heureka

$$\begin{array}{rcl}
3y+ax &=& 1 \qquad | \qquad x=\dfrac{1}{a-3}\\\\
3y + a\left( \dfrac{1}{a-3} \right) &=& 1\\\\
3y + \dfrac{a}{a-3} &=& 1 \qquad | \qquad \cdot(a-3)\\\\
3y \cdot(a-3) + a &=& a-3 \\\\
3y \cdot(a-3) + a &=& a-3 \qquad | \qquad -a\\\\
3y \cdot(a-3) + a-a &=& a-a-3 \\\\
3y \cdot(a-3) &=& -3 \qquad | \qquad :3\\\\
y \cdot(a-3) &=& -1 \qquad | \qquad : (a-3)\\\\
y &=& \dfrac{-1}{ a-3 }\\\\
y &=& \dfrac{1}{ -(a-3) }\\\\
y &=& \dfrac{1}{ -a+3 }\\\\
\mathbf{y} & \mathbf{=} & \mathbf{\dfrac{1}{ 3-a } }
\end{array}$$

find a number divide by 1 2 3 4 5 6 7 8 10 and also divided by 13 17 and 19

$$2^3 \cdot 3 \cdot 5 \cdot 7 \cdot 13 \cdot 17 \cdot 19 =3~527~160$$

5b4c5 x 6b3c2 =

$$\begin{array}{rcl}
\mathbf{5b^4c^5 \cdot 6b^3c^2 }\\
&=& 5\cdot 6 b^4b^3c^5c^2\\
&=& 30 \cdot b^{4+3}\cdot c^{5+2}\\
\mathbf{5b^4c^5 \cdot 6b^3c^2 }& \mathbf{=} & \mathbf{30 b^7c^7}\\
\hline
\\
\end{array}$$

Write the explicit formula that represents the geometric sequence -2, 8, -32, 128

$$\small{\text{$
\begin{array}{rcl}
t_1 &=& -2\qquad r = \dfrac{8}{-2}=-4\\\\
t_n &=& t_1 \cdot r^{n-1}\\\\
t_n &=& (-2) \cdot (-4)^{n-1}\\\\
t_n &=& (-2) \cdot (-4)^n\cdot (-4)^{-1}\\\\
t_n &=& (-2)(-4)^{-1} \cdot (-4)^n\\\\
t_n &=& \frac{-2}{-4} \cdot (-4)^n\\\\
\mathbf{t_n} & \mathbf{=} & \mathbf{ \frac{1}{2} \cdot (-4)^n }
\end{array}
$}}$$

In scientific notation all numbers are written in the form $$a \cdot 10^b$$

Any given integer can be written in the form $$a \cdot 10^b$$ in many ways: for example, 350 can be written as $$3.5\cdot 10^2$$ or $$35\cdot 10^1$$ or $$350 \cdot 10^0$$.

In normalized scientific notation, the exponent b is chosen so that the absolute value of a remains at least one but less than ten (1 ≤ |a| < 10). Thus 350 is written as $$3.5\cdot 10^2$$

4.   $$1300 = 1.3 \cdot 10^3\\
\hline$$

5.   $$962000 = 9.62 \cdot 10^5$$

6.   $$63000000 = 6.3\cdot 10^7$$

10. $$14600 = 1.46\cdot 10^4$$

11. $$8900000 = 8.9 \cdot 10^6$$

12. $$593000000=5.93\cdot 10^8$$

Hallo sabi92,

$$\small{\text{Your answer:~~
$
\mathbf{x=-\dfrac{a}{1-a} }
$}}\\ \\
\small{\text{now you can write:~~
$
\mathbf{x=\dfrac{a}{-(1-a)} }
$}}\\\\
\small{\text{and:~~
$
\mathbf{x=\dfrac{a}{-1-(-a)} }
$}}\\\\
\small{\text{finally:~~
$
\mathbf{x=\dfrac{a}{-1+a} }
$}}\\\\
\small{\text{or:~~
$
\mathbf{x=\dfrac{a}{a-1} }
$}}\\\\$$

$$\begin{array}{rcl}
10 &=& x^{\frac{1}{3}}y^{\frac{2}{3}} \qquad | \qquad 1^3\\\\
10^3 &=& x^{\frac{3}{3}}y^{2\cdot\frac{3}{3}} \\\\
10^3 &=& xy^2 \\\\
xy^2 &=& 10^3 \\\\
xy^2 &=& 10^2\cdot 10 \qquad | \qquad :x \\\\
y^2 &=& 10^2\cdot \dfrac{10}{x} \qquad | \qquad \sqrt{} \\\\
\mathbf{y} & \mathbf{=} & \mathbf{10\cdot \sqrt{ \dfrac{10}{x} }}\\\\
\end{array}$$

$$\small{\text{$
\begin{array}{rcl}
\dfrac{a+x}{ax} &=& \dfrac{1}{a}+\dfrac{a}{a+x} \\\\
\dfrac{a}{ax}+\dfrac{x}{ax} &=& \dfrac{1}{a}+\dfrac{a}{a+x} \\\\
\dfrac{1}{x}+\dfrac{1}{a} &=& \dfrac{1}{a}+\dfrac{a}{a+x} \\\\
\dfrac{1}{x} &=& \dfrac{a}{a+x} \\\\
a+x &=& xa \\
xa &=& a+x \\
xa-x &=& a\\
x(a-1) &=& a \\\\
\mathbf{x} & \mathbf{=} & \mathbf{\dfrac{a}{a-1}}
\end{array}
$}}$$

please help !!! In an arithmetic sequence, t(8) = 145 and t(50) = 607. Find t(129).

$$\small{\text{$
\begin{array}{lrcl}
& t_8 = a+(8-1)d &=& a+7d \\
& t_{50} =a + (50-1)d &=& a+49 d\\
\\
(1) & \dfrac{ t_{50}+t_8 }{2} &=& a + 28d \\
\\
& t_{50}-t_8 &=& 42d \\
(2) & ( t_{50}-t_8 )\cdot \dfrac{100}{42} &=& 100d\\
\\
& t_{129} &=& a + (129-1)d \\
& t_{129} &=& a + 128d \\
& t_{129} &=& a + 28d + 100d \qquad | \qquad a + 28d = \dfrac{ t_{50}+t_8 }{2} \qquad 100d =( t_{50}-t_8 )\cdot \dfrac{100}{42} \\
& t_{129} &=& \dfrac{ t_{50}+t_8 }{2} + ( t_{50}-t_8 )\cdot \dfrac{100}{42}\\\\
& t_{129} &=& \dfrac{607+145}{2} + (607-145)\dfrac{100}{42}\\\\
& t_{129} &=& 376 + 1100 \\
&\mathbf{ t_{129} }& \mathbf{=} & \mathbf{ 1476 }\\
\end{array}
$}}$$

$$\sum \limits_{k=0}^N {N \choose k} \cdot (P^k)(1-P)^k
=[1+p\cdot(1-p)]^N$$