heureka

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

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