heureka

sqrt(sqrt(x-5) + x) = 5        x = 21

$$\sqrt{ \sqrt{x-5 } +x } = 5 \quad | \quad x = 21\\\\
\sqrt{ \sqrt{21-5 } +21 } = 5 \\\\
\sqrt{ \sqrt{16 } +21 } = 5 \\\\
\sqrt{ 4 +21 } = 5 \\\\
\sqrt{ 25 } = 5 \\\\
5 \stackrel{!}{=} 5 \\$$

How to find integral of logarithmic functions:

$$\boxed{\int{log_a(x) \ dx} = x * log_a(x) - \dfrac{x}{\ln{(a)}} + c\ }$$

$$\begin{array}{lcrclcrccccl}
(u & * & v)' & = & u' &*& v & + & u &*& v' &\\
(x & * & log_a{(x)})' & = & 1 & *& log_a{(x)} & + & x & * & \dfrac{1}{x*\ln{(a)}} & \quad | \quad \textcolor[rgb]{1,0,0}
{
( log_a{(x)} )' = \dfrac{1}{x*\ln{(a)}}
}\\ \\
(x & * & log_a{(x)})' & = &&& log_a{(x)} & + &&& \dfrac{1}{\ln{(a)}} & \quad | \quad \int{}\ dx \\ \\
x & * & log_a{(x)} & = &&& \int{ log_a{(x)} \ dx} & + &&& \dfrac{1}{\ln{(a)}} \int{\ dx} & \\ \\
x & * & log_a{(x)} & = &&& \int{ log_a{(x)} \ dx} & + &&& \dfrac{x}{\ln{(a)}} & \\ \\
\int{ log_a{(x)} \ dx} & & & = &&& x * log_a{(x)} & - &&& \dfrac{x}{\ln{(a)}} &
\end{array}$$

Example:

$$basic \ a = e: \\
\int{ log_e{(x)} \ dx} =x * log_e{(x)} - \dfrac{x}{\ln{(e)}} \\
\int{ \ln{(x)} \ dx} =x * ln{(x)} - \dfrac{x} {1} \\
\int{ \ln{(x)} \ dx} =x * ln{(x)} - x\\$$

$$basic \ a = 10: \\
\int{ log_{10}{(x)} \ dx} =x * log_{10}{(x)} - \dfrac{x}{\ln{(10)}} \\
\int{ \log{(x)} \ dx} =x * log{(x)} - \dfrac{x}{\ln{(10)}}$$

(2222^5555+5555^2222)(mod7)=0

sin(12^12^14)

Hi Melody,

here is the solution from WolframAlpha:

I think, the argument (12^12)^14 is to big for our calculator. But the mod - function is correct and departs from the argument multiple from 360 degrees.

The formula is  $$\sin(\alpha) = \sin(\alpha \pm n*360\ensurement{^{\circ}} )$$

x+y=10 x-y=2 solve using elemination(allgerbra 2) ?

$$\\ \frac{(x+y)\textcolor[rgb]{1,0,0}{+}(x-y)}{2} = x = \frac{10+2}{2} = \frac{12}{2} = 6 \\
\frac{(x+y)\textcolor[rgb]{1,0,0}{-}(x-y)}{2} = y = \frac{10-2}{2} = \frac{8}{2} = 4$$

sin 105 - sin 15

$$\sin{( 105\ensurement{^{\circ}} )} - \sin{( 15\ensurement{^{\circ}} ) } \\
=\sin{( 90\ensurement{^{\circ}} + 15\ensurement{^{\circ}} )} - \sin{( 15\ensurement{^{\circ}} ) } \\
= \underbrace{\sin{( 90\ensurement{^{\circ}} ) }}_{=1}\cos{( 15\ensurement{^{\circ}} ) }+\underbrace{\cos{( 90\ensurement{^{\circ}} ) }}_{=0} \sin{( 15 \ensurement{^{\circ}}) }- \sin{( 15\ensurement{^{\circ}} ) }\\
=\cos{(\textcolor[rgb]{1,0,0}{15\ensurement{^{\circ}}} )} - \sin{( \textcolor[rgb]{1,0,0}{15 \ensurement{^{\circ}}}) } \\ \\
\boxed{
=\sqrt{2}\cos{(\textcolor[rgb]{1,0,0}{15\ensurement{^{\circ}}} +45\ensurement{^{\circ}} )}\\
=-\sqrt{2}\sin{(\textcolor[rgb]{1,0,0}{15\ensurement{^{\circ}}} -45 \ensurement{^{\circ}})} \\
=\frac{\sqrt{2}}{2} = 0.70710678119
}$$

Find the value of  x such that the area of a triangle whose vertices have coordinates (6, 5), (8, 2) and ( x, 11) is 15 square units.

$$area = \pm15 =
\dfrac{
\left|
\left[
\left(\begin{array}{c}
x\\
11
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\times\left[
\left(\begin{array}{c}
8\\
2
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\right|
}
{2}
\\\\
\pm30 =
\left|
\left[
\left(\begin{array}{c}
x\\
11
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\times\left[
\left(\begin{array}{c}
8\\
2
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\right|
} \\
\pm30 =
\left|
\left(\begin{array}{c}
x-6\\
11-5
\end{array} \right)
\times
\left(\begin{array}{c}
8-6\\
2-5
\end{array} \right)
\right|
} \\
\pm30 =
\left|
\left(\begin{array}{c}
x-6\\
6\end{array} \right)
\times
\left(\begin{array}{c}
2\\
-3\end{array} \right)
\right|
} \\\\
x_1 =?\\
30 =
\left|
\left(\begin{array}{c}
x_1-6\\
6\end{array} \right)
\times
\left(\begin{array}{c}
2\\
-3\end{array} \right)
\right|
} \\
(x_1-6)(-3)-6*2=30\\
(x_1-6)(-3)=42\\
(x_1-6)=-14\\
\textcolor[rgb]{1,0,0}{x_1 =-8}\\\\$$

$$x_2 =?\\
-30 =
\left|
\left(\begin{array}{c}
x_2-6\\
6\end{array} \right)
\times
\left(\begin{array}{c}
2\\
-3\end{array} \right)
\right|
} \\
(x_2-6)(-3)-6*2=-30\\
(x_2-6)*3+12=30\\
(x_2-6)*3=18\\
x_2-6=6\\
\textcolor[rgb]{1,0,0}{ x_2=12 }$$

a. is okay

$$\small\text{
$
\sin{ ( 12^{12^{14}} \ensurement{^{\circ}} ) } \\
=\sin{ ( 12^{ 168 } \ensurement{^{\circ}} ) } \\
= \sin{ ( 216 \ensurement{^{\circ}} ) } \\
= -0.587785252292
$
}}$$