Recent content by Bernie Hunt

Excellent idea. I used a framing square from the shop and got 23.25 feet. Thanks! Bernie

This may be off topic, but a lot of smart and practical people hang out here so maybe someone can help. I need to find the height of my chimny in my back yard. I'm not wild about heights, so I'd rather not get out a long tape measure and a longer ladder to climb to the top. All the word...

Thanks for your reply Hurkyl. I haven't had DE yet, so I can't really comment on your reply. Bernie

The Problem; Given H = U + PV and dU = TdS - PdV Find dH in terms of T, S, P, V My Solution; H = U + PV dH = dU + PdV + VdP dH = (TdS - PdV) + PdV + VdP dH = Tds + VdP My Question Am I missing a step between the first and second steps? I'm taking the derivative of both sides...

Thanks Matt! I got it figured out. Along the way I also figured out there are many wrong definitions on the net for perfect squares, hahahaha. Bernie

Matt, Thanks for the reply. One bit of clarification; a and b are integers and a^2 = b Is a the perfect square? So if I was asked to find the perfect square of b, then the answer would be a? Sorry for all the questions, but I'm struggling with the english syntax of this problem. I...

Homework Statement X mod m is the remainder when x is divided by m. This value is called a residue. Find all perfect squares from the set of residues mod 16. The Attempt at a Solution There was a suggestion that this would become clearer when the definition of perfect square was...

The proof is definitely <=, but I think I found the justification. I don't have my notes handy, but I think it's disjunctive addtion justifies that if p then (p or q). Also if (p or q) and (~q) then p. Bernie

OK, here we go; Prove: If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ). Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0. (a+b+c)^2...

StatusX, Cancel out which like terms? Expanding out the square gives; ( a + b + c )^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca Which of these terms can be canceled? Bernie

ACM, Thanks for the quick reply. I think there is a flaw in your logic. The statements workout to; 2(a + b + c)^2 > (a + b + c)^2 > 2(ab + bc + ac) 2(a + b + c)^2 > 4(ab + bc + ac) > 2(ab + bc + ac) Therefore 4(ab + bc + ac) > (a + b + c)^2 Which is saying that: x > y x > z...

Wouldn't this solve easier with u substituion? u = 1+y^2 du = 2y dy Solution is 1/3(1+y^2)^3/2 If you want integration by parts, multiply it out and make u to be y so that du is 1. By trig you'll need to substitute tan because the problem is + Y^2. Bernie

The Problem; If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ). Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0. I haven’t make...

Argh, Thud, thud, thud ... (The sound of beating my head on the desk again.( That's the second time I made a units mistake last night. My montra for today will be "Check the units!" Bernie

Argh! Thud, thud, thud ... That's the sound of me banging my head against the wall! Thanks for the quick reply. Bernie