ((4/3)a+10) Sorry for the slow reply, I've been very busy with school.

Homework Statement Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', 'a = ', or 'a ≠', then specify a value or comma-separated list of values. x1−2x2+2x3 = −1...

The answer in my course is 2x-1 but it could be wrong, it is easily the worst course I've done. Thank you for all your help though.

Now I got my answer I'm just curious where the highlighted 2 came from.

I'm still not understanding how to get the answer. If ##\frac {1}{6} \ln(2)-0 = C## then C ≈ 0.1155... or at least that is what my calculator says.

Sorry, I'm still a little confused. After integrating should it look like this then? ##\frac {1}{6} \int \frac {1}{u}## ##= \frac {1}{6} \ln |1+y^6| = \ln|x| + C##

Oh I see. Would the integral look like this then ##\ln |6(1+y^6)|##? Or do I put the ##\frac{1}{6}## on the outside of the natural log?

Yes I definitely messed up on the ##\ln |1| = 0## but I don't understand where this ##\frac {1}{6}## is coming from.

Homework Statement Solve for y: ##\frac {dy}{dx} = \frac {1+y^6}{xy^5}## , where y(1) = 1. Answer ## y = \sqrt[6] {2x-1}## Homework EquationsThe Attempt at a Solution ##\frac {dy}{dx} = \frac {1+y^6}{xy^5}## ##\frac{dy (y^5)}{1+y^6} = dx \frac {1}{x}## u= 1+y6 ##\frac {du}{y^5}=dx## ##\int...

That's weird. When I do 3/2pi it gives me 4.71 but if I do 3/6.28 I get my answer. Thanks for the help, I wish I had thought of trying that earlier!

Homework Statement A stone is dropped into some water and a circle of radius r is formed and slowly expands. The perimeter of the circle is increasing at 3 m/s. At the moment the radius is exactly 2m, what rate is the radius of the circle increasing? Answer ## \frac {dr} {dt} = 0.48 m/s##...

I swear I put that in my calculator a few times but this last time it worked... Thanks a lot anyways :)