heureka

Find the most left and right tangent

$$\boxed{y=a*x^2\quad a=3.48}\\
\mbox{Point }(x_p,y_p) \text{ : } \left( x_p = -5.67 \mbox{ , }y_p =-44.75 \right)\\
\text{Find the tanget Point } (x_t, y_t) \\ \\
y_t = a*x_t^2\\
\text{Slop of } y=a*x^2 \text{ is } y'= 2a*x_t \\
\text{Slop of the line through Point p is } y' = \dfrac{y_t-y_p}{x_t-x_p}\\
\text{the slops must be equal: } y' = 2a*x_t = \dfrac{y_t-y_p}{x_t-x_p}\\\\
\Rightarrow 2a*x_t*(x_t-x_p)=y_t-y_p \quad | \quad y_t=a*x_t^2 \\
2a*x_t*(x_t-x_p)=a*x_t^2-y_p \\
2a*x_t^2 - 2a*x_t*x_p=a*x_t^2-y_p \\
2a*x_t^2 -a*x_t^2 - 2a*x_t*x_p + y_p = 0 \\
\boxed{a*x_t^2 - 2a*x_t*x_p + y_p = 0 }\\$$

$$\Rightarrow \boxed{ x_{t_{1,2}}=x_p\pm\sqrt{x_p^2-\dfrac{y_p}{a}}
\qquad y_{t_{1,2}}=a*x_{t_{1,2}}^2}$$

$$x_{t_1}=-5.67+\sqrt{(-5.67)^2-\left(\dfrac{-44.75}{3.48}\right) }\\
x_{t_1}=-5.67+6.70880730103 = 1.03880730103 \\
y_{t_1}=3.48*1.03880730103 ^2 = 3.75533971815$$right most tangent

$$x_{t_2}=-5.67-\sqrt{(-5.67)^2-\left(\dfrac{-44.75}{3.48}\right) }\\
x_{t_2}=-5.67-6.70880730103 = -12.3788073010\\
y_{t_2}=3.48*(-12.3788073010)^2 = 533.257348282$$ left most tangent 

Hi Melody,

thank you for the rules.

(1)       2,8,25,50,75,100  Magic number is 431

five Numbers: 

$$\left(
\dfrac{75}{100}+50
\right)*8+25=431$$

(2)       4,6,25,50,75,100   Magic number is 821

five Numbers: 

6*(100+50) - 75 - 4 = 821  ( your solution )

6*(100+25) + 75 - 4 = 821 ( your solution )

Hi Melody,

6(100+50) - 75 - 4 = 821

But what about Number 25  (4,6,25,50,75,100) ?

(100+25)*6 + 75 - 4 = 821

But what about Number 50  (4,6,25,50,75,100)  ?

1^3 + 2^3 + 3^3 + 4^3 + 5^3 +.... 2014^3   ? 

$$s_1 = 1+2+3+ ... + n$$

$$\begin{array}{llcl}
& (1+i)^2 &=& i^2+2i+1 \\
& (1+i)^2 - 1 &=& 1+2i \\
\hline
i=1 & \not{2^2}-1^2 &=& 1+ 2*1\\
i=2 & \not{3^2}-\not{2^2} &=& 1+ 2*2 \\
i=3 & \not{4^2}-\not{3^2} &=& 1+ 2*3 \\
... &... &...& ... \\
i=n & (1+n)^2-\not{n^2} &=& 1+2n\\
\hline
\Sigma & (1+n)^2-1&=&n+2*(\underbrace{1+2+3+...+n}_{s_1})
\end{array}\\
(1+n)^2-1&=&n+2s_1\\
n+n^2=2s_1\\
\boxed{s_1=\dfrac{n(n+1)}{2}}$$

$$s_2= 1^2+2^2+3^2+ ... + n^2$$

$$\begin{array}{llcl}
& (1+i)^3 &=& i^3+3i^2+3i+1 \\
& (1+i)^3 - i^3 &=& 1+3i^2+3i \\
\hline
i=1 & \not{2^3}-1^3 &=& 1+ 3*1^2+ 3*1 \\
i=2 & \not{3^3}-\not{2^3} &=& 1+ 3*2^2+ 3*2 \\
i=3 & \not{4^3}-\not{3^3} &=& 1+ 3*3^2+ 3*3 \\
... &... &...& ... \\
i=n & (1+n)^3-\not{n^3} &=& 1+ 3*n^2+ 3n \\
\hline
\Sigma & (1+n)^3-1&=&n+3*(\underbrace{1^2+2^2+3^2+...+n^2}_{s_2})+3*(\underbrace{1+2+3+...+n}_{s_1})
\end{array}\\
(1+n)^3-1 = n+3s_2+3s_1\\
(1+n)^3-1 = n+3s_2+3*\frac{n*(n+1)}{2}\\
s_2=n*\left( \frac{1}{6}+\frac{n}{2}+\frac{n^2}{3}\right)\\
s_2=\frac{n}{6}*(1+3n+2n^2)\\
\boxed{s_2=\dfrac{n}{6}(n+1)(2n+1) }$$

$$s_3= 1^3+2^3+3^3+ ... + n^3$$

$$\begin{array}{llcl}
& (1+i)^4 &=& i^4+4i^3+6i^2+4i+1 \\
& (1+i)^4 - i^4 &=& 1+4i^3+6i^2+4i\\
\hline
i=1 & \not{2^4}-1^4 &=& 1+ 4*1^3+ 6*1^2+ 4*1 \\
i=2 & \not{3^4}-\not{2^4} &=& 1+ 4*2^3+ 6*2^2+ 4*2 \\
i=3 & \not{4^4}-\not{3^4} &=& 1+ 4*3^3+ 6*3^2+ 4*3 \\
... &... &...& ... \\
i=n & (1+n)^4-\not{n^4} &=& 1+ 4*n^3+ 6*n^2+ 4*n\\
\hline
\Sigma & (1+n)^4-1&=&n+4*(\underbrace{1^3+2^3+3^3+...+n^3}_{s_3})
+6*(\underbrace{1^2+2^2+3^2+...+n^2}_{s_2}) +4*(\underbrace{1+2+3+...+n}_{s_1})
\end{array}\\
(1+n)^4-1 = n+4s_3+6s_2+4s_1\\
(1+n)^4-1 = n+4s_3+6*\frac{n}{6}(n+1)(2n+1)+4*\frac{n*(n+1)}{2}\\
4*s_3=n^4+4n^3+6n^2+4n-n-2n^3-3n^2-n-2n^2-2n\\
4*s_3=n^4+2n^3+n^2\\
\boxed{s_3=\dfrac{n^2*(n+1)^2}{4}}$$

$$1^3 + 2^3 + 3^3 + 4^3 + 5^3 +.... 2014^3 = \left(
\dfrac{2014*2015}{2}
\right)^2 \\
= 2029105^2\\
=4\;117\;267\;101\;025$$

(2)       4,6,25,50,75,100   Magic number is 821

$$\dfrac{(6+4*50)*100-75}{25} = 821$$

$$\left( 50+6 \right)
\left( \dfrac{4*100}{25}\right)-75 = 821$$

$$\left( 50-6 \right)
\left( \dfrac{25*75}{100}\right)-4 = 821$$

$$\left( \dfrac{25*75}{50}+100\right)*6-4 = 821$$

$$\dfrac{( 75-25)*100-50 }{6}-4 = 821$$

$$\dfrac{( 50*100+25-75}{6}-4 = 821$$

$$50-4+25*(6-75+100) = 821$$

$$75-4+6*(100+50-25) = 821$$

$$100-4-25+6*(50+75) = 821$$

(1)       2,8,25,50,75,100  Magic number is 431

$$\boxed{\dfrac{\left(50-{\dfrac{100}{25}}\right)*75-2}{8} = 431}$$

$$\boxed{\dfrac{50*75-2-100}{8}-25 = 431}$$

$$\boxed{\left( 100 + \dfrac{75}{50}
\right)*\left( \dfrac{8}{2}\right) = 431}$$

$$c^2=a^2+b^2\\\\
\Rightarrow a^2 = c^2 -b^2 = (c-b)(c+b)=(85-53)(85+53)=32*138\\
=16*2*2*69\\\\
a=\sqrt{16*4*69}=\sqrt{16}\sqrt{4}\sqrt{69}=4*2*\sqrt{69}=8\sqrt{69}\approx66.5$$

 {*{20}c}    ?

Hi Melody,

I will answer your question:  the meening of {*{20}c} 

Generally :

*{count}{form}

Example:

{*{20}c}   means  {cccccccccccccccccccc}    "c" twenty times

{*{3}{|r}}  means  {|r|r|r}                        "|r" three times

1) 4 / √32   ?

$$\boxed{\dfrac{4 }{ \sqrt{32} }} \\ \\
=\dfrac{\sqrt{16} }{ \sqrt{32}} \\ \\
=\sqrt{ \dfrac{16 }{ 32} } \\ \\
=\sqrt{ \dfrac{1 }{ 2} } \\ \\
=\dfrac{1}{\sqrt{2} } * \dfrac{\sqrt{2}}{\sqrt{2}}\\\\
\boxed{=\dfrac{\sqrt{2}}{2}}$$

Ahmad and Bryan saved the same amount of money. Bryan spent $114 buying a birthday gift for her friend. In the end, Ahmad had 3 times as much as Bryan. How much did Ahmad have    ?

Amount of money = M

Bryan = M - $114

M = Ahmad = 3 * Bryan

Bryan = M - $114

Bryan = 3 * Bryan - $114

2*Bryan = $114

$$\boxed{\text{Bryan} = \$\;57}$$

Ahmad = 3 * Bryan

Ahmad = 3 * $57

$$\boxed{Ahmad = \$\;171}$$