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

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