jak sie dodaje te pierwiastki [tex]2\sqrt{3} + 4\sqrt{2}[/tex]

Ciąg (1,0,1,0,1,0,1,0,1....) z definicji odpowiada funkcji tworzącej

[tex]f(x) = 1 + 0x + x^2 + 0x^3 + x^4 + 0x^5 + x^6 + ...[/tex]

Co jest równe sumie

[tex]\sum_{n=0}^{\infty}x^{2n}[/tex]

A to z kolei jest szeregiem geometrycznym o [tex]q = x^{2}[/tex]

czyli

[tex]f(x) = \sum_{n=0}^{\infty}x^{2n} = \frac{a_{0}}{1-q} = \frac{1}{1-x^2}[/tex]

Skarbonka - die Sparbüchse

Odpowiedź:

In my opiniom school fan be very fun. I have a lot of fun at school. I like to learn new things. The teachers at my school are very nice and make sure pupils have some fun while studying. I’m sure that every student has fun learning interesting things in their favourite class. For example, my favourite subject is chemistry. I always have fun in our school’s chemistry lab. I love learning about famous scientists. Although I don’t think that I will discover a new element or make a revolutionary discovery I want to become a scientist too. In some classes we always have fun. When we have PE at my school we always play basketball, soccer, tennis or volleyball. Everyone thinks it’s the best class. There are also bad things about school that are not fun at all. Standardised tests stress student out. Too much homework and exams make student fell tired. Almost every school has a group of bullies. As a person who is bullied I can say that it’s very annoying, bad and makes my mood worse. To come to a close, school  isn’t a perfect environment, but it is a place where I have a lot of fun.

Wyjaśnienie:

W tym zdaniu jest napisane że inni dokuczają ci w szkole. Jeśli go nie chcesz to omiń. Tekst zachowa sens i spójność.

Cewka obecnie posiada temperaturę około 324,8K, czyli 51,8°C.

Zależność rezystancji od temperatury możemy opisać wzorem: [tex]R_{T}=R_{0}(1+\alpha \Delta T)[/tex]

Dane:

[tex]R_{T}=1,28*45\Omega=57,6\OmegaR\Omega\\R_{0}=46\Omega\\\alpha =3,91*10^{-3} \frac{1}{K} \\T_{0}=12^{o}C=285K\\\Delta T= T_{K}- T_{0}\\ T_{K}=?[/tex]

Przekształcamy wzór, aby obliczyć [tex]\Delta T[/tex]:

[tex]R_{T}=R_{0}(1+\alpha \Delta T)/:R_0\\\frac{R_T}{R_0}=1+\alpha\Delta T/-1\\ \frac{R_T}{R_0}-1=\alpha\Delta T/:\alpha\\\frac{ \frac{R_T}{R_0}-1}{\alpha} =\Delta T[/tex]

[tex]\Delta T[/tex] możemy rozpisać jako [tex]T_K-T_0[/tex] , a więc:

[tex]\frac{ \frac{R_T}{R_0}-1}{\alpha} =\Delta T \\\frac{ \frac{R_T}{R_0}-1}{\alpha} =T_K-T_0/+T_0\\\frac{ \frac{R_T}{R_0}-1}{\alpha} +T_0=T_K[/tex]

Podstawiamy dane do wzoru i obliczamy temperaturę końcową:

[tex]\frac{ \frac{52\Omega}{45\Omega}-1}{3,91*10^{-3}} +285K=T_K\\\\324,8K=T_K[/tex]