Joebevo
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 Apr12 comment Differential equation by series solution method: equating coefficients to zero I think I got it. n=1,2,3.. and upwards already invoke n=0, which is set to 0. In other words, $2a_2$ is the coefficient of $x^0$, and we have to set coefficients of all powers of $x$ equal to zero. Mar31 comment How do you integrate $\int x(x+3)^ \left( 1/3 \right) dx$ The answer that I got, which is shown as correct is : $\frac{3}{7}(x+3)^{\frac{7}{3}}-\frac{9}{4} (x+3)^{\frac{4}{3}}$ Mar31 comment How do you integrate $\int x(x+3)^ \left( 1/3 \right) dx$ I apologize for not being clear. I did substitute the $x$ with $u-3$, but the form of the equation seems similar. It didn't occur to me that the new form can be distributed, while the old one cannot. Feb2 comment What is the intuitive meaning of the basis of a vector space and the span? Is there an upper bound on the number of matrices that can be used as a basis for $\mathbb R ^2$, like the one you have shown above? Jan22 comment Given $a=bc$ and $c\geq 1$ and $b\neq 0$, which is greater: $a$ or $(b+c)$? $b$ is nonzero. Sep10 comment How do I calculate the quartiles for this problem? There is some confusion between the answers by EpicGuy and Sim. My source agrees with the former, while the quartiles calculator at alcula.com/calculators/statistics/quartiles agrees with the latter. I would appreciate it if someone could resolve this. Sep10 comment How do I calculate the quartiles for this problem? BTW, I am the OP. I was finally able to log into my own account. Sep10 comment How do I calculate the quartiles for this problem? Your answer doesn't agree with the other answer posted by EpicGuy. It also doesn't agree with my original source from where I took the problem. I like your method, but the other answer is the correct one. I'd like to know how to get the correct answer using this method. Could you please double-check this? Thanks. Aug10 comment Intersection point of 2 circles Sorry, what I meant was, you substitute out whatever's possible using the 2nd equation. So, for example, you could substitute using $x^2+y^2=100$ in the remaining equation, and the $x^2$ and $y^2$ terms disappear. Jul1 comment Regression towards the mean v/s the Gambler's fallacy @GerryMyerson: Interesting. That's how Bayesian reasoning works, doesn't it? Jul1 comment Regression towards the mean v/s the Gambler's fallacy "The only regression is that the coin is likely not to give such weird results in the next bunch of tosses."...Ah. Got it! Apr25 comment Should the sign be reversed if I square both sides of an inequality? Similarly, I'd like to know how to square \$x