# Zack

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 Dec17 comment Derive a formula for the volume of the wedge in terms of the constants a, b, c. I'm having trouble connecting it to the 3rd dimension. Another similar triangle? Dec17 comment Derive a formula for the volume of the wedge in terms of the constants a, b, c. It is the vertical height from the $a$ line, parallel to $c$ Dec17 comment Derive a formula for the volume of the wedge in terms of the constants a, b, c. Yes I believe so. I see the way I'm doing it now is not working Dec13 comment What is $\sum_{n=1}^{\infty} \frac{n}{2^{\sqrt{n}}}$? well maybe forget L'Hôpital Dec13 comment What is $\sum_{n=1}^{\infty} \frac{n}{2^{\sqrt{n}}}$? Could L'Hôpital's rule be applied? It is infinity over infinity. ALso, according to wolfram alpha "lim x->infinity sum (x/(2^sqrt(x)))" returns $51.919191....$ Dec6 comment The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ Oops. It was for an applied optimization problem. It didn't explicitly say "maximum," but I didn't want to include the extra info, just the derivative. I will edit Sep15 comment $\displaystyle \frac{d}{dx}2^x$ where $x=0$ @Shahar Yes {need more characters} Sep15 comment $\displaystyle \frac{d}{dx}2^x$ where $x=0$ @user164587 Haha, I hadn't ever noticed that. I'm in Calc 1 and we still use log as base 10 if not specified. And most, if not all calculators, naturally use base 10 Sep14 comment Prove rigorously: $\displaystyle \lim_{x\rightarrow 2}x^2+5x=14$ So it's like squaring both sides? (But multiplying by equivalents) Sep13 comment Prove rigorously: $\displaystyle \lim_{x\rightarrow 2}x^2+5x=14$ How did you find $|x-2||x+7|<(9+ \delta)\delta$? Sep12 comment Prove rigorously: $\displaystyle \lim_{x\rightarrow 2}x^2+5x=14$ $\delta>|x-2|$? Sep11 comment Derivative at 4, when $f(x)=\frac{1}{\sqrt{2x+1}}$ There was an error in my question (wrong function). This is relevant. Sorry for the confusion Sep11 comment Derivative at 4, when $f(x)=\frac{1}{\sqrt{2x+1}}$ yes there is an error. will fix in a moment Sep7 comment Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$ I know this an old question, but if you had simply changed your √5/5 to a √5/√5, you would've got 3√5÷5/5 and gotten your answer. The whole point of multiplying a complex faction by a number to simplify it is to times it by 1 (√5/√5 in this case) because multiplying anything by 1 is the same thing. Aug24 comment verify $\lim_{x\rightarrow0}(4x^2+2x+5)=5$ ok I think $|x||4x+2|$ is the answer my professor is looking for Aug23 comment verify $\lim_{x\rightarrow0}(4x^2+2x+5)=5$ We haven't gotten to delta-epsilon proofs yet Aug23 comment verify $\lim_{x\rightarrow0}(4x^2+2x+5)=5$ Actually that is in the next section of our book, on limit laws. We are not supposed to use those. But I see that that is an acceptable answer otherwise Aug23 comment verify $\lim_{x\rightarrow0}(4x^2+2x+5)=5$ That wouldn't be a "proof" Aug19 comment Suppose that $|x-4|\leq1$ Ah I hadn't thought of solving it that way. I'm sure that will save me some time in the future Aug18 comment Suppose that $|x-4|\leq1$ Did you see my edit? I think I did it right. Oh the absolute bars