Macrophage
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Actually, you are on the right track! Now let's complete it with only elementary calculus. $$\frac{a+b+\frac{b^2}{4a}}{b-a}=\frac{4a^2+4ab+b^2}{4a(b-a)}=\frac{(2a+b)^2}{4a(b-a)}$$ Knowing that $b>... View answer Accepted answer 4 votes First, I assume that$\alpha,\beta$in your problem are independent, so $$(P+Q)_{max}=P_{max}+Q_{max}$$ Let$\cos^2x=k$. Hence, $$P=k^2+(1-k)^2+1$$ and $$Q=k^3+(1-k)^3+1$$ Since we know$k\in[0,1]$we ... View answer Accepted answer 4 votes So basically it's because for equivalent infinitesimal expressions of a function, the limit of its ratio with the original function as x approach 0 can be proved to be 1. (sin x/x for example. Sorry, ... View answer 3 votes It seems that you should get to sum$32k+30$, not$8k+20$, and then simplify. View answer 3 votes I'll try to solve the first problem. Since$A$is singular, there exist$u\in \mathbb R^n$s.t.$uA^T=0$. Thus, $$u^TA1_n=0=u^T\cdot (s1_n)=s(u^T\cdot 1_n)=0$$($s$is the sum of elements in each row) ... View answer Accepted answer 3 votes Recognizing that$n$th coefficient in the Taylor series of$\sqrt{x+1}$(a binomial with power$\frac{1}{2}$) is given by$a_n=\frac{\prod_{n=0}^{\infty}{(\frac{1}{2}-n)}}{n!}$. You can use ratio test ... View answer Accepted answer 3 votes Using the lower and raise command should give you a desirable result like this:$\overset{\lower{0.5ex}{\cdot}}{\underset{\raise{1ex}{\cdot}}{\sim}}$. Hope you find this useful, thanks! View answer Accepted answer 3 votes Here is step-by-step hints to do it: For the composition of$f(g(x))$, focus on$g(x)$first, it has a range of$\mathbb R$and since$f(x)$has a domain of$\mathbb R$too, everything is well. ... View answer 2 votes Instead of stacking together two tildes, you could put a dot above and a dot below the tilde to indicate an approximate distribution(similar to a notation for approximate equal with a dot above and ... View answer Accepted answer 2 votes The functions corresponding to two curves are inverse functions. To see this, let$f(x)=e^{x^2}$and$g(x)=\sqrt{\ln x}$$$f(g(x))=e^{\ln x}=x \quad \Rightarrow\quad g(x)=f^{-1}(x)$$ Hence $$\int_{e}^{... View answer 2 votes Start with area formula for a circle:$$A=\pi r^2$$And you differentiate both sides with respect to time(now imagine area and radius are functions of time!),$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$In ... View answer 2 votes Since you have horizontal dashed lines drawn on the graphing area and the explicit form of the cubic function is not given, you just need to put k equal to the different options and check if that ... View answer 2 votes Put yourself in the complex plane and rewrite the original equation as z^4=10000, z\in \mathbb C. Expressing z as re^{i\theta} leads to r=10, \theta=\frac{k\pi}{2}, k\in \mathbb Z. Thus, you ... View answer Accepted answer 2 votes Ae^{-kt} with k=\frac{ln(2)}{\tau} is equivalent to Ae^{kt} with k=\frac{ln(0.5)}{\tau} So, you are doing nothing wrong! :) View answer Accepted answer 2 votes 1.The theorem is mainly used for establish the divergence of a series. (Contrapositive to the original). 2.Your statement is true. Actually, whenever the limit is not going to 0, you can conclude ... View answer Accepted answer 2 votes Using the equation for points on spheres: \qquad(x-a)^2+(y-b)^2+(z-c)^2=r^2 Using coordinates of the four points provided, we have four simultaneous equations to solve for a, b, c, d. \begin{... View answer 2 votes Both cases above are monotonic because the functions \sec^{-1} x and \tan^{-1} x are themselves monotonic. The sum of two monotonic functions is not necessarily monotonic. Try out f(x)=\sin x-x,... View answer Accepted answer 1 votes The left hand side of the first equation is a matrix multiplied by a number so you multiply each entry of the matrix. Hence, the systems of equations represented is$$\begin{cases} &w\cdot N_sE=-... View answer Accepted answer 1 votes I don't think that's the case for$0$because you don't usually write a number less than$100$by adding zero(s) to the front. However, as Mjiig pointed out, you can allow$0$if you want to, though$...

$$\iint_S\vec F\cdot \vec ndS=\iiint_V\nabla\cdot \vec FdV=3\iiint_V(x^2+y^2+z^2)dV$$ $$=3\iiint_Vr^2dV=3\int_0^14\pi r^4dr=\frac{12}{5}\pi$$

Multiply each number in your list by $(\max-\min)$ and then add min to it. :)

Hint: Use partial fraction decomposition. The gist is that you have to factor the denominator into linear terms$(ax+b)$ or irreducible quadratic terms$(ax^2+bx+c)$. Next, you expand the fraction into ...

A vector field $V$ on $\mathbb R^2$ maps a point p to tangent vectors $V($p$)$. It's not necessary that this tangent vector has greater magnitude with p further from the origin. Let $V:(x,y)\... View answer 1 votes $$\frac{1}{n}\geq\frac{1}{n!}>0, \forall n \in \mathbb R^+$$ and $$\lim_{x\rightarrow\infty}\frac{1}{n}=\lim_{x\rightarrow\infty}0=0$$ so $$\lim_{x\rightarrow\infty} \frac{1}{n!}=0$$ by squeeze ... View answer Accepted answer 1 votes By definition a power series$\Sigma a_n(x-x_0)^n$is centered at$x_0$. So in your problem it's indeed centered at -1 View answer 1 votes$0$is not an accumulation point for$\{\frac{1}{a_n}\} \iff \lim_{n\rightarrow\infty}\frac{1}{a_n}\neq0 \iff \lim_{n\rightarrow\infty}{a_n}\neq\infty$, which says the sequence is bounded. Then you ... View answer Accepted answer 1 votes Thanks for the helpful comments to the original problem. The number of possible strings can be described by the Calatan numbers:$1, 1, 2, 5, 14, 42, 132...$. The following is what I found on OEIS: ... View answer Accepted answer 1 votes For$a_n$, it's the maximum value of$f(x)$on interval$I=[0,1]$. Knowing that$f'(\frac{1}{n+1})=0$(the only critical point),$f(0)=0, f(1)=0$,$\ f''(\frac{1}{n+1})<0$, you can conclude$f(...
The minute hand passes $6°$ per minute and the hour hand pass $\frac{1}{2}°$ per minute. Since the clock hands run clockwise, and from 3:30 ($\theta_1=75°$), the minute hand is in front of and runs ...