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3

This equation has no solution. Change sides to see that $$\frac{z}{z-5}+(1/3)+\frac{5}{5-z}=0 \\ Or,\frac{z}{z-5}+(1/3)+\frac{-5}{z-5}=0 \\ Or, \frac{z-5}{z-5}+(1/3)=0$$ Which is impossible.


3

David Mitra has answered the question in a comment (which I upvoted), but it might be useful for you if I indicate how one might approach the problem, rather than just "seeing" the answer. I'd think about how one might prove that $\pi_1(S)$ is closed; the proof will fail, but how it fails will give a hint about a counterexample. So suppose I try to prove ...


2

I always try to complete the squares: $$6xy+8y^2 −12x−26y+11=0$$ $$\left[ 8y^2+2.2 \sqrt{2}y\left(\frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}}\right)\right]-12x+11=0 $$ $$\left[ 8y^2+2.2 ...


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I'm not sure if I grasp exactly what you are asking, but I hope the following information can help. Your initial assumption that the ration 2:3 is equal to $2/3$ is false. In fact, the ratio 2:3 means that, out of a total of $2+3=5$ objects, that there are 2 of one and 3 of another. Thus, given a ratio, you add the two numbers to get the total value, and ...


2

Let $A$ and $X(t)$ be the two matrices $$ A = \begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 &-\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix} \quad\text{ and }\quad X(t) = \begin{bmatrix} x_1 (t) \\ x_2 (t) \\ x_3 (t) \end{bmatrix}$$ and $\omega = \sqrt{\omega_1^2 + \omega_2^2 + \omega_3^2} \ge 0$. When $A$ is ...


2

Factor the $dx$ to the right-hand side and you can then integrate both sides. $$ \tan y\, dy = \frac{1}{x}\,dx$$ However, the $\int \tan y\, dy$ isn't something that I remember off the top of my head, though I would recommend $u$-substitution in the form of: $$\int \tan y\, dy = \int \frac{\sin y}{\cos y}dy$$ LET: $u=\cos y, du = -\sin y\, dy$ $$\int ...


1

Have you tried to map the semiplane to the disk? Actually, the function $$z\mapsto\frac{z-i}{z+i}$$ sends the real line precisely to the unit disk. If $Im\ z>0$, then one sees that $|z-i|<|z+i|$, and the reverse inequality holds for $Im\ z>0.$ A quick way to prove the result is to map everything to the unit disk, and then you simply deal with ...


1

Let $v = (1,0,0)$ and $w = (a,b,0)$. $$ v\cdot w = \mid\mid v \mid\mid.\mid\mid w\mid\mid\cos \dfrac{2\pi}{3} \quad \Rightarrow \quad -2a = \sqrt{a^2 + b^2} \ \Rightarrow \ 3a^2 = b^2 \quad (1) $$ But, $$ \sqrt{3} = \mid\mid v \mid\mid.\mid\mid w\mid \mid \sin \dfrac{2\pi}{3} \quad \Rightarrow \quad \sqrt{3} = \sqrt{a^2 + b^2}.\dfrac{\sqrt{3}}{2} \quad ...


1

Indeed, $$ f(t) = L^{-1}\biggl[\dfrac{1}{(s^2 + 1)^2}\biggr] = L^{-1}\biggl[\dfrac{1}{s^2 + 1} - \dfrac{s^2}{(s^2 + 1)^2}\biggr] = \sin t - L^{-1}\biggl[\dfrac{s^2}{(s^2 + 1)^2}\biggr] \ (1) $$ But, $$ \dfrac{d}{ds}\biggl(\dfrac{s}{s^2 +1}\biggr) = \dfrac{1}{(s^2 + 1)^2} - \dfrac{s^2}{(s^2 + 1)^2} \quad \Rightarrow $$ $$ ...


1

You can use a standard symmetric mollification kernel. Such a function is compactly supported but this is in agreement with your conditions. http://en.wikipedia.org/wiki/Mollifier More precisely: Take $$ \varphi: x\mapsto C \exp\left(-\frac{1}{1 - x^2}\right) $$ ($x\in (-1,1)$ and $\varphi=0$ else) where $C$ is chosen such that $$ ...


1

Take any decreasing function of the norm such that the integral in (3) converges, and use the value of the integral as a normalization factor. The simplest is the box function, $f(||u||)=1$ for $||u||\le\sigma$ and $0$ elsewhere. The normalization factor will be the volume of the hypersphere $\frac{\pi^{n/2}\sigma^n}{\Gamma(n/2+1)}$. If you convolve this ...


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Find an equation for $f(x)$ by integration Find the value of the constant using $(1,6)$ Evaluate $f$ at $x=2$


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If you move everything to one side. z-5 and 5-z eliminate each other. There is no letter z left. As you said yourself z=5 has no solution. That's the reason why this equation can not be solved.


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You made a sign error when you tried to clear denominators. If you multiply the entire equation by $3(z-5)$, the right-hand side should be $3(z-5)\left(-\frac{5}{5-z}\right)=-15\left(\frac{z-5}{5-z}\right)=15$. If you fix this mistake, you'll be left with a simplified equation whose only solution is $z=5$, which doesn't work in the original equation.


1

If you are given a ratio for a line, then the line is split into segments, whose lengths are in the given ratio. For example, if you had the ratio $2:3$, you are right in thinking that the line can scale to $2$ segments, each of length $2$ and $3$. Obviously the total is $5$ so they each amount to $\frac{2}{5}$ and $\frac{3}{5}$ of the total length. But if ...



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