Recent content by kLPantera

I typed the problem word for word, I'll include the multiple choice answers in this post: a) 3 b) 15/4 c)4 d)9 e)12

Homework Statement If theta increases at a constant rate of 3 radians per minute, at what rate is x increasing in units per measure at the instant when x equals 3 units? The Attempt at a Solution I drew the triangle and I came to the conclusion that I needed to use 5sin(theta)=x...

Homework Statement Need to find the derivative: y=logx/2-logx=logx*(2-log(x))^-1 The Attempt at a Solution The derivative of log(x) is 1/xln(10) and the derivative of (2-log(x))^-1 is -(2-log(x))^-2*1/xln(10)? This ended as 1/xln(10)(2-logx) - logx/xln(10)(2-log(x))^2...

So how did the numerator become: (2-cos(3x)-cos(4x)) = (1-cos(3x)) + (1-cos(4x))?

lim as x-> 0 (2-cos3x-cos4x)/(x). I'm not sure how the numerator became (1-cos3x)(1-cos4x)/(x) What am I missing? Could someone please point it out? Is it a trignometric factoring formula I'm not thinking of?

Homework Statement limi as x-> positive infinity x+r((x^2)+2x) The Attempt at a Solution multiply by conjugate x-r((x^2)+2x) I get (x^2)-(x^2)+2x/x-r((x^2)+2x) Which becomes 2x/x-r((x^2)+2x) Which I end up with 2/1-r(1-(2/x)) But I go wrong somewhere because I end up...

http://math.colorado.edu/~jkeller/math1300/lectures/L8limits3.pdf Example 8, the 2nd to 3rd step... where did the negative go? It's (x^2 - 3x) - x^2 then it became 3x in the numerator. Where'd the negative go? I don't see any other sign changes so could someone please tell me?

Ah I understand it now lol

For the first integral no I did not.

I'm confused as to when to change the limits on a definite integral. Ex. Integral with the limits a=1, b=5, 3/(x+1)dx I set u = x+1 and du = dx I used u-substitution and everything worked out fine. However for this one... Ex. Integral with the limits a = 0, b = 2...

There are only 2 graphs: f'(x) and g'(x). I understand the one for g(x) now however... For the f'(x) graph, f'(x) is always above the x-axis. That means the slope is always positive?

Homework Statement When given the graph of f '(x) the graph sort of oscillates above the x-axis and the graph of g '(x) which starts in Quadrant III sort of curves up, goes through point (0,0) keeps going up then begins to even out. Homework Equations How can you tell how many...

Thanks again!

First we fired at 0 degrees, which we then used to find Vo Vix and Viy were found after he gave us the angle of degree to fire it at. For Delta x we had a Delta x for when it was 0 degrees. We then were given an angle to fire at. We have to find the Delta x for when it will be fired at 12...

Fired at 0 degrees: Delta x: For 1 click: 1.28 meters For 2 clicks: 1.68 meters For 3 clicks: 2.28 meters Note* clicks means how much we compressed the spring within the cannon Delta y = 0.87 meters Each of the three rings are set at equal intervals of 1/4 of delta x.