Projects And Homeworks of a school Student

 Here are the first 10000 digits of pi calculated by my own program You can also use it and calculate by your own to custom decimal places from  https://projectandwork.blogspot.com/2023/05/blog-post.html?m=1 Enjoy!!! 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729...

 Python program to calculate π Chundnovsky's formulae: `\frac{426880\sqrt{10005}}{\pi}=` `\sum _{k=0}^{\infty}\frac{(6k)!(13591409+545140134k)}{(3k)!(k!)^3(-640320)^{3k}}` #made by Raxit Gupta import decimal  import math  #we use Chundnovsky's formulae def compute_pi(n):   decimal.getcontext().prec = n + 3   C = 426880 * decimal.Decimal(10005).sqrt()   K = decimal.Decimal(6)   M = decimal.Decimal(1)   X = decimal.Decimal(1)   L = decimal.Decimal(13591409)   S = L   # For better precision, we calculate to n+3 and remove the last two digit   for i in range(1, n+3):     M = decimal.Decimal(M* ((1728*i*i*i)-(2592*i*i)+(1104*i)-120)/(i*i*i))     L = decimal.Decimal(545140134+L)     X = decimal.Decimal(-262537412640768000*X)     S += decimal.Decimal((M*L) / X)     return str(C/S)[:-2] n=int(input("Enter decimal places:")) P=compute_pi(n) print(P)           ...

Run the following program and enter the desired decimal places of `\pi`. You would get your result .

In a ∆ABC, CA=CB. D and E are points on AB such that AD=DE=EB. If C=120° find `\angle CED` First the figure is CA=CB and AD=DE=EB: Now, let's construct a line segment CP perpendicular to AB to C: Now, by the property of isoscleles triangle, its altitude is its median. So CP divided angle C in two halfs and also bisects AB. So AP=PB Now, in `\triangle CDP` and `\triangle CEP`: CP=CP (given) `\angle CPD `=`\angle CPE`=90° DP=PE (as AP=PB and AD=EB. So AP-AD=AD-EB hence DP=PE) `\therefore  \triangle CDP \cong \triangle CEP`...........(i) But DP+PE=DE Now let AD=DE=EB=2x Then DP=PE=x Also, `\angle PCA =\angle PCB =60`° `also ~ \angle PCD=\angle PCE=\alpha`......from(i) Now, let's focus on `\triangle PCB` as follows: Here, `tanC=\frac{PB}{PC}` `\rightarrow tan60`°=`\frac{3x}{PC}` `\rightarrow \frac{3x}{\sqrt{3}}=PC` Now in `\triangle PCE=` `\rightarrow tan \alpha =\frac{PE}{PC} ` `\rightarrow  tan \alpha =\frac{x}{\frac{3x}{\sqrt{3}}}` `\rightarrow tan \alpha =\frac{\sqrt{3} x}{3x}...

 T rex dino game   write you scores in comment!! .

  AKASH CLASS 10 MODULE MATHEMATICS (1+2) PDF Hello guys, Here I am posting Akash's module book PDF of class 10 Mathematics . This book is really very high-level book and information for foundations, NTSE, Olympiads and CBSE board exams.  The PDF is below of this page. You can also  Akash module of Biology, Chemistry , Physics and Social Science Free   PDF in my blog  https://projectandwork.blogspot.com  . If you like this, please just leave a nice comment:)

  AKASH CLASS 10 MODULE CHEMESTRY (1+2) PDF Hello guys, Here I am posting Akash's module book PDF of class 10 Chemistry. This book is really very high-level book and information for foundations, NTSE, Olympiads and CBSE board exams.  The PDF is below of this page. You can also  Akash module of Biology, Maths, Physics and Social Science Free   PDF in my blog  https://projectandwork.blogspot.com  . If you like this, please just leave a nice comment:)

  AKASH CLASS 10 MODULE PHYSICS (1+2) PDF Hello guys, Here I am posting Akash's module book PDF of class 10 Physics. This book is really very high-level book and information for foundations, NTSE, Olympiads and CBSE board exams.  The PDF is below of this page. You can also  Akash module of Biology, Maths, Chemistry and Social Science Free   PDF in my blog  https://projectandwork.blogspot.com  . If you like this, please just leave a nice comment:)

AKASH CLASS 10 MODULE BIOLOGY (1+2) PDF Hello guys, Here I am posting Akash's module book PDF of class 10 Biology. This book is really very high-level book and information for foundations, NTSE, Olympiads and CBSE board exams.  The PDF is below of this page. You can also Akash module of Physics, Maths, Chemistry and Social Science Free   PDF in my blog https://projectandwork.blogspot.com  . If you like this, please just leave a nice comment:) Thank you...

Euler's continued Fraction Formula: If `r_i` are complex numbers and `x` is defined by: ` x=1+\sum_{i=1}^{\infty}r_{1}r_{2}\cdot r_(i)=1+\sum_{i=1}^\infty  (\prod_{j=1}^{i}r_{j})` Then this equality can be proved by induction:  `x=\frac{1}{1-\frac{r_1}{1+r_1-\frac{r_2}{1+r_2-\frac{r_3}{1+r_3-\ddots}}}}` Its prove will be posted later...

  Find   x , if \displaystyle \begin{array}{|l}4^{\frac{x}{y}+\frac{y}{x}}=32\\ \log_3(x-y)+\log_3(x+y)=1\end{array} 4 y x ​ + x y ​ = 32 lo g 3 ​ ( x − y ) + lo g 3 ​ ( x + y ) = 1 ​ Solution: Checking if the system is defined for two variables is a hard task, so we shall find the eventual solutions to the system and check directly if the system is defined for them. We shall only write  \displaystyle \begin{array}{|l}x+y>0\\x-y>0\end{array} x + y > 0 x − y > 0 ​  for now. \displaystyle \begin{array}{|l}\frac{x}{y}+\frac{y}{x}=log_432\\log_3(x^2-y^2)=1\end{array} y x ​ + x y ​ = l o g 4 ​ 32 l o g 3 ​ ( x 2 − y 2 ) = 1 ​ \displaystyle \begin{array}{|l}\frac{x}{y}+\frac{y}{x}=\frac{1}{2}log_232\\x^2-y^2=3\end{array} y x ​ + x y ​ = 2 1 ​ l o g 2 ​ 32 x 2 − y 2 = 3 ​ \displaystyle \begin{array}{|l}\frac{x^2+y^2}{xy}=\frac{5}{2}\\x^2-y^2=3\end{array} x y x 2 + y 2 ​ = 2 5 ​ x 2 − y 2 = 3 ​ \displaystyle \begin{array}{|l}x^2+y^2=\frac{5}{2}xy\\x^2-y^2=3\end{a...