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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 Equations The 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 :)

Homework Statement Find the point "c" that satisfies the Mean Value Theorem For Derivatives for the function ## f(x) = \frac {x-1} {x+1}## on the interval [4,5]. Answer - c = 4.48 Homework Equations ##x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a}## ##f'(c) = \frac { f(b) - f(a)} {b-a}## The Attempt...

Alright, well I just got both. Can't hurt to have more.

So I got the wrong one? Crap

I'm doing an online course called "Studyforge" or at least created by them. The course lacks detailed solutions for many of their problems which is hard for me because I learn more easily from seeing something done. I ordered...

I did not know of any books available. The course I'm doing is not very good. Thanks for the info I'm more than willing to purchase a book if it'll help.

Yes, I've been trying to master the basics. I'm hoping that khan academy can help with that.

Ohhh ok so it should look something like this? ##\frac {d}{dx}(xy^2+\frac 2 y )= \frac d {dx}(4)## ##\frac {d}{dx}(xy^2)+\frac d{dx}\frac 2 y=0## ##(x\frac d{dx}⋅2xy)+(1⋅y^2)+(\frac d{dx}\frac 2 y )= 0## ## 2xy⋅\frac {dy}{dx}+y^2-\frac 2 {y^2}⋅\frac {dy}{dx}=0## ## \frac {dy}{dx}(2xy- \frac 2...

Ok so the chain rule is ##f'((g(x))⋅g'(x)##, correct? I'm a little confused as to what I am subbing in where.

Ok, let me see if I've done this properly. The product rule - ##(x\frac{dy}{dx}⋅2xy)+(1⋅y^2)##. Now for the ##\frac 2 y ## do I use the quotient rule?

I think I'll just have to watch more videos on implicit differentiation because I'm lost.

##\frac {d} {dx} x⋅y^2+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 {y^2}= 0## then I would get ##-\frac {y^2} {2xy-\frac 2 {y^2}}## I thought that I did the product rule. f(x) = x f'(x) = 1 g(x) = y2 g'(x) = 2y I just did the product rule as f'g⋅g'f. After the product rule I add the ##\frac 2...

Ohhhh, I should have done this? ##\frac {d} {dx} (1⋅y^2)+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 y^2= 0## then I would get ##-\frac {y^2} {2xy-\frac 2 {y^2}}## ?

So ##\frac d {dx}(xy^2)+\frac d {dx}(\frac 2 y) = 0## should turn into ##x\frac {d} {dx} (y^2)+ x⋅\frac d {dx} y^2- \frac 2 {y^2}(\frac {dy} {dx}) = 0## because I don't see the derivative of ##\frac 2 y##?

##2xy+x⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 0## is this supposed to be ##x +2xy⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 0##?

Homework Statement Find the equation of the tangent line to the curve ##\ xy^2 + \frac 2 y = 4## at the point (2,1). Answer says ##\ y-1 = -\frac 1 2(x-2)## And with implicit differentiation I should have gotten ##\frac {dy} {dx}= -\frac {y^2} {2xy-\frac {2} {y^2}}## Homework Equations ##\...

The only other question I have to compare to is f(x) = (x2-x)/(x2-1) which I've factored and eliminated discontinuity. (x(x-1))/(x-1)(x+1). I take out the (x-1) so x ≠ 1 and I am left with x/(x+1). The question above is the first I've encountered with a radical and it has thrown me off...

Do I simply state that it has a discontinuity at x = -2 but isn't removable then because otherwise I'm at a loss.

No it can not...

So before I do anything I should state the restrictions on the denominator? Which are x ≠ 0 and x ≡ -2, I believe.

Sorry was meant to be the second one. By the way how do you write questions in that form? Would be much simpler for people to understand what I'm trying to ask if I could display them properly.

Homework Statement Where are the following functions discontinuous? f(x) = (x+2)/√((x+2)x) Homework Equations The Attempt at a Solution f(x) = (x+2)/√((x+2)x) = (x+2)/x√(2x) multiply both denominator and numerator by √(2x) = (x√2+2√x)/(x(2x)) Can I leave it like this and state that x ≠ 0...

Ok, I'll try it. I'm going to have to practice these quite a bit more I think.

So I should start with 2sin2xcosx and work backwards? Thank you for all the help by the way.

2cosx-2cos3x = 2sin2xcosx 2cosx(1-cos2x) = 2sin2xcosx 2cosx(sin2x) = 2sin2xcosx 2cosxsin2x = 2sin2xcosx Does this work?

Woops. I think I should have gotten 2sin2xcosx 2cosx-2cos3x = 2sin2xcosx Does this look right?

I thought I could cancel a cosx, maybe I cannot. I tried to eliminate the denominator like you said. Here's what I got. (secx+1)(2cosx-2cos2x) = (Sin2x)(tanx) ((1+cosx)/cosx)(2cosx-2cos2x)=(2sinxcosx)(sinx/cosx) Expanding ((2cosx-2cos2x+2cos2x-2cos3x)/cosx) = 2 Moving the cosx under left side...

Homework Statement (secx+1)/(sin2x) = (tanx)/2cosx-2cos2x) Homework Equations The Attempt at a Solution Left Side ((1+cosx)/cosx)/2sinxcosx ((1+cosx)/cosx) x (1/2sinxcosx) cancel the a cosx from both to get (1/2sinxcosx) This is all I could manage with left side so I tried right side Right...

Awesome thanks for the help. Now I just hope I can reproduce this result with another question!

I found that t10=(13C9)(x2)13-9)(-x)9 t10=(715)(x2)4)(-x)9 =(715)(x8)(-x)9 =-715x17?

*sigh* I figured as much. Something just isn't clicking for me on this and I just don't understand what I'm doing wrong.

tk+1=13C4(x2)13-4(-1)4 t5=13C4(x2)9(-1)4 t5=(715)(x18)(-1)4 =715x18 This is what I've come up with. Am I supposed to factor out the negative first before putting it all together?

Whoops! Sorry I meant to put n = 13!

So all I need is the x8 and xn-5? Then bases are the same so - n-5 = 8 ----> n=3?

Homework Statement The sixth term of the expansion of (x-1/5)n is -1287/(3125)x8. Determine n. Homework Equations tk+1=nCkan-kbk The Attempt at a Solution tk+1=nCkan-kbk t5+1=nC5(x)n-5(-1/5)5 This is where I'm stuck. Do I sub in -1287/(3125)x8 to = t6? If so what do I do from here...

tk+1=10C3(2x)10-3(3)3. This might be what you were talking about for #2? When I did it like this I got the needed answer but I'm still at a loss for #1.