heureka

y= ln(13+x)+3

$$\small{\text{
$
\begin{array}{rcll}
y &=& ln(13+x)+3 & \quad | \quad -3\\
y-3 &=& ln(13+x) & \quad | \quad e^x\\
e^{y-3} &=& e^\ln{(13+x)}\\
13+x &=& e^{y-3} & \quad | \quad -13\\
x &=& e^{y-3} - 13
\end{array}
$
}}$$

write the chain of unit fractions for each 

5/8

7/8

4/12

6/12

8/12

4/6

2/6

a=2i-j+k ; b =3k-i-2j find (2a-b, 2a+b)

$$\small{\text{
$
\begin{array}{rcl}
2a-b &=& 5i-k\\
2a+b &=& 3i-4j+5k
\end{array}
$
}}$$

intgral : t*sin(w*t) dt           w=omega=konst.   ?

$$\int t\sin(\omega t) \ dt\\\\
\boxed{\int uv' = uv-\int u'v}\\\\
\begin{array}{cl}
u=t & v= \int \sin(\omega t) \ dt
= -\frac{1}{\omega}\cos(\omega t)\\
u'= 1 & v' = \sin(\omega t)
\end{array}
\\
\int \underbrace{t}_{u}\underbrace{\sin(\omega t)}_{v'} \ dt=
\underbrace{ t}_{u}\underbrace{ (-\frac{1}{\omega}\cos(\omega t) )}_{v}
-\int \underbrace{1}_{u'}\times \underbrace{(-\frac{1}{\omega}\cos(\omega t)) }_{v}\ dt
\\
\int t\sin(\omega t) \ dt= -\frac{t}{\omega}\cos(\omega t)
+\frac{1}{\omega}\int \cos(\omega t)\ dt
\\\\
\int \cos(\omega t) \ dt
= \frac{1}{\omega}\sin(\omega t)
\\\\
\int t\sin(\omega t) \ dt= -\frac{t}{\omega}\cos(\omega t)
+\frac{1}{\omega^2}\sin(\omega t)
\\\\
\textcolor[rgb]{1,0,0}{\int t\sin(\omega t) \ dt=\dfrac{\sin(\omega t)-\omega t \cos(\omega t) } {\omega^2}} + c$$

what is the explicit equation for the terms (1,6) (2,18) (3,54)

$$\boxed{y=ax^2+bx+c}\\
\begin{array}{lrcl}
(1): & \quad 6 &=& a*1^2+b*1 +c \\
(2): & \quad 18 &=& a*2^2+b*2 +c \quad | \quad : 2 \\
(3): & \quad 54 &=& a*3^2+b*3 +c \quad | \quad : 3 \\
\hline
(1): & \quad 6 &=& 1a+b + c \\
(2): & \quad 9 &=& 2a+b +\frac{1}{2} c \\
(3): & \quad 18 &=& 3a+b +\frac{1}{3} c \\
\hline
I: (2)-(1): & 9-6=3 &=& 2a+b +\frac{1}{2}c-(1a+b+c) = a -\frac{1}{2} c\\
II: (3)-(2): & 18-9=9 &=& 3a+b +\frac{1}{3}c-(2a+b+\frac{1}{2}c)= a -\frac{1}{6}c\\
\hline
II-I: & 9-3=6 &=& a-\frac{1}{6}c-(a-\frac{1}{2}c) = \frac{1}{3}c \\
\textcolor[rgb]{1,0,0}{c} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{18}\\
\hline
I: & 3 &=& a - \frac{1}{2}18 = a-9 \\
\textcolor[rgb]{1,0,0}{a} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{12}\\
\hline
(1): & 6 &=& 12 + b + 18 = 30 + b \\
\textcolor[rgb]{1,0,0}{b} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{-24}\\
\end{array}
\\
\boxed{y=12x^2-24 x+18 }\\$$

normalcdf(-2.7,1.3)

normalcdf (-2.7, 1.3, 0, 1 ) =  0.89973254161135

$$P.S. \qquad \lambda= \frac{ \frac{1}{2}\vec{a}(\vec{c}+\vec{a} )
} { (\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a})
} =
\frac{ \frac{1}{2}\vec{a}\vec{c} + \frac{1}{2}a^2 }{ c^2 + \vec{c}\vec{a} + \frac{1}{4}\vec{a}\vec{c} +\frac{1}{4}a^2 }
=\frac{0 + \frac{1}{2}a^2 }{ a^2+0+0+ \frac{1}{4}a^2 } =
\frac{ \frac{1}{2}a^2 }{\frac{5}{4}a^2 } = \frac{2}{5}$$

$$\small{\text{
$ \vec{a}*\vec{c} =0 \quad \vec{a} $ and $ \vec{c} $ are perpendicular and $ a^2=c^2 $ is a square.
}}$$

$$\lambda= \frac{2}{5}$$

$$\vec{OP}=\frac{2}{5}\vec{c}+\frac{3}{5}\vec{a}$$

Sorry flflvm97 you have a missing term in your equation:

$$\vec{MA}+\vec{AP}-\vec{MP}=0 \\
\vec{MP}=\vec{MA}+\vec{AP} \\
\lambda(\frac{1}{4}\vec{a}+\vec{c})=\textcolor[rgb]{1,0,0}{\frac{1}{2}\vec{a}} +\mu(-\vec{a}+\vec{c})$$

Prove the Identity  4cot(4x)= 2cot(2x)-2tan(2x)

$$\\\small{\text{
If $ t = \tan{(2x)} $ than $ \cot(4x)=\frac{1-t^2}{2t}
$ and $ \cot{(2x)}=\frac{1}{t}$ .
}
$\\$
\small{\text{
So $ 4* ( \frac{1-t^2}{2t} ) = 2\frac{1}{t} - 2t
$ or $ \frac{1-t^2}{t} = \frac{1}{t} - t = \frac{1-t^2}{t}
$
}$$

2x+3y=15

If this is a Diophantine equation: x = -15 - 3n   and y = 15 + 2n     n is a Integer

what is the greatest common factor of 7983 and 100000

I. Euclid

$$\small{\text{
$
\begin{array}{rcrrr}
& && q& r\\
100000 & : & 7983 & 12 & 4204\\
7983 & : & 4204& 1 & 3779\\
4204& : & 3779& 1 & 425\\
3779& : & 425& 8 & 379\\
425& : & 379& 1 & 46\\
379& : & 46& 8 & 11\\
46& : & 11& 4 & 2\\
11& : & 2& 5 & \textcolor[rgb]{1,0,0}{1}\\
2& : & 1& 2 & 0
\end{array}
$
}}$$

The greatest common factor of 7983 and 100000 = 1 .    They are relatively prime.

II. Prime factorization

$$\small{ \text{
$
\begin{array}{rrcrl}
100000 \quad & 2^5 &*& 5^5 & \qquad *(3^0*887^0)\\
7983 \quad & 3^2 &*& 887^1 & \qquad *(2^0*5^0)\\
\end{array}
$
}}$$

take the smallest exponents: $$2^0*3^0*5^0*887^0 = \textcolor[rgb]{1,0,0}{1}$$