heureka

Jenny has 8 stamp books that each contain 42 pages. Each page in her books contains 6 stamps. Jenny decides to reorganize her stamp books such that each page contains 10 stamps. This will give her more space to collect stamps without having to buy new books. Under her new system, Jenny fills up 4 complete books, still with 42 pages per book. Her fifth book now contains 33 pages filled with 10 stamps per page and 1 last page with the remaining stamps. How many stamps are on that last page ?

$$\small{\text{$
\begin{array}{lrcl}
\\
\mathrm{all~ stamps}: & 8~ \mathrm{books} \cdot 42 ~\mathrm{pages}\cdot 6 ~\mathrm{stamps}&=& 2016 ~\mathrm{stamps.} \\\\
\mathrm{now}: & x~ \mathrm{books} \cdot 42 ~\mathrm{pages}\cdot 10 ~\mathrm{stamps}&=& 2016 ~\mathrm{stamps.} \\\\
& x &=& \dfrac{2016}{420}\\\\
&\mathbf{x} & \mathbf{=} & \mathbf{4.8 ~\mathrm{books}}\\
\end{array}
$}}\\$$

Jenny has 4 complete books. Remaining a 0.8 book    -     0.8 * 42 = 33.6 pages

Jenny has 33 complete pages. Remaining a 0.6 page   -    0.6 * 10 = 6 stamps

Jenny has 6 stamps on that last page

A 9x12 rectangle is inscribed in a circle. What is the radius of the circle ?

$$\small{\text{$
\begin{array}{rcl}
r^2 &=& \left(\dfrac{9}{2}\right)^2+ \left(\dfrac{12}{2}\right)^2\\\\
r^2 &=& 4.5^2+6^2\\\\
r^2 &=& 56.25 \\\\
r &=& \sqrt{56.25 }\\\\
\mathbf{r} & \mathbf{=} & \mathbf{7.5}
\end{array}
$}}$$

$$\small{\text{
What integer $n$ satisfies $0\le n<9$ and $-1111\equiv n\pmod 9~?$
}}$$

$$\small{\text{
$-1111 \pmod 9 \equiv -4 \equiv -4+9 \equiv 5$
}}$$

n= 5

$$\small{\text{Evaluate $(2009)^2 - (2008)(2010). $}}$$

I.

$$\small{\text{$
\begin{array}{rcl}
(2009)^2 - (2008)(2010) \\
&=& 2009^2-(2009-1)(2009+1)\\
&=& 2009^2-(2009^2-1)\\
&=& 2009^2 - 2009^2 + 1 \\
&=& 1
\end{array}
$}}$$

II.

$$\small{\text{$
\begin{array}{rcl}
(2009)^2 - (2008)(2010) \\
&=& (2008+1)^2 - 2008\cdot 2010 \\
&=& 2008^2 + 2\cdot 2008 + 1 - 2008\cdot 2010\\
&=& 2008^2 - 2008 (-2+2010) + 1 \\
&=& 2008^2 - 2008^2 + 1 \\
&=& 1
\end{array}
$}}$$

III.

$$\small{\text{$
\begin{array}{rcl}
(2009)^2 - (2008)(2010) \\
&=& (2010-1)^2 - 2008\cdot 2010 \\
&=& 2010^2 - 2\cdot 2010 + 1 - 2008\cdot 2010\\
&=& 2010^2 - 2010 (2+2008) + 1 \\
&=& 2010^2 - 2010^2 + 1 \\
&=& 1
\end{array}
$}}$$

The ratio of the mass of the sugar in packet A to the mass of the sugar in packet B is 5 is to 6.After the mass of the sugar in packet A is increase by 30 percent. By what percentage must the mass of the sugar in packet B be decrease so that the total mass of sugar in packet A and B remain the same ?

$$\\\small{\text{
$A:B= 5:6 $ ~~or~~ $B=\dfrac{6}{5}A$ ~~or ~~ $A=\dfrac{5}{6}B$
}}\\\\
\small{\text{
total mass of sugar in packet A and B: $\qquad A+B= A+ \dfrac{6}{5}A = \dfrac{11}{5}A$
}}\\\\
\small{\text{
increase $A = a \qquad a = 1.3\cdot A$
}}\\
\small{\text{
decrease $ B = b$
}}\\
\small{\text{
total mass of sugar in packet a and b: $\qquad a+b=A+B=\dfrac{11}{5}A$
}}\\
\small{\text{$
\begin{array}{rcl}
a+b&=&\dfrac{11}{5}A \\\\
1.3\cdot A +b &=&\dfrac{11}{5}A \\\\
b &=& \dfrac{11}{5}A - 1.3\cdot A\\\\
\mathbf{b} & \mathbf{=} & \mathbf{0.9 \cdot A}\qquad | \qquad A=\dfrac{5}{6}B \\\\
b &=& 0.9 \cdot \dfrac{5}{6}B \\\\
\mathbf{b} & \mathbf{=} & \mathbf{0.75 \cdot B }
\end{array}
$ }}\\$$

sugar in packet B be decrease  25 %

2. if p = 2-a, then find the value of a^3 + 6ap + p^3 -8 

$$\small{\text{$
\begin{array}{rclcl}
a^3 + 6ap + p^3 -8 \\
&=& a^3 + p^3 + 6ap -8 \qquad &|& \qquad (a^3+p^3) = (a+p)(a^2-ap+p^2)\\
&=& (a+p)(a^2-ap+p^2)+ 6ap -8 \qquad &|& \qquad p = 2-a \\
&=& (a+2-a)(a^2-ap+p^2)+ 6ap -8 \\
&=& 2(a^2-ap+p^2)+ 6ap -8 \\
&=& 2\left[ a^2-ap+p^2+ 3ap -4\right] \\
&=& 2( a^2+2ap+p^2 -4) \qquad &|& \qquad (a^2+2ap+p^2) = (a+p)^2 \\
&=& 2\left[ (a+p)^2 -4 \right] \qquad &|& \qquad p = 2-a \\
&=& 2\left[ (a+2-a)^2 -4) \right]\\
&=& 2( 2^2 -4)\\
&=& 2( 4 -4)\\
&=& 2\cdot 0\\
&=& 0
\end{array}
$}}$$

1.  factorize 9(2a-b)^2 - 4(2a-b)^2 -13 

$$\small{\text{$
\begin{array}{rcl}
9x^2 - 4x^2 -13 &=& 5x^2 - 13 \qquad | \qquad x= 2a-b \\\\
5x^2 - 13 &=& 0\\
5x^2 &=& 13 \\\\
x^2 &=& \dfrac{13}{5}\\\\
x^2 &=& 2.6 \\
\mathbf{x_{1,2} } & \mathbf{=} & \mathbf{\pm\sqrt{2.6}}\\\\
5x^2 - 13 &=& 5\cdot(x-\sqrt{2.6})(x+\sqrt{2.6})\qquad | \qquad x= 2a-b \\\\
9(2a-b)^2 - 4(2a-b)^2 -13 = 5(2a-b)^2 - 13 &=&
5\cdot\left[(2a-b)-\sqrt{2.6}\right] \left[(2a-b)+\sqrt{2.6} \right] \\\\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\boxed{~~ax^2+bx+c=0 \qquad
x_{1,2} = \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} ~~}
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
a=m ~~b=-1~~c=-\dfrac{m-2}{4}\\\\
x_{1,2} &=& \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1^2-4m\left(-\dfrac{m-2}{4}\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m\left(m-2\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m^2-2m} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1-2m+m^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{(1-m)^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 1 + (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 2-m }{2m} \\\\
x_1 &=& \dfrac{ 2 }{2m} - \dfrac{ m }{2m} \\\\
\mathbf{x_1} & \mathbf{=} & \mathbf{ \dfrac{ 1 }{m} - \dfrac{ 1 }{2} } \\\\\\
x_2 &=& \dfrac{ 1 - (1-m) }{2m} \\\\
x_2 &=& \dfrac{ 1 - 1 + m)}{2m} \\\\
x_2 &=& \dfrac{ m }{2m} \\\\
\mathbf{x_2} & \mathbf{=} & \mathbf{\dfrac{ 1 }{2}}
\end{array}
$}}$$

 if a jelly bean machine has 16 pink jelly beans, 34 blue jelly beans, 24 black jelly beans, and 26 purple jelly beans, what is the probability that a jelly bean chosen at random will be black?

$$\small{\text{probability ~}} =
\dfrac{ 24 \text{ black} }
{
16 \small{\text{ pink } }
+ 34 \small{\text{ blue }}
+ 24 \small{\text{ black }}
+ 26 \small{\text{ purple }}
} = \small{\dfrac{24}{100}
= 0.24 = 24 \% }$$

$$\\ a) \text{
Compute x+y and} \sqrt{x^2+y^2} \text{ when } x=5 \text{ and } y=12\\\\
b) \text{ When is } \left[x+y=\sqrt{x^2+y^2}~?\right]\\
\text{ When is } \left[x+y\neq\sqrt{x^2+y^2}~?\right]$$

a)

$$\small{\text{$
\begin{array}{rcl}
x+y &=& 5 + 12\\
x+y &=& 17 \\\\
\sqrt{x^2+y^2}&=& \sqrt{5^2+12^2}\\
\sqrt{x^2+y^2}&=& \sqrt{25+144}\\
\sqrt{x^2+y^2}&=& \sqrt{169}\\
\sqrt{x^2+y^2}&=& \sqrt{13^2}\\
\sqrt{x^2+y^2}&=& 13\\
\end{array}
$}}$$

b)

$$x+y \stackrel{\textcolor[rgb]{1,0,0}{?}} = \sqrt{x^2+y^2} \qquad \text{ squaring } \\\\
(x+y)^2 \stackrel{\textcolor[rgb]{1,0,0}{?}} = x^2+y^2\\\\
x^2+2xy+y^2 \stackrel{\textcolor[rgb]{1,0,0}{?}}= x^2+y^2 \\\\
\boxed{~~
x^2+\underbrace{\textcolor[rgb]{1,0,0}{2xy}}_{=0}+y^2 = x^2+y^2
~~}\\\\
\boxed{~~
x^2+\underbrace{\textcolor[rgb]{1,0,0}{2xy}}_{ \neq0}+y^2 \neq x^2+y^2
~~}$$