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According to your solution, X=(v_1\ v_2)=\begin{pmatrix}3e^{-2t}-2e^{-3t}&-6e^{-2t}+6e^{-3t}\\ e^{-2t}-e^{-3t}&-2e^{-2t}+3e^{-3t}\end{pmatrix}.

A gradient vector field means the vector field is the gradient of a scalar function. For example, if ##\vec V## is a gradient vector field, then there is a scalar function ##\phi## such that ##\vec V=\nabla\phi.## If ##\vec V=v_1\vec i+v_2\vec j##, then...

Try the punctured disc with boundary ##C_{\epsilon}\cup[\epsilon,R]\cup C_R.##

There are number of ways, for conics, some traditional ways are: For y=x2; x=t, y=t2. For x2+y2=1; x=cost, y=sint.

Ω:|z|<1 is the open unit circular disc, center at the origin. f(Ω) for f(z)=z-i, translate Ω to -i. The unit circular disc, center at -i. f(Ω) for f(z)=2z+3i, dilation by 2, then translate to 3i. The circular disc with radius 2, center at 3i.

The only trapezium can be formed is a square.

b) P(A|2 germinate) =P(2 germinate|A)P(A)/P(2 germinate) =\frac{\frac12P_a^2}{\frac12(P_a^2+P_b^2)}=\frac{P_a^2}{P_a^2+P_b^2}.

All quadrilaterals formed must be symmetric about AB. You can't create quadrilaterals that have 3 unequal sides or its sides crossed.

Statement correct for n=1. For ##n\ge2,## \sum_{i=1}^n\frac1{\sqrt i}\ge\int_1^{n+1}\frac1{\sqrt x}\ dx =2\sqrt{n+1}-2\ge\sqrt n.

It should be: ##C_1:\vec r=(1-t,t,0),0\le t\le1.## ##C_2:\vec r=(0,1-t,t),0\le t\le1.## ##C_3:\vec r=(t,0,1-t),0\le t\le1.##

##\begin{cases}(s-4)X+2Y=2&...(1)\\5X-(s-2)Y=2&...(2)\end{cases}## ##(s-2)\times(1)+2\times(2):((s-4)(s-2)+10)X=2(s-2)+4,##X=\frac{2s}{s^2-6s+18}. ##5\times(1)-(s-4)\times(2):(10+(s-2)(s-4))Y=10-2(s-4),##Y=\frac{18-2s}{s^2-6s+18}.

(s-4)X+2Y=2,5X-(s-2)Y=2. X=\frac{2s}{s^2-6s+18},Y=-\frac{2s-18}{s^2-6s+18}.

L^{-1}\frac s{(s+b)^2+a^2}=e^{-bx}(\cos ax-\frac ba\sin ax)

Use Open Mapping Theorem.

If W contains S, since W is a subspace, any linear combinations of the vectors in S will also in W, hence W contains span(S).