heureka

wenn ich dir eine karte gebe dann haben wir gleich viele karten aber wenn du mir eine karte gibst hab ich doppelt so viele wie du.

$$\small{\text{ \begin{array}{lccl} 1. & ich - 1 & = & du + 1 \\ 2. & ich + 1 & = & 2 * du \\ \end{array} }}$$

(1)  ich = du + 2

(2)  ich = 2 * du - 1

gleichsetzen:

$$\begin{array}{rcl} ich = du + 2 & = & 2 * du - 1 \\ du + 2 & = & 2 * du - 1 \\ 3 & = & du \\ du &=& 3 \end{array}$$

$$\begin{array}{rcl} ich & = & du + 2\\ ich & = & 3 + 2\\ ich & = & 5 \end{array}$$

"Ich" hat 5 Karten und "du" hat 3 Karten

5(ich) -1 = 3(du) + 1

           4 = 4

5(ich) +1 = 2 * 3(du)

            6 = 2 * 3

The number 81 is 3 raised to which power :  3^4 = 81

$$\small{\text{ \begin{array}{rcl} 3^x &=& 81 \quad | \quad \ln{()} \\\\ \ln{(3^x)} &=& \ln{(81)} \\\\ x*\ln{(3)} &=& \ln{(81)} \\\\ x &=& \frac{ \ln{(81)} } {\ln{(3)} } \\\\ x &=& 4 \end{array} }}$$

$$\boxed{3^4 = 81}$$

3 3/8 x 2 2/3 = 9

I.

$$\small{\text{ \begin{array}{rcl} 3 \frac{3}{8} \times 2 \frac{2}{3} &=& \left( \frac{3*8+3}{8} \right) \times \left( \frac{2*3+2}{3} \right)\\\\ &=& \left( \frac{24+3}{8} \right) \times \left( \frac{6+2}{3} \right)\\\\ &=& \left( \frac{27}{8} \right) \times \left( \frac{8}{3} \right)\\\\ &=& \frac{27}{ \textcolor[rgb]{1,0,0}{\not}{8}} *\frac{ \textcolor[rgb]{1,0,0}{\not}{8}}{3} \\\\ &=& \frac{27}{3}\\\\ &=& \frac{\textcolor[rgb]{1,0,0}{\not}{3}*9}{\textcolor[rgb]{1,0,0}{\not}{3}}\\\\ &=& 9 \end{array} }}$$

II.

$$\small{\text{ \begin{array}{rcl} 3 \frac{3}{8} \times 2 \frac{2}{3} &=& \left( 3+ \frac{3}{8} \right) \times \left( 2+\frac{2}{3} \right)\\\\ &=& 3*2+\textcolor[rgb]{1,0,0}{\not} {3}*\frac{2}{\textcolor[rgb]{1,0,0}{\not}{3}} +\frac{3}{8}*2+\frac{\textcolor[rgb]{1,0,0}{\not}3}{8}*\frac{2}{\textcolor[rgb]{1,0,0}{\not}3} \\\\ &=& 6+2+\frac{6}{8}+\frac{2}{8}\\\\ &=& 8+\frac{8}{8}\\\\ &=& 8+1\\\\ &=& 9 \end{array} }}$$

Express in the form a+bi where a&b are real numbers.

1. (2+5i)-(3-4i): 2 + 5i - 3 +4i = -1 +9i

2. (4-3i)(2+i): 4*2 +4i -6i -3i*i = 8 -2i -3(-1) = 11 - 2i       | i*i = -1 !

3. i (3-2i)^2: i(3*3-2*3*2i + 4i*i) = i(9-12i-4) = i(5-12i) = 5i -12i*i = 5i + 12 = 12 + 5i

4. i^52: (i*i)^26 = (-1)^26 = 1

5. 1+5i / 1-2i = $$\small{\text{ $ \frac{1+5i} {1-2i} = \frac{(1+5i)(1+2i)} {(1-2i)(1+2i)} = \frac{1*1+7i+10i*i}{1-4i*i}=\frac{1+7i-10}{1+4} =\frac{-9+7i}{5} =-\frac{9}{5}+\frac{7}{5}i $ }}$$

6. 7+3i / 4i = $$\small{\text{ $ \frac{7+3i} {4i} = \frac{(7+3i)(-4i)} {(4i)(-4i)} = \frac{-28i-12i*i}{-16i*i}=\frac{-28i+12}{16} =\frac{ -7i+3 }{4} =\frac{3}{4}-\frac{7}{4}i $ }}$$

7. 1 / a+2i = $$\small{\text{ $ \frac{1} {a+2i} = \frac{1*(a-2i)} {(a+2i)(a-2i)} = \frac{ a-2i }{a^2-4i*i}=\frac{a-2i}{a^2+4} =\frac{a}{a^2+4}-\frac{2}{a^2+4}i $ }}$$

how many factors of 72 are even

1. Prime Factorization: $$72 =2^3*3^2$$

2. how many factors of 72 are even: 

   $$\small{\text{ $ \begin{array}{lrll} 2.1 & 2^{ \textcolor[rgb]{1,0,0}{3} } &:& \qquad { \textcolor[rgb]{1,0,0}{3} } $ factors: $ 2^1 = 2 \quad 2^2 = 4 \quad 2^3 = 8\\\\ 2.2& 2^{ \textcolor[rgb]{1,0,0}{3} }*3^{ \textcolor[rgb]{1,0,0}{2} } & :& \qquad { \textcolor[rgb]{1,0,0}{3}* \textcolor[rgb]{1,0,0}{2} } =6 $ factors: $ 2^1*3^1 = 6 \quad 2^2*3^1 = 12 \quad 2^3*3^1 = 24\\ &&&\qquad $ and $ 2^1*3^2 = 18 \quad 2^2*3^2 = 36 \quad 2^3*3^2 = 72 \end{array} $ }}$$  

wie berechne ich subfakultät  (!n)

berechnet.

Beispiel !6:

Beispiel !7:

$$\small{\text{ $ \begin{array}{rcl} !7 &=& 7! * \left( 1 -\frac{1}{1!} +\frac{1}{2!} -\frac{1}{3!} +\frac{1}{4!} -\frac{1}{5!} +\frac{1}{6!} -\frac{1}{7!} \right)\\\\ &=& 5040* \left(\not{1}-\not{1} +\frac{1}{2} -\frac{1}{6} +\frac{1}{24} -\frac{1}{120} +\frac{1}{720} -\frac{1}{5040} \right) \\\\ &=& \frac{5040}{2} -\frac{5040}{6} +\frac{5040}{24} -\frac{5040}{120} +\frac{5040}{720} -\frac{5040}{5040}\\\\ &=& 2520-840+210-42+7-1\\\\ &=& 1854 \end{array} $ }}$$

449,00 -1,9%=

$$\small{\text{ $ \begin{array}{rcl} 449,00 -1,9\% = \\\\ &=& 449,00 - \frac{1,9}{100}*449,00\\\\ &=& 449,00*(1-\frac{1,9}{100})\\\\ &=& 449,00*(\frac{100-1,9}{100})\\\\ &=& 449,00*(\frac{98,1}{100})\\\\ &=& 4,49*98,1\\\\ &=& 440.469 \end{array} $ }}$$

Hi Melody,

I copy it from somewhere.

88^9 mod 85 =?

$$\small{\text{ $ 88^9=(85+3)^9= \begin{array}{rcl} 1 &*& \underbrace{ 85^9 }_{=Rest\ 0}\\ +9 &*& \underbrace{ 85^8 }_{=Rest\ 0}*3^1\\ +36 &*& \underbrace{ 85^7}_{=Rest\ 0} *3^2\\ +84 &*& \underbrace{ 85^6}_{=Rest\ 0}*3^3\\ +126 &*& \underbrace{ 85^5}_{=Rest\ 0}*3^4\\ +126 &*& \underbrace{ 85^4}_{=Rest\ 0}*3^5\\ +84 &*& \underbrace{ 85^3}_{=Rest\ 0}*3^6\\ +36 &*& \underbrace{ 85^2}_{=Rest\ 0}*3^7\\ +9 &*& \underbrace{ 85^1}_{=Rest\ 0}*3^8\\ +1 &*& 3^9\\ \end{array} $ }} \\ \small{\text{ $ \boxed{88^9\mod\ 85 \equiv 3^9\mod 85 \equiv 19683\mod 85 = 48} $ }}$$

Ist Dir 19683 noch zu groß, dann rechne weiter:

$$\small{\text{ $ 3^9=3*3^8=3*81^2=3*(85-4)^2= \begin{array}{rcl} 3&*& \underbrace{ 85^2 }_{=Rest\ 0}\\ -3* 2 &*& \underbrace{ 85 }_{=Rest\ 0}*4\\ +3 &*& 4^2\\ \end{array} $ }} \\ \small{\text{ $ \boxed{88^9\mod\ 85 \equiv 3^9\mod 85 \equiv 3*4^2\mod 85 \equiv 48\mod85 = 48} $ }}$$

(i) ratio among their surface areas is 4:1.

$$r_a=2r_b$$

$$\begin{array}{rcl} Surface_A &=& 4\pi(2r_B)^2 \\ Surface_B &=& 4\pi(r_B)^2 \\ \hline \\ \frac{Surface_A}{Surface_B} &=& \frac{4\pi(2r_B)^2}{4\pi(r_B)^2 } \\\\ \frac{Surface_A}{Surface_B} &=& \frac{4\pi (2)^2 (r_B)^2}{4\pi(r_B)^2 }=2^2 = 4= \frac{4}{1} \\\\ \end{array}$$

(ii) the ratio among their volumes is 8:1.

$$\begin{array}{rcl} V_A &=& \frac{4}{3}\pi(2r_B)^3 \\\\ V_B &=& \frac{4}{3}\pi(r_B)^3 \\\\ \hline \\ \frac{V_A}{V_B} &=& \frac{\frac{4}{3}\pi(2r_B)^3}{\frac{4}{3}\pi(r_B)^3 } \\\\ \frac{V_A}{V_B} &=& \frac{\frac{4}{3}\pi(2)^3(r_B)^3}{\frac{4}{3}\pi(r_B)^3 } = 2^3 = 8 = \frac{8}{1} \\\\ \end{array}$$