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I see it now. Thank you.

Homework Statement Does the series from n=1 to infinity of (2)/(n^2-1) converge or diverge? If it converges, find the sum. Homework Equations The Attempt at a Solution I can see right away that the series converges by a limit comparison test by looking at the series. However, to find the...

Thank you, I will try that.

Homework Statement Does the series ∑[n=1,∞) sin4n / 4^n converge or diverge? [h2]Homework Equations[/h2] Ratio Test lim n->∞ | a_n+1 / a_n | The Attempt at a Solution By Ratio Test. Let a_n = sin(4n) / 4^n So, lim n->∞ | (sin (4n+1) / 4^n+1) / (sin 4n / 4^n) | Skipping a few steps...

Alright, thank you for your help.

So I would have to treat it as a Harmonic Series?

Homework Statement Is the series from n=1 to infinity of 3/n converging or diverging? Homework Equations The Attempt at a Solution Since 3/n is not a geometric series, my guess is that we can just use the Test for Divergence and take it's limit to see if it's converging or diverging. As...

Homework Statement Solve ln(2x+1)=2-ln(x) for x. Homework Equations The Attempt at a Solution e^(ln(2x+1)) = e^((2-ln(x)) 2x + 1 = e^(2/x) 2x^2 + x = e^2 2x^2+x-e^2 = 0 At this point, I know you're supposed to use the quadratic equation. But, my problem is with how...

It's meant to be a Calculus problem, or at least it's on my Calculus homework. Alright, I went back and re-did this problem and got (3/2)r^2(sqrt(3)). I solved for the area of one equilateral triangle using r as one of the sides, multiplied it by six, and then reduced it to what is above.

I'm sorry, I should have said "inscribed". The picture looks like this - http://pic.blog.plover.com/math/pi-2/inscribed-hexagon.gif

Homework Statement A circle of radius r is impressed in a hexagon. Find the area of the hexagon. Homework Equations Area of a triangle = (1/2)bh The Attempt at a Solution The hexagon can be split up into six triangles, and with the formula for the area of a triangle, becomes...

Would I factor the bottom into (1+x^2)(1+x^2)?

1. Find the local max and min of x/(1+x^2) The Attempt at a Solution Take the derivative of x/(1+x^2) and set it equal to zero. f'(x)= (1-x^2)/(1+x^2)^2 = 0 This is where I am stuck. My first guess was to factor the top, resulting in (1-x)(1+x)/(1+x^2)^2 = 0, and then cancel the 1+x...

Haha, I really don't know where to go with it. Substituting the coordinates of the other point on the line and setting it equal to (dy/dx) is my first guess. 0-y1/3-x1=-(2x/14y). But even then I don't know where to go with that. Is this in anyway similar to solving for two tangent lines of a...

Shouldn't it be the (dx/dy) in the slope, not (dy/dx)? But my guess for that would be plug in the points into the equation and you get 0=3(6/0)+b, which, you can't do..If you did it (dx/dy), you'd end up with b=0?

I'm pretty sure the point isn't on the ellipse. I'm sorry I guess I should have made that more clear.

When you solve for dx/dy, you get 2x(dx/dy)+14y=0, 2x(dx/dy)=-14y, (dx/dy)=-14y/2x, (dx/dy)=-14(0)/2(3), (dx/dy)=0/6, (dx/dy)=0. So you end with the tangent line being y=0. There's still supposed to be at least one more line. This, again, is what is confusing me. How would I go about finding...

x^2+7y^2=8 (3,0) The derivative is 2x + 14y(dy/dx)=0. To get dy/dx by itself, first subtract 2x on both sides and you get 14y(dy/dx) = -2x. Then divide by 14y on both sides and you get (dy/dx)=-2x/14y. Plugging in the points (3,0) into the equation gives you 6/0, which is undefined. This is...

Find the tangent lines to the ellipse x^2 + 7y^2 = 8 at the point (3,0) Slope-intercept form: y=mx+b I know you have to differentiate the equation implicitly to get the slope, but you come across a zero in the denominator and that has me stumped.

From studying to the thinking process when solving problems. Thanks for any help and advice.