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1. ### Linear Algebra - REF with another variable

1) No solution 2) Infinitely many solutions 3) One unique solution I got the a=15 by making 2a-16 = (4/3a-6)
2. ### Linear Algebra - REF with another variable

2a-16 = ((4/3)a-6) a=15 No solution = Never Unique solution ---- a≠15 Infinitely many solutions--- a=15?
3. ### Linear Algebra - REF with another variable

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)
4. ### Linear Algebra - REF with another variable

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.
5. ### Linear Algebra - REF with another variable

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...
6. ### Modeling epidemics - solving differential equation

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.
7. ### Modeling epidemics - solving differential equation

Now I got my answer I'm just curious where the highlighted 2 came from.
8. ### Modeling epidemics - solving differential equation

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.
9. ### Modeling epidemics - solving differential equation

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##
10. ### Modeling epidemics - solving differential equation

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?
11. ### Modeling epidemics - solving differential equation

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

Homework Statement Solve for y: ##\frac {dy}{dx} = \frac {1+y^6}{xy^5}## , where y(1) = 1. Answer ## y = \sqrt {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...
13. ### Related Rates stone drop

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!
14. ### Related Rates stone drop

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##...
15. ### Find Point c that satisfies the Mean Value Theorem

I swear I put that in my calculator a few times but this last time it worked... Thanks a lot anyways :)
16. ### Find Point c that satisfies the Mean Value Theorem

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...
17. ### Find the equation of the tangent line of the curve

Alright, well I just got both. Can't hurt to have more.
18. ### Find the equation of the tangent line of the curve

So I got the wrong one? Crap
19. ### Find the equation of the tangent line of the curve

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...
20. ### Find the equation of the tangent line of the curve

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.
21. ### Find the equation of the tangent line of the curve

Yes, I've been trying to master the basics. I'm hoping that khan academy can help with that.
22. ### Find the equation of the tangent line of the curve

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...
23. ### Find the equation of the tangent line of the curve

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.
24. ### Find the equation of the tangent line of the curve

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?
25. ### Find the equation of the tangent line of the curve

I think I'll just have to watch more videos on implicit differentiation because I'm lost.
26. ### Find the equation of the tangent line of the curve

##\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...
27. ### Find the equation of the tangent line of the curve

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}}## ?
28. ### Find the equation of the tangent line of the curve

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##?
29. ### Find the equation of the tangent line of the curve

##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##?
30. ### Find the equation of the tangent line of the curve

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 ##\...
31. ### Finding discontinuities in functions

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...
32. ### Finding discontinuities in functions

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

No it can not...
34. ### Finding discontinuities in functions

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

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.
36. ### Finding discontinuities in functions

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...
37. ### Proving Reciprocal Identities

Ok, I'll try it. I'm going to have to practice these quite a bit more I think.
38. ### Proving Reciprocal Identities

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

2cosx-2cos3x = 2sin2xcosx 2cosx(1-cos2x) = 2sin2xcosx 2cosx(sin2x) = 2sin2xcosx 2cosxsin2x = 2sin2xcosx Does this work?
40. ### Proving Reciprocal Identities

Woops. I think I should have gotten 2sin2xcosx 2cosx-2cos3x = 2sin2xcosx Does this look right?
41. ### Proving Reciprocal Identities

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...
42. ### Proving Reciprocal Identities

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...
43. ### Solve Binomial Theorem Homework: Find Coefficients of Degree 17 & x7

Awesome thanks for the help. Now I just hope I can reproduce this result with another question!
44. ### Solve Binomial Theorem Homework: Find Coefficients of Degree 17 & x7

I found that t10=(13C9)(x2)13-9)(-x)9 t10=(715)(x2)4)(-x)9 =(715)(x8)(-x)9 =-715x17?
45. ### Solve Binomial Theorem Homework: Find Coefficients of Degree 17 & x7

*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.
46. ### Solve Binomial Theorem Homework: Find Coefficients of Degree 17 & x7

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?
47. ### Binomial Theorem - Determine n

Whoops! Sorry I meant to put n = 13!
48. ### Binomial Theorem - Determine n

So all I need is the x8 and xn-5? Then bases are the same so - n-5 = 8 ----> n=3?
49. ### Binomial Theorem - Determine n

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...
50. ### Solve Binomial Theorem Homework: Find Coefficients of Degree 17 & x7

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.