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Dec
17
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?
Dec
17
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 $c$
Dec
17
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
Dec
13
comment What is $\sum_{n=1}^{\infty} \frac{n}{2^{\sqrt{n}}}$?
well maybe forget L'Hôpital
Dec
13
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 $51.919191....$
Dec
6
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
Sep
15
comment $\displaystyle \frac{d}{dx}2^x$ where $x=0$
@Shahar Yes {need more characters}
Sep
15
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
Sep
14
comment Prove rigorously: $\displaystyle \lim_{x\rightarrow 2}x^2+5x=14$
So it's like squaring both sides? (But multiplying by equivalents)
Sep
13
comment Prove rigorously: $\displaystyle \lim_{x\rightarrow 2}x^2+5x=14$
How did you find $|x-2||x+7|<(9+ \delta)\delta $?
Sep
12
comment Prove rigorously: $\displaystyle \lim_{x\rightarrow 2}x^2+5x=14$
$\delta>|x-2|$?
Sep
11
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
Sep
11
comment Derivative at 4, when $f(x)=\frac{1}{\sqrt{2x+1}}$
yes there is an error. will fix in a moment
Sep
7
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.
Aug
24
comment verify $\lim_{x\rightarrow0}(4x^2+2x+5)=5$
ok I think $|x||4x+2|$ is the answer my professor is looking for
Aug
23
comment verify $\lim_{x\rightarrow0}(4x^2+2x+5)=5$
We haven't gotten to delta-epsilon proofs yet
Aug
23
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
Aug
23
comment verify $\lim_{x\rightarrow0}(4x^2+2x+5)=5$
That wouldn't be a "proof"
Aug
19
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
Aug
18
comment Suppose that $|x-4|\leq1 $
Did you see my edit? I think I did it right. Oh the absolute bars