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Beng, Chandra and Danial have some stickers. Chandra and Danial have 7/10 of the stickers. Beng and Danial have 6/7 of the stickers. Beng and Chandra have a total of 620 stickers. How many more stickers did Danial collect than Beng

$$\small{\text{ All Stickers $=x$ }}\\\\ \begin{array}{lcccc} (1) & b+c+d=x & \quad c+d=x-b &\quad b+d=x-c &\quad b+c=x-d\\\\ \hline \\ (2) & c+d = \frac{7}{10}x &\frac{7}{10}x = x-b \\\\ (3) & b+d = \frac{6}{7}x & &\frac{6}{7}x = x-c \\\\ (4) & b+c = 620 & && 620 = x-d \\\\ \hline \\ & b+c+d=x & b=\frac{3}{10}x & c=\frac{1}{7}x & d=x-620\\\\ \end{array}\\ \begin{array}{rrcl} & \frac{3}{10}x + \frac{1}{7}x + x-620 &=& x\\ \\ & x &=& 1400 \end{array}$$

$$\\b=\frac{3}{10}*1400 = 420\\\\ c=\frac{1}{7}*1400 = 200\\\\ d=1400-620 = 780\\\\ d-b=780-420 = 360$$

How many more stickers did Danial collect than Beng?  360

10 pails of water can fill 5/8 of a tank. Another 4 pails and 5 jugs of water are needed to fill the tank to its brim. How many jugs of water can the tank hold ?

$$\dfrac{ \frac{5}{8}\ \mathrm{tank} } { 10\ \mathrm{pails}} * 14 \ \mathrm{pails} = \frac{5*14}{8*10} \ \mathrm{tank} = \frac{7}{8}\ \mathrm{tank} \\\\ \frac{7}{8}\ \mathrm{tank} + 5\ \mathrm{jugs} = 1\ \mathrm{tank} \\ 5\ \mathrm{jugs} = 1\ \mathrm{tank}- \frac{7}{8}\ \mathrm{tank} =\frac{1}{8}\ \mathrm{tank} \quad | \quad * 8\\ 5*8\ \mathrm{jugs} = \frac{8}{8}\ \mathrm{tank} \\ 40\ \mathrm{jugs} = 1\ \mathrm{tank} \\$$

If a and b are the roots of the quadratic equation x^2+4x-k=0 and a>b, find the value of a-b

$$\\\small{\text{ $ x_{1,2} =\dfrac{-4\pm\sqrt{16-4*(-k)}}{2} =\dfrac{-4\pm\sqrt{16+4k}}{2} =\dfrac{-4\pm\sqrt{4(4+k)}}{2} $ }}\\\\ \small{\text{ $ x_{1,2}=\dfrac{-4\pm 2\sqrt{4+k}}{2} =-2\pm \sqrt{4+k} $ }}\\\\ \small{\text{ $ a=-2+ \sqrt{4+k} \quad | \quad a > b $ }}\\ \small{\text{ $ b=-2- \sqrt{4+k} $ }}\\ \small{\text{ $ a-b=-2+ \sqrt{4+k} -( -2- \sqrt{4+k} )=-2+ \sqrt{4+k} + 2 + \sqrt{4+k} )=2\sqrt{4+k} $ }}\\ \small{\text{ $ a-b =2\sqrt{4+k} $ }}$$

The area of a circle is 36cm^2 how do I find the radius ?

$$\small{\text{The area of a circle is }} A = \pi r^2 \\ A = 36\ cm^2, \\ \small{\text{ then the radius } r = \sqrt{ \frac{A }{\pi} } = \sqrt{ \frac{36\ cm^2 }{\pi} } =\frac{ 6 } {\sqrt{\pi} }\ cm = 3.38513750129 \ cm \approx 3.385\ cm }$$

How do you find the geometric mean of a pair of numbers and put it in the simplest radical form? (e.g. 3 and 9)

For instance, the geometric mean of two numbers, say 2 and 8,

is just the square root of their product; that is $$\sqrt{2\cdot 8}=4$$ .

$$\small{\text{ The geomeric mean of the pair of numbers 3 and 9 is: $ \sqrt[2]{3*9}= \sqrt[2]{3}* \sqrt[2]{9}= \sqrt[2]{3}* 3 $ }}$$

3a=5b=6c,  then

$$\dfrac{(a+b)}{(b-c)}\ =\ ?$$  

$$\small{\text{ $ \begin{array}{lcl} &3a=5b=6c \\ (1) & 3a=5b\\\\ (2) & 3a=6c & \quad |\quad a=\frac{6}{3}c = 2c\\\\ (3) & 5b=6c & \quad |\quad b=\frac{6}{5}c \end{array} $ }}\\\\\\ \small{\text{ $ \dfrac{(a+b)}{(b-c)} = \dfrac{ 2c +\frac{6}{5}c }{\frac{6}{5}c -c } = \dfrac{ c\left(2 +\dfrac{6}{5}\right) }{ c\left(\dfrac{6}{5} -1\right) } =\dfrac{ 2 +\dfrac{6}{5} }{ \dfrac{6}{5} -1 } =\dfrac{ \dfrac{16}{5} }{ \dfrac{1}{5} } =\dfrac{16}{5} * \dfrac{5}{1} = 16 $ }}$$

How do you convert Joules to Hertz ?

$$\small{\text{ Energy, E:\quad $ \boxed{E = h\nu} \quad $ Planck constant $ h = 4.135667516\cdot 10^{-15}\ eV\cdot s, \quad $ frequency $ \nu $ in Hz $(s^{-1})$ }}\\ \boxed{1\ \mathrm{eV} = 1{,}602 \, 176 \, 565 \cdot 10^{-19}~\mathrm {J} }$$

$$\small{\text{ Herz to Joules: }}\\ \small{\text{ We have 50000 Hz: }}\\ \small{\text{ $ 50000 Hz\ *\ 4.13566733 \cdot 10^{-15}\ eV s\ *\ 1.6021765 \cdot 10^{-19} \frac{J}{eV}= 3.3130345 \cdot 10^{-29} J $ }}$$

$$\small{\text{ Joules to Herz: }}\\ \small{\text{ We have $3.3130345 \cdot 10^{-29} J $: }}\\\\ \small{\text{ $ \dfrac{3.3130345 \cdot 10^{-29} J}{ 4.13566733 \cdot 10^{-15}\ eV s\ *\ 1.6021765 \cdot 10^{-19} \frac{J}{eV} }= 50000 Hz$ }}$$

   $$\small{\text{ c and a ? $ \boxed{ \; ac = \dfrac{2A}{\sin{(B)}} \qquad a=4-c \qquad A= \frac{3}{4}\sqrt{3} \qquad \sin{(B)}=\frac{\sqrt{3}}{2}\; } $ }}\\\\\\ \small{\text{ $ (4-c)c = 2*\dfrac{3\sqrt{3}}{4}*\dfrac{2}{\sqrt{3}}= 3$ }}\\\\ \small{\text{ $ c^2-4c+ 3 =0 \qquad $ }} \small{\text{ $ c_{1,2}=\dfrac{ 4\pm\sqrt{16-4*3 } } {2} = \dfrac{ 4\pm \sqrt{4} } {2} = \dfrac{ 4\pm 2 } {2} =2\pm 1 $ }}\\\\ \small{\text{ $ c_1= 2+1 \quad \boxed{c_1=3} \qquad c_2= 2-1 \quad \boxed{c_2= 1}$ }}\\\\ \small{\text{ $ a_1=4-c_1 \qquad a_1=4-3 \qquad \boxed{a_1=1}$ }}\\ \small{\text{ $ a_2=4-c_2 \qquad a_2= 4-1 \qquad \boxed{a_2=3}$ }}$$

(2) if b=square root 13 ,and a+c=4 what is the area of the triangle ABC

$$\boxed{\small{\text{(1) The shape of the triangle by Heron's formula:$\quad A^2=s(s-a)(s-b)(s-c) \quad s = \frac{a+b+c}{2}$ }} }\\\\ \small{\text{$s =\frac{a+c+b}{2} \quad | \quad a+c=4 $ }}\\ \small{\text{ $s = \frac{4+b}{2} = 2 + \frac{b}{2}\qquad \boxed{s=2 + \frac{b}{2} } $ }}\\\\ \small{\text{ $ A^2=s(s-a)(s-b)(s-c) = \textcolor[rgb]{1,0,0}{s(s-b)} (s-a)(s-c) $ }} }}\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{s(s-b)}=( 2+\frac{b}{2} )(2+\frac{b}{2}-b) =(2+\frac{b}{2})(2-\frac{b}{2})=4-\frac{b^2}{4} \quad | \quad b^2=13 $ }}\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{s(s-b)} = 4-\frac{13}{4} =\frac{16-13}{4} = \frac{3}{4}$ }}\\\\ \small{\text{ $ A^2=\textcolor[rgb]{1,0,0}{s(s-b)} (s-a)(s-c) = \textcolor[rgb]{1,0,0}{ \frac{3}{4} } (s-a)(s-c) $ }}\\ \small{\text{ $ A^2=\frac{3}{4}(s-a)(s-c)= \frac{3}{4}(s^2-sc-sa+ac) =\frac{3}{4}(s^2-s(a+c)+ac) \quad | \quad a+c=4 $ }}\\ \small{\text{ $ A^2=\frac{3}{4}( \textcolor[rgb]{1,0,0}{s^2-4s} +ac)$ }}\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{s^2-4s} = (2+ \frac{b}{2} )^2-4(2+\frac{b}{2} )=4+2b+ \frac{b^2}{4} -8-2b=-4+\frac{b^2}{4} =\frac{-16+13}{4}= -\frac{3}{4} $ }}\\\\ \small{\text{ $ A^2=\frac{3}{4}( \textcolor[rgb]{1,0,0}{ -\frac{3}{4} } +ac) $ }}\\ \small{\text{ $ A^2*\frac{4}{3} -ac + \frac{3}{4} = 0$ }}\\$$

$$\boxed{\small{\text{(2) The shape of the triangle :$\quad A= \dfrac {ac \sin{(B)} } {2} \qquad ac = \dfrac{2A}{\sin{(B)}} $ }} }\\\\ \small{\text{ $ A^2*\frac{4}{3} -ac + \frac{3}{4} = 0$ }} }\\ \small{\text{ $ A^2*\frac{4}{3} -\frac{2A}{\sin{(B)}} + \frac{3}{4} = 0$ }} }\\\\ \small{\text{ $ A_{1,2}= \dfrac { \frac{2}{\sin{(B)}} \pm\sqrt{ \left( \frac{2}{ \sin{(B)} } \right)^2-4*\frac{4}{3}*(\frac{3}{4} ) } } { 2*\frac{4}{3} } = \dfrac { \frac{2}{\sin{(B)}} \pm 2\sqrt{ \frac{1}{ \sin^2{(B)} }-1 } } { \frac{8}{3} } = \dfrac { \frac{2}{\sin{(B)}} \pm 2 \frac{ \cos{(B)} } { \sin{(B)} } } { \frac{8}{3} } $ }}\\ \small{\text{ $ A_{1,2} = \frac{3}{4} \left( \frac{1}{\sin{(B)}} \pm \frac{ \cos{(B)} } { \sin{(B)} } \right) = \frac{3}{4} \left( \frac{1\pm\cos{(B)} }{\sin{(B)}} \right) $ }}\\\\ \small{\text{ $ A_{1} = \frac{3}{4} \left( \frac{1+\cos{(B)} }{\sin{(B)}} \right) \quad | \quad \cos{(B)} = - \frac{1}{2} \quad \sin{(B)} =\sqrt{ 1-\cos^2{ (B) }} = \frac{\sqrt{3}}{2} $ }}\\ \small{\text{ $ A_{1}=\frac{3}{4}*\frac{1+ (-\frac{1}{2}) }{ \frac{\sqrt{3}}{2} } =\frac{3}{4}*\frac{2}{ \sqrt{3} } *\frac{1}{2} =\frac{3}{4}*\frac{2*\sqrt{3}}{3} *\frac{1}{2} = \frac{1}{4}*\sqrt{3} $ }}\\ \small{\text{ $ \boxed{A_{1}= \frac{1}{4}\sqrt{3} $ no solution!$} $ }}\\\\ \small{\text{ $ A_{2} = \frac{3}{4} \left( \frac{1-\cos{(B)} }{\sin{(B)}} \right) \quad | \quad \cos{(B)} = - \frac{1}{2} \quad \sin{(B)} =\sqrt{ 1-\cos^2{ (B) }} = \frac{\sqrt{3}}{2} $ }}\\ \small{\text{ $ A_{2}=\frac{3}{4}*\frac{1- (-\frac{1}{2}) }{ \frac{\sqrt{3}}{2} } =\frac{3}{4}*\frac{2}{ \sqrt{3} }*\frac{3}{2} =\frac{3}{4}* \frac{2*\sqrt{3}}{3} *\frac{3}{2} = \frac{3}{4}*\sqrt{3} $ }}\\ \small{\text{ $ \boxed{A_{2}= \frac{3}{4}\sqrt{3}} $ }}$$

In a triangle ABC, a, b , c are the opposite side of angle A,angle B ,and angle C, and

  $${\frac{{\mathtt{cosB}}}{{\mathtt{cosc}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{b}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{c}}\right)}}$$

(I) find the angle B

$$\boxed{ \dfrac{ \cos{(B)} }{\cos{(C)}}=-\dfrac{b}{(2a+c)} } \small{\text{ $\quad $For any triangle: $ a=b\cos{(C)}+c\cos{(B)}. $ }}\\\\ \small{\text{ $ (2a+c)\cos{(B)} = -b\cos{(C)} \quad | \quad -b\cos{(C)}= c\cos{(B)} - a $ }}\\ \small{\text{ $ (2a+c)\cos{(B)} = c\cos{(B)} - a $ }}\\ \small{\text{ $ (2a)\cos{(B)} = - a $ }}\\ \small{\text{ $ 2\cos{(B)} = - 1 $ }}\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{\cos{(B)} = - \dfrac{1}{2} \qquad B = 120\ensurement{^{\circ}} }$ }}$$