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...
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##
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##...
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...
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.
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...
##\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##?
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...
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...
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...
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?
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.