heureka

The limit, as x approaches 4, [5+(sqrt(x))]/[(sqrt(5))+x]

$\begin{array}{rcll} \lim \limits_{x\to 4} { \left( \frac{ 5+\sqrt{x} } { \sqrt{5}+x } \right) }= \ ?\\ \hline \\ \lim \limits_{x\to 4} { \left( \frac{ 5+\sqrt{x} } { \sqrt{5}+x } \right) } &=& \left( \frac{ 5+\sqrt{4} } { \sqrt{5}+4 } \right)\\ &=& \left( \frac{ 5+2 } { \sqrt{5}+4 } \right)\\ &=& \left( \frac{ 7 } { \sqrt{5}+4 } \right)\\ &=& 1.12250219614 \end{array}$

When fnding a square root  of 64x4y16 do you minus the powers by two or divide them by two

$\begin{array}{rcll} 64x^4y^{16} &=& 8^2x^4y^{16} \\\\ \sqrt{ 64x^4y^{16} } &=& \sqrt{ 8^2x^4y^{16} }\\ \sqrt{ 64x^4y^{16} } &=& \left( 8^2x^4y^{16} \right)^{\frac12}\\ &=& \left( 8^2\right)^{\frac12} \cdot \left( x^4 \right)^{\frac12} \cdot \left( y^{16} \right)^{\frac12}\\ &=& \left( 8 \right)^{2\cdot \frac12} \cdot \left( x \right)^{4\cdot \frac12} \cdot \left( y \right)^{16\cdot \frac12}\\ &=& \left( 8 \right)^{\frac22} \cdot \left( x \right)^{\frac42} \cdot \left( y \right)^{\frac{16}{2}}\\ &=& \left( 8 \right)^{1} \cdot \left( x \right)^{2} \cdot \left( y \right)^{8}\\ &=& 8^1 \cdot x^2 \cdot y^8 \\ &=& 8 \cdot x^2 \cdot y^8 \\ \end{array}$

You divide the powers by two.

what is the coefficient of the simplified expression of -5n+3(6+7n)

$\begin{array}{rcll} -5n+3(6+7n) &=& -5\cdot n+3\cdot 6 + 3\cdot 7 \cdot n \\ &=& -5\cdot n+18 + 21 \cdot n \\ &=& 21 \cdot n -5\cdot n+18 \\ &=& 16 \cdot n +18\\ &=& 18 + 16 \cdot n\\ \end{array}$

50 bacteria land on your tooth and they quadruple in number every hour, what is the equation?

$\begin{array}{|l|lcl|} \hline \text{now} & 50 \ \text{ bacteria} &=& 50 \cdot 4^0 \ \text{ bacteria}\\ \text{after } 1 \text{ hour} & 50\cdot 4 \ \text{ bacteria} &=& 50 \cdot 4^1 \ \text{ bacteria}\\ \text{after } 2 \text{ hours} & (50\cdot 4^1)\cdot 4 \ \text{ bacteria} &=& 50 \cdot 4^2 \ \text{ bacteria}\\ \text{after } 3 \text{ hours} & (50\cdot 4^2)\cdot 4 \ \text{ bacteria} &=& 50 \cdot 4^3 \ \text{ bacteria}\\ \text{after } 4 \text{ hours} & (50\cdot 4^3)\cdot 4 \ \text{ bacteria} &=& 50 \cdot 4^4 \ \text{ bacteria}\\ \text{after } 5 \text{ hours} & (50\cdot 4^4)\cdot 4 \ \text{ bacteria} &=& 50 \cdot 4^5 \ \text{ bacteria}\\ \cdots & \cdots&&\cdots\\ \text{after } n \text{ hours} & &=& 50 \cdot 4^n \ \text{ bacteria}\\ \hline \end{array}$

There is a 4-digit number such as: abcd. When you raise each number to the power of itself and add up their products, thus: a^a + b^b + c^c + d^d, you get the original number that you started with, namely: abcd!. What is this unique 4-digit number? Thanks and have a good day.

$\begin{array}{rcll} a^a + b^b + c^c + d^d &=& abcd \\ 3^3+4^4+3^3+5^5 &=& 3435 \end{array}$

Kann mir jemand eben erklären wie ich an diese aufage ran gehen muss? 

$\begin{array}{rcll} \left( \frac{49}{25} \right)^{\frac{3}{2}} &=& \left( \frac{7^2}{5^2} \right)^{\frac{3}{2}} \\ &=& \left( \frac{7}{5} \right)^{\frac{2\cdot 3}{2}} \\ &=& \left( \frac{7}{5} \right)^3 \\ &=& \left( 1,4 \right)^3 \\ \mathbf{ \left( \frac{49}{25} \right)^{\frac{3}{2}} }&\mathbf{=}& \mathbf{2,744} \\ \end{array}$

(2^n)^n x (2^n)^3 x 4 =1

$\begin{array}{rcll} (2^n)^n \cdot (2^n)^3 \cdot 4 &=&1 \\ (2^n)^n \cdot (2^n)^3 \cdot 2^2 &=& 2^0 \\ 2^{n^2} \cdot 2^{3n} \cdot 2^2 &=& 2^0 \\ 2^{n^2+3n+2} &=& 2^0 \\ n^2+3n+2 &=& 0 \\ \end{array}$

$\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}$

$\begin{array}{rcll} n^2+3n+2 &=& 0 \quad | \quad a=1 \quad b = 3 \quad c = 2 \\ n &=& \dfrac{-3 \pm \sqrt{3^2-4\cdot 1 \cdot 2} }{2\cdot 1} \\ n &=& \dfrac{-3 \pm \sqrt{9-8} }{2} \\ n &=& \dfrac{-3 \pm \sqrt{1} }{2} \\ n &=& \dfrac{-3 \pm 1 }{2} \\ \\ n_1 &=& \dfrac{-3 + 1 }{2} \\ n_1 &=& \dfrac{-2 }{2} \\ \mathbf{n_1} & \mathbf{=} & \mathbf{-1} \\ \\ n_2 &=& \dfrac{-3 - 1 }{2} \\ n_2 &=& \dfrac{-4 }{2} \\ \mathbf{n_2} & \mathbf{=} & \mathbf{-2} \end{array}$

6^(2x+4)=3^(3x)*2^(x+8)

$\begin{array}{rcll} 6^{2x+4} &=& 3^{3x}\cdot 2^{x+8} \\ 6^{2x}\cdot 6^4 &=& 3^{3x}\cdot 2^x \cdot 2^8 \\ 6^{2x}\cdot (2\cdot 3)^4 &=& 3^{3x}\cdot 2^x \cdot 2^8 \\ 6^{2x}\cdot 2^4 \cdot 3^4 &=& 3^{3x}\cdot 2^x \cdot 2^8 \qquad & | \qquad : 2^4\\ 6^{2x}\cdot 3^4 &=& 3^{3x}\cdot 2^x \cdot 2^4 \\ (2\cdot 3)^{2x}\cdot 3^4 &=& 3^{3x}\cdot 2^x \cdot 2^4 \\ 2^{2x}\cdot 3^{2x}\cdot 3^4 &=& 3^{3x}\cdot 2^x \cdot 2^4 \qquad & | \qquad : 2^x\\ 2^{2x-x}\cdot 3^{2x}\cdot 3^4 &=& 3^{3x} \cdot 2^4 \\ 2^x\cdot 3^{2x}\cdot 3^4 &=& 3^{3x} \cdot 2^4 \qquad & | \qquad : 3^{2x}\\ 2^x \cdot 3^4 &=& 3^{3x-2x} \cdot 2^4 \\ 2^x \cdot 3^4 &=& 3^x \cdot 2^4 \\ \frac{2^x} {3^x} &=& \frac{ 2^4 } {3^4} \\ \left(\frac23 \right)^x &=& \left(\frac23 \right)^4 \\ \mathbf{ x } & \mathbf{=} & \mathbf{4} \\ \end{array}$

Für eine Kiste mit 12 Flaschen Saft muss man 14,28 € bezahlen. Wie viel Euro muss man für 4 Flaschen bezahlen?

$\begin{array}{rcll} \frac{ 14,28\ € } {12\ \text{Flaschen} } \cdot 4\ \text{Flaschen} &=& \frac{ 14,28\ € } {3} \\ &=& 4,76\ € \end{array}$

Für 4 Flaschen  muss man 4,76 € bezahlen.