heureka

Harvey’s expresso express, a drive through coffer stop, is famous for its great house coffee,

a blend of Colombian and mocha java beans.

Their archrival, JoJo’s java, sent a spy to steal their ratio for blending beans.

The spy returned with a torn part of an old receipt that showed only the total number of pounds and the total cost, 18 pounds for $92.07.

At first JoJo was angry, but then he realized that he knew the price per pound of each kind of coffee

($4.89 for Colombian and $5.43 for mocha java).

Show how he could use equations to figure out how many pounds of each type of beans Harvey’s used.

Let x = pounds of Colombian coffee
Let y = pounds of mocha jave coffee

$\begin{array}{|rcll|} \hline x + y &=& 18\ \text{pounds} \qquad \text{ or } \qquad x= 18\ \text{pounds} - y\\ x\cdot \dfrac{$4.89}{\text{1 pound}} + y\cdot \dfrac{$5.43}{\text{1 pound}} &=& $92.07 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 4.89x+5.43y&=& 92.07 \quad &| \quad x = 18 - y \\ 4.89\cdot (18 - y) + 5.43y&=& 92.07 \\ 4.89\cdot18 -4.89y + 5.43y&=& 92.07 \\ 88.02 + 0.54y&=& 92.07 \quad &| \quad -88.02 \\ 0.54y&=& 92.07 -88.02 \\ 0.54y&=& 4.05 \quad &| \quad : 0.54 \\ y&=& \frac{4.05}{ 0.54 } \\ \mathbf{y}&\mathbf{=}& \mathbf{7.5\ \text{pounds } } \\\\ x &=& 18-y \\ x &=& 18 - 7.5 \\ \mathbf{x} &\mathbf{=}& \mathbf{10.5\ \text{pounds }} \\ \hline \end{array}$

Harvey’s used 10.5 pounds of Colombian coffee and 7.5 pounds of mocha jave coffee

x² + y² – 6x + 8y = 144

The equation of a circle in the xy-plane is shown above. 

What is the diameter of the circle?

Let d = diameter of the circle

Let r = radius of the circle

Let d = 2r

$\begin{array}{|rcll|} \hline x^2 + y^2 -6x+8y &=& 144\\ (x^2-6x) + (y^2+8y) &=& 144 \\ \left(x-\frac{6}{2}\right)^2 -\left(\frac{6}{2}\right)^2 +\left(y+\frac{8}{2}\right)^2 -\left(\frac{8}{2}\right)^2 &=& 144 \\ \left(x-3\right)^2 - 3^2 +\left(y+4\right)^2 -4^2 &=& 144 \\ \left(x-3\right)^2 - 9 +\left(y+4\right)^2 -16 &=& 144 \\ \left(x-3\right)^2 +\left(y+4\right)^2 &=& \underbrace{144+16 + 9}_{=r^2}\\\\ r^2 &=& 144+16 + 9 \\ r^2 &=& 169 \\ r &=& 13 \quad & | \quad \cdot 2 \\ 2r &=& 26 \quad & | \quad d = 2r \\ \mathbf{d} &\mathbf{=}& \mathbf{26} \\ \hline \end{array}$

The diameter of the circle is 26.

The center of the circle is ( 3, -4 ).

What is the rule if x is 6 and y is 7. Then x is 5 and y is 5.

Table

IN(X)      6    5    4    3      2      1        0

OUT(Y)  7    5    3     1     -1     -3      -5

arithmetical sequence:

$\begin{array}{|rcll|} \hline a_n &=& a_1 + (n-1)\cdot d \quad & | \quad d = +2 \qquad a_1 = -3 \\ a_n &=& -3 + (n-1)\cdot 2 \\ a_n &=& -3 + 2n-2 \\ \mathbf{a_n} &\mathbf{=}& \mathbf{-5 + 2n} \quad & | \quad n = IN(X) \qquad a_n = OUT(Y) \\\\ \mathbf{OUT(Y)} &\mathbf{=}& \mathbf{-5 + 2\cdot IN(X)} \\ \hline \end{array}$

1, 4, 8, 9, 75 and  100              Magic number is   273

$\begin{array}{|rcll|} \hline 273 &=& 100+\dfrac{8+9\cdot(1+75)}{4} \\\\ 273 &=& 75 + \dfrac{100\cdot (9-1) - 8 }{4} \\\\ 273 &=& 75 + \dfrac{1+8\cdot 100 -9 }{4} \\\\ 273 &=& 75 + 9\cdot \left ( \dfrac{100-8 }{4} - 1\right ) \\\\ 273 &=& 1+8\cdot ( 100-75+9) \\\\ 273 &=& 1+8+4\cdot (75-9) \\\\ 273 &=& 9 + 4\cdot(75-8-1) \\\\ 273 &=& 75 +\dfrac{8\cdot (100-1)}{4} \\\\ 273 &=& 1+ 8\cdot( 9 +\dfrac{100}{4}) \\\\ 273 &=& 4\cdot 75 \cdot \left( 1-\dfrac{9}{100} \right) \\\\ \hline \end{array}$

roots of equation x^3+3x^2+3x+3

$\small{ \begin{array}{|lrcll|} \hline & x^3+3x^2+3x+3 &=& 0 \quad &|\quad x^3+3x^2+3x+1 = (x+1)^3 \\ & (x+1)^3 +2 &=& 0 \\ & (x+1)^3 &=& -2 \\ & x+1 &=& \sqrt[3]{-2} \\ & x &=& \sqrt[3]{-2} -1 \\\\ (1) & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot (\cos(\frac{\pi}{3}) + i \cdot \sin(\frac{\pi}{3})) \quad &|\quad \tan{\varphi} = \frac{0}{-2}~ \Rightarrow ~ \varphi = \pi \\ & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot (\frac12+ i \cdot \frac{\sqrt{3}}{2} ) \\ & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot \frac12+ i \cdot \frac{\sqrt{3}}{2} \cdot \sqrt[3]{2} \\ & x_1 = \sqrt[3]{-2}-1 &=& \sqrt[3]{2} \cdot \frac12 - 1 + i \cdot \frac{\sqrt{3}}{2} \cdot \sqrt[3]{2} \\ & \mathbf{x_1} &\mathbf{=}& \mathbf{-0.37003947505 + i \cdot 1.09112363597 }\\\\ (2) & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot (\cos(\frac{\pi}{3}+\frac23 \pi) + i \cdot \sin(\frac{\pi}{3}+\frac23 \pi)) \\ & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot (\cos(\pi) + i \cdot \sin(\pi)) \\ & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot (-1+ i \cdot 0 ) \\ & \sqrt[3]{-2} &=& -\sqrt[3]{2} \\ & x_2 = \sqrt[3]{-2}-1 &=&-\sqrt[3]{2} - 1 \\ & \mathbf{x_2} &\mathbf{=}& \mathbf{-2.25992104989} \\\\ (3) & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot (\cos(\frac{\pi}{3}+2\cdot \frac23 \pi) + i \cdot \sin(\frac{\pi}{3}+2\cdot \frac23 \pi)) \\ & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot (\cos(\frac53 \pi) + i \cdot \sin(\frac53 \pi)) \\ & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot (\frac12- i \cdot \frac{\sqrt{3}}{2} ) \\ & \sqrt[3]{-2} &=& \sqrt[3]{2} \cdot \frac12- i \cdot \frac{\sqrt{3}}{2} \cdot \sqrt[3]{2} \\ & x_3 = \sqrt[3]{-2}-1 &=& \sqrt[3]{2} \cdot \frac12 - 1 - i \cdot \frac{\sqrt{3}}{2} \cdot \sqrt[3]{2} \\ & \mathbf{x_3} &\mathbf{=}& \mathbf{-0.37003947505 - i \cdot 1.09112363597} \\\\ \hline \end{array} }$

wer kann diese Zahlenreihe lösen und mir den Lösungsweg erklären ?

14   15   5   6   2   3   1

$\begin{array}{|rcll|} \hline 14 & & 15 & & 5 & & 6 & & 2 & & 3 & & 1 \\ &+1& &:3& &+1& &:3& &+1& &:3& \\ \hline \end{array}$

What quadratic equation has the roots 2 - 3^1/2 and 2 + 3^1/2

$\begin{array}{|rcll|} \hline [~x-(2 - \sqrt{3}) ~]\cdot[~x-(2 + \sqrt{3}) ~] &=& 0 \\ [~(x-2) - \sqrt{3}) ~]\cdot[~(x-2) + \sqrt{3}) ~] &=& 0 \\ (x-2)^2-(\sqrt{3})^2 &=& 0 \\ x^2-4x+4-3 &=& 0 \\ x^2-4x+1 &=& 0 \\\\ \mathbf{y=x^2-4x+1} \\ \hline \end{array}$

John and Nat were given some money.
If John spends $50 and Nat spends $100 each day,
John would still have $2500 left while Nat would have spent all her money.
If John spends $100 and Nat spends $50 each day,
John would still have $1000 left while Nat would have spent all her money.

How much were John and Nat given each?

Let j = Johns money at start

Let n = Nats money at start

Let x = days until Nat spent all her money first run.

Let y = days until Nat spent all her money second run.

1. Run

$\begin{array}{|rcll|} \hline j - x\cdot 50 &=& 2500 \\ j &=& 2500+50x\\\\ n - x\cdot 100 &=& 0 \\ n &=& 100\cdot x \\ \hline \end{array}$

2. Run

$\begin{array}{|rcll|} \hline j - y\cdot 100 &=& 1000 \\ j &=& 1000+100y \\\\ n - y\cdot 50 &=& 0 \\ n &=& 50y \\ \hline \end{array}$

We set equal:

$\begin{array}{|rcll|} \hline n = 50y &=& 100x \\ y &=& 2x \\\\ 1000+100y &=& 2500+50x \quad & | \quad -1000\\ 100y &=& 1500+50x \\ 100\cdot(2x) &=& 1500+50x \\ 200x &=& 1500+50x \quad & | \quad -50x\\ 150x &=& 1500 \quad & | \quad :150\\ x &=& \frac{1500}{150} \\ x &=& 10 \\\\ n &=& 100x \\ n &=& 100\cdot 10 \\ \mathbf{ n } & \mathbf{=} & \mathbf{$ 1000 }\\\\ j &=& 2500+50x \quad & | \quad x = 10\\ j &=& 2500+50\cdot 10 \\ j &=& 2500+500 \\ j &=& 3000 \\ \mathbf{ j } & \mathbf{=} & \mathbf{$ 3000 }\\\\ \hline \end{array}$

John were given $3000 and Nat were given $1000

If you fold up the sides the container holds exactly 1L. {nl} Whats the length of a?

$\begin{array}{|rclrcl|} \hline B_{\triangle} &=& \frac{a}{2}\cdot h_{\triangle} & \quad h_{\triangle}^2 + (\frac{a}{2})^2 &=& a^2\\ & & & \quad h_{\triangle}^2 &=& a^2 - (\frac{a}{2})^2 \\ & & & \quad h_{\triangle}^2 &=& a^2 - \frac{a^2}{4} \\ & & & \quad h_{\triangle}^2 &=& \frac{3}{4}a^2 \\ B_{\triangle} &=& \frac{a}{2}\cdot \frac{a}{2}\sqrt{3} & \quad \mathbf{ h_{\triangle} } & \mathbf{=} & \mathbf{\frac{a}{2}\sqrt{3}} \\ \mathbf{B_{\triangle}} &\mathbf{=}& \mathbf{\frac{a^2}{4}\cdot \sqrt{3}} \\ \hline \end{array}$

$\begin{array}{|rclrcl|} \hline \sin(60^{\circ}) = \frac{a}{2r} &=& \frac{\sqrt{3}} {2} \\ \frac{a}{2r} &=& \frac{\sqrt{3}} {2} \\ \frac{a}{r} &=& \sqrt{3} \\ \frac{r}{a} &=& \frac{1} { \sqrt{3} } \\ \mathbf{r} &\mathbf{=}& \mathbf{\frac{a} { \sqrt{3} }} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline r^2+h^2 &=& a^2 \\ h^2 &=& a^2-r^2 \\ h^2 &=& a^2 -\left(\frac{a} { \sqrt{3} }\right)^2 \\ h^2 &=& a^2 -\frac{a^2} { 3 } \\ h^2 &=& \frac{2}{3}a^2 \\ \mathbf{h} &\mathbf{=}& \mathbf{\frac{\sqrt{2}} {\sqrt{3}}a} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline V &=& \frac{1}{3} \cdot B_{\triangle} \cdot h \qquad &| \qquad \mathbf{ B_{\triangle} = \frac{a^2}{4} \sqrt{3}} \qquad \mathbf{ h=\frac{\sqrt{2}} {\sqrt{3}}a } \\ V &=& \frac{1}{3} \cdot \left( \frac{a^2}{4} \sqrt{3} \right) \cdot \left( \frac{\sqrt{2}} {\sqrt{3}}\cdot a \right) \\ V &=& \frac{1}{3} \cdot \frac{a^2}{4} \cdot \sqrt{2}\cdot a \\ V &=& \frac{\sqrt{2}}{12} \cdot a^3 \\\\ \frac{\sqrt{2}}{12} \cdot a^3 &=& V \\ a^3 &=& \frac{12} {\sqrt{2}} \cdot V \\ a^3 &=& \frac{12} {\sqrt{2}} \cdot V \cdot \frac{\sqrt{2}}{\sqrt{2}} \\ a^3 &=& \frac{12\cdot \sqrt{2} } {2} \cdot V \\ a^3 &=& 6\cdot \sqrt{2} \cdot V \\ \mathbf{a} &\mathbf{=}& \mathbf{\sqrt[3]{6\cdot \sqrt{2} \cdot V}} \qquad &| \qquad V = 1l = 1\ dm^3\\ a & =& \sqrt[3]{6\cdot \sqrt{2} \cdot 1\ dm^3} \\ a & =& \sqrt[3]{8.48528137424\ dm^3} \\ a & =& 2.03964890266\ dm \\ \hline \end{array}$

The length of a is 2.04 dm

What is 59/15 in base 60? (Babylonian Division)

$\begin{array}{rcll} \dfrac{59}{15} &=& 3 + \dfrac{14}{15} \\ 3 + \dfrac{14}{15} &=& 3\cdot 60^0 + \left[\dfrac{14}{15}\cdot 60\right]\cdot 60^{-1} \\ 3 + \dfrac{14}{15} &=& 3\cdot 60^0 + 56\cdot 60^{-1} \\ 3 + \dfrac{14}{15} &=& 3.56_{60}\\ \end{array}$