Recent content by davemoosehead

If the first 6 are not selected, then you would have (n+r+1-6 C n+1) ? I'm having a hard time following what you are saying.

If we lined up all the n+r+1 things, the first one of {n+1 \choose n} would be the one after the last one of {n+1 \choose 1}

I've been thinking about this all afternoon, and still cannot come up with anything. It's given me a new way of looking at it though which makes it look less intimidating. So you're saying think of it in terms of what we're not choosing (n+1), rather than what we are choosing (r). So.. {n...

Homework Statement {n \choose 0} + {n+1 \choose 1} + {n+2 \choose 2}+...+{n+r \choose r} = {n+r+1 \choose r} We have to prove by counting both sides in a different way. For example, consider {n \choose 0}^2 + {n \choose 1}^2+...+{n \choose n}^2 = {2n \choose n} The RHS can be described as a...

Homework Statement Which of the following is not an eigenvector for T \left( \left[ {\begin{array}{cc} x \\ y \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} x + y \\ x+ y \\ \end{array} } \right] ? A) v = [-2 -2]T B) v = [1 -1]T C) v = [1 2]T D)...

Homework Statement Our class exercise in World Prehistory was to figure out the volume of the building we were in and answer the following: How many a) basketloads, b) mud bricks, c) stone blocks would it take to build structure? How many worker-hours would it take? How many...

so, pi/12 + sqrt(3)/2 is a local max and 5pi/12 - sqrt(3)/2 is a local min? f(0) = 1 is a global min and f(2pi) = pi + 1 is a global max?

critical numbers: pi/6, 5pi/6 I drew out this table: Interval | Test Value | Sign | Behavior (0, pi/6) | pi/10 | + | Inc (pi/6, 5pi/6)| pi/2 | - | Dec (5pi/6, 2pi) | pi | + | Inc f(pi/6) = pi/12 + cos pi/6 = pi/12 + 6sqrt(3)/12 = pi + 6sqrt(3)...

Homework Statement Classify the local extrema of f(x) = x/2 + cos(x), 0 <= x <= 2pi. Give the exact values of the critical numbers and extrema. Find the absolute (or global) maximum and minimum values of the function. The Attempt at a Solution So i need to find the critical numbers...

ok...hows this look? ln(L) = lim(x->0) { ln(1-10x)/x } = 0/0 so.. ln(L) = lim(x->0) { (-10)/(1-10x) } ln(L) = -10 L = e^-10

Homework Statement lim (1-10x)^(1/x) x->0 evaluate the limit Homework Equations L'hostpital's ruleThe Attempt at a Solution take derivative: lim (-10+100x)/x x->0 can't divide by zero so take the derivative again but x goes away: lim 100 x->0 is 100 the limit? is there a limit? now that...

Thanks for your help For 3) I ended up getting lim x-> 0 (sin x /x)+(sin x /x) / cos 2x = 2

2) f'(t) = sec²(sin t²)(cos t²)(2t)

1) Since y is a function of x -xy'sin(y) + cos(y) - ysin(x) + y'cos(x) = 0 y'(-xsin(y)) + y'cos (x) = ysin(x) - cos(y) y' = (ysin (x) - cos (y)) / (-xsin(y) + cos(x))

1) Working... 2) How far off am I? All you do is apply the chain rule right? 3) Is there a way to solve this without l'hospital's rule? I don't think we've learned it yet...