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Aug
10
comment Getting the quadratic function given the vertex and one point.
Excuse me, I'm not very familiar with the concept of derivatives yet. Of course, if I ignore the concept and solve the equations I will get the expected answer, but I'm not sure what you meant by $f'(x) = 2ax+b$
Jul
13
comment Finding the acute angle between the planes $x-3y+2z=14$ and $-x+y+z=10$
What do you think about @Timbuc's comment? About $180 - 107$?
Jul
12
comment Finding the intersection point between two lines using a matrix
Even if I only need two rows, if a third one happens to have an inconsistency, will that cause the equation to have no solution?
Oct
25
comment Does the line $(2,1,1)+t(-3,1,5)$ live within the plane $31x+3y+18z=62$?
@Galc127: The formula you posted, is that for calculating the orthogonal projection? In that case, don't I have to multiply the result by $(31,3,18)$, and then calculate the magnitude of the result?
Oct
24
comment Why isn't the orthogonal vector to a direction vector of a plane not necessarily perpendicular to such plane?
Would the same apply parallel vectors? I mean, if I find a vector that is parallel to that line's direction vector, would it also be parallel to the plane?
Oct
1
comment Why $(-1 \cdot h) = -1$ when $h$ approaches $0$?
By the way, if $h \not = 0$, how come the denominator is $(x-1)^2$? I had said that "$h$ is practically $0$" but I'm not that convinced anymore.
Sep
20
comment Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
@AndréNicolas I see now. Thanks. By the way, you picked $\frac{1}{4}$ because it made arithmetic easier - but how can I determine the "largest acceptable value" that can replace $\frac{1}{4}$? (out of curiosity)
Sep
20
comment Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
@EWHLee I see. I wonder if $\frac{1}{4x-9}$ should have been $\frac{1}{|4x-9|}$ instead... Wait, no, that doesn't work either. Goddammit. What should I have done instead?
Sep
20
comment Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
@AndréNicolas Ahhhh yes, that's right. By the way, isn't $|(x-3)| < \delta < \frac{9}{4}$ enough to keep $|4x-9|$ away from $0$ since $\frac{9}{4}$ is the solution to $4x-9$?
Sep
19
comment Proving $\lim_{x\to1}(x^2+3)=4$
Hm I'm not very sure how can $|x-1| < \delta$ lead to $|x+1|\le 3$. If you increase $|x-1|$ by $2$ to reach $|x+1|$, shouldn't it be $|x+1| \le 2$ rather than $3$?
Sep
19
comment Why does $|x-1|^2+3|x-1| < \epsilon \implies |x-1|^2 < \frac{\epsilon}{2} \ \ \land \ \ 3|x-1| < \frac{\epsilon}{2}$?
@Oleg567: What if $a = b = \epsilon/2$?
Sep
7
comment For what values of $p$ and $b$ is the vector $(b,8,b+7)$ a solution of this system?
Thanks, that's right (although $b$ yields $-5$).
Sep
7
comment For what values of $p$ and $b$ is the vector $(b,8,b+7)$ a solution of this system?
@Adriano: Huh, it seems it should be $b = -5$. The exercise's answer seems to be wrong haha.
Sep
1
comment Is $A$ invertible if $ABAB^2 = I$?
Only of the same size?
Aug
21
comment Factoring $x^3-8$ by grouping
@Winther that works - but how did you figure that? Just made up?
Aug
21
comment Factoring $x^3-8$ by grouping
Sorry, how did you realize that $x^3 - 8 = (x^3 + 2x^2 + 4x) + (-2x^2 - 4x - 8)$?
Jul
5
comment Calculating an orthonormal base given another base.
Aha, this worked! I'll keep on testing it. Do you know why my particular approach didn't work?
Jul
5
comment Calculating an orthonormal base given another base.
By "second vector" you mean $(a,b,c)$?
Jul
3
comment Finding the smallest $x$ given a set of congruence conditions.
Did you mean $123$ instead of $133$?
Jun
22
comment How can a subspace have a lower dimension than its parent space?
Yes, that's exactly it! Thank you.