Majid

Thank you so much Alan ^^ I'll try to read the books that you recommended

thanks for your answer ^^

thanks Emily! I just want to know how it can be useful ^^

${k}^{2} -17k+16=0$

delta b²-4ac = (-17)²-4*1*16 = 225

${k}_{1}= \frac{17+15}{2} = 16$

${k}_{2}=\frac{17-15}{2} = 1$

${k}^{2} -17k+16= (k-16)(k-1)$

Thanks Alna, for your answer! Yes, I'm sorry for the mistake, I just didn't notice it, it's actually :h=-T(dv/dT) at constant p.

Here's my answer for this question: 

$H=U+pv$(H is the enthalpy)

$H=Q+W+pv$(U=Q+W)

$dH=dQ+dW+d(pv)$

knowing that $dQ={C}_{v}dT+hdp$

$dH={C}_{v}dT+hdp-pdv+pdv+vdp$

$dH={C}_{v}dT+hdp+vdp$

$dH={C}_{v}dT+(h+v)dp$

H is a state function, that means it's a total  differential:

$\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T}=\begin{pmatrix} \frac{\partial({h+v})} {\partial T} \end{pmatrix}_{p}$

$\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T} =\begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} + \begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p}$(1)

On the other hand we have:

$dQ={C}_{v}dT+hdp$$*(\frac{1}{T})$

$dS=\frac{{C}_{v}}{T}dT+\frac{h}{T}dp$

S is a state function, that means it's a total  differential:

$\begin{pmatrix} \frac{\partial({C}_{v}/T)} {\partial p} \end{pmatrix}_{T} = \begin{pmatrix} \frac{\partial({h/T})} {\partial T} \end{pmatrix}_{p}$

$\frac{1}{T}\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T} = \frac{ {T} \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} - \frac{h}{T²}}{T²}$

$\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T} = \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} -\frac{h}{T}$(2)

(1)=(2)

$\begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} + \begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} = \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} - \frac{h}{T}$

$\begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} = - \frac{h}{T}$

$h= -T\begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p}$

when we have the coordinates of two points, and that we want to calculate the distance between them, we do the following: 

1-calculate the victor that links the two points ( in this case MZ )

2-we calcutate the magnitude of the vector MZ 

$MZ= \begin{pmatrix} -1-6\\ 14-16 \end{pmatrix} = \begin{pmatrix} -7\\ -2 \end{pmatrix}$ this is the vector MZ 

Now its magnitude:

$MZ = \sqrt{(-7)²+(-2)²}=\sqrt{53} = 7.2801..=7.3$

thanks Emily for the answer, I was wrong in a smal thing:

$\lim_{z\rightarrow i}\frac{z^3+2}{z^9+8}$

$z\rightarrow i$

not x

${8}^{-2}=\frac{1}{{8}^{2}} =\frac{1}{64} =0.015625$

$x²-3x-10=0$

I could see that x=5 is a root for this equation (if we replace x by 5 we get 0) 

That means I can factorize this equation like that: 

$x²-3x-10 = (x-5)(x+2)$

$\Rightarrow (x-5)(x+2)=0$

$x-5=0 \Rightarrow x=5$or

$x+2=0 \Rightarrow x=-2$

Emily, isn't it 2 left to the power -4? You and Max made 2 left to the power 4 

$\sqrt[3]{{(2{n}^{0})}^{-4}.2.{m}^{2}.{n}^{4}.{m}^{-1}.{n}^{3}\over {m}^{-4}.{n}^{4}}$

$=\sqrt[3]{{2}^{-4}.2.{m}^{2}.{m}^{4}.{n}^{4}.{m}^{-1}.{n}^{3}.{n}^{-4}}$

$=\sqrt[3]{{2}^{-3}.{m}^{2}.{m}^{4}.{m}^{-1}{n}^{4}.{n}^{3}.{n}^{-4}}$

$= \sqrt[3]{{2}^{-3}.{m}^{5}.{n}^{3}}$

$= \sqrt[3]{{n}^{3}\over{2}^{3}}*\sqrt[3]{{m}^{5}}$

$= \frac{n}{3}.{m}^{\frac{5}{3}}$

It's right Cphill :)