heureka

Compute:  85 choose 82

\(\begin{array}{|rcll|} \hline && \mathbf{\binom{85}{82}} \\\\ &=& \binom{85}{85-82} \\\\ &=& \binom{85}{3} \\\\ &=& \dfrac{85}{3}\cdot \dfrac{84}{2}\cdot \dfrac{83}{1} \\\\ &=& 85\cdot \dfrac{84}{2\cdot 3}\cdot 83 \\\\ &=& 85\cdot \dfrac{84}{6}\cdot 83 \\\\ &=& 85\cdot 14 \cdot 83 \\\\ &\mathbf{=}&\mathbf{ 98770 } \\ \hline \end{array}\)

please explain q.22

\(\text{CAYB is a quadrilateral. } \\ \vec{CA} = 3\vec{a} \\ \vec{CB} = 6\vec{b} \\ \vec{BY} = 5\vec{a}-\vec{b} \)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\vec{CY}} {\vec{CX}}} \\\\ &=& \dfrac{\vec{BC}+\vec{YB}} {\vec{AC}+\vec{XA}} \\\\ &=& \dfrac{6\vec{b}+5\vec{a}-\vec{b}} {3\vec{a}+\frac13\vec{AB}} \\\\ &=& \dfrac{5\cdot(\vec{a}+\vec{b})} {3\vec{a}+\frac13(\vec{CB}-\vec{CA})} \\\\ &=& \dfrac{5\cdot(\vec{a}+\vec{b})} {3\vec{a}+\frac13(6\vec{b}-3\vec{a})} \\\\ &=& \dfrac{5\cdot(\vec{a}+\vec{b})} {3\vec{a}+2\vec{b}-\vec{a}} \\\\ &=& \dfrac{5\cdot(\vec{a}+\vec{b})} {2\cdot(\vec{a}+\vec{b})} \\\\ &=& \mathbf{\dfrac{5}{2}} \\ \hline \end{array}\)

i dont understand the last two steps can u pls explain

\(\begin{array}{|rcll|} \hline && \dfrac{{\color{red}(x-2)(x+2)} + 2x} {3x(x+2)-12} \quad & | \quad {\color{red}(x-2)(x+2)} = x^2 + 2x-2x- 2\cdot 2 = x^2 -4 \\\\ &=& \dfrac{x^2-4 + 2x } {3x^2+6x-12} \\\\ &=& \dfrac{x^2-4 + 2x } {{\color{red}3}x^2+2\cdot {\color{red}3}\cdot x-({\color{red}3}\cdot 4)} \\\\ &=& \dfrac{x^2-4 + 2x } { {\color{red}3}\cdot ( x^2+2\cdot x-( 4))} \\\\ &=& \dfrac{x^2-4 + 2x } { {\color{red}3}\cdot ( x^2-4+2x)} \\\\ &=& \dfrac{(x^2-4 + 2x) } { {\color{red}3}\cdot ( x^2-4+2x)} \quad &| \quad \dfrac{(x^2-4 + 2x) } { ( x^2-4+2x)} = 1 \\\\ &=& \dfrac{1} { {\color{red}3}} \cdot 1 \\\\ &=& \dfrac{1} { {\color{red}3}} \\\\ \hline \end{array}\)

if i needed 80 people to go to the zoo, $6 for an adult and $3 for children and $336 was made,

how many children and adults when to the zoo?

Let x = adults
Let y = children

\(\begin{array}{|rcll|} \hline & 6x + 3y &=& 336 \quad & | \quad : 3 \\ (1) & 2x + y &=& 112 \\ (2) & x + y &=& 80 \\\\ \hline (1)-(2): & (2x + y) - (x + y) &=& 112-80 \\ & 2x + y - x - y &=& 112-80 \\ &\mathbf{x} &\mathbf{=}& \mathbf{32} \\\\ & x + y &=& 80 \\ & 32 + y &=& 80 \\ & y &=& 80 - 32 \\ &\mathbf{y} &\mathbf{=}& \mathbf{48} \\ \hline \end{array}\)

32 adults and 48 children went to the zoo.

Determinant of a matrix

 3  -5  -2  2

-4   7   4  4

 4  -9  -3  7

 2  -6  -3  2

1.  tridiagonal matrix, e. g. Gauß
\(\begin{pmatrix} 3 & -5 & -2 & 2 \\ 0 & \frac{1}{3} & 1\ \frac{1}{3} & 6\ \frac{2}{3} \\ 0 & 0 & 9 & 51 \\ 0 & 0 & 0 & 3 \\ \end{pmatrix} \)

2.  Determinant: multiply the diagonally elements

The determinant of this matrix is \(3 \cdot \frac{1}{3}\cdot 9\cdot 3 = \mathbf{27 }\)

How would I solve this?

((x-2)/(2x)+(1)/(x+2))/((3)/(2)-(6)/(x^(2)+2x))

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{ \dfrac{x-2}{2x} + \dfrac{1}{x+2} } { \dfrac{3}{2}- \dfrac{6} {x^{2}+2x} } } \quad & | \quad x^{2}+2x = x(x+2) \\\\ &=& \dfrac{ \dfrac{x-2}{2x} + \dfrac{1}{x+2} } { \dfrac{3}{2}- \dfrac{6} {x(x+2)} } \\\\ &=& \dfrac{ \dfrac{(x-2)(x+2) + 1\cdot 2x}{2x(x+2)} } { \dfrac{3x(x+2)-2\cdot 6}{2x(x+2)} } \\\\ &=& \dfrac{\left[~(x-2)(x+2) + 1\cdot 2x ~\right]}{2x(x+2)} \cdot \dfrac{2x(x+2)} {\left[~ 3x(x+2)-2\cdot 6 ~\right] } \\\\ &=& \dfrac{(x-2)(x+2) + 1\cdot 2x} {3x(x+2)-2\cdot 6} \\\\ &=& \dfrac{(x-2)(x+2) + 2x} {3x(x+2)-12} \\\\ &=& \dfrac{x^2-4 + 2x } {3x^2+6x-12} \\\\ &=& \dfrac{1}{3} \cdot \dfrac{\left(x^2+2x-4\right)} {\left(x^2+2x-4\right)} \\\\ &\mathbf{=}& \mathbf{ \dfrac{1}{3}} \\ \hline \end{array}\)

bitte hilfe

how do i get the answer to: 5.478 to the power 967?

\(5.478^{967} = 1.7680762752393208561357751278000043204596021262381226... \times 10^{714} \)

\(\begin{array}{|rcll|} \hline z &=& 5.478^{967} \quad & | \quad \log_{10} \\ \log_{10}(z) &=& \log_{10} ( 5.478^{967} ) \\ \log_{10}(z) &=& 967 \cdot \log_{10} ( 5.478 ) \\ \log_{10}(z) &=& 967 \cdot 0.73862202792 \\ \log_{10}(z) &=& 714.247500997 \quad & | \quad 10^{()} \\ z &=& 10^{714.247500997} \\ &=& 10^{714+0.247500997} \\ &=& 10^{714}\cdot 10^{0.247500997} \\ &=& 10^{0.247500997} \cdot 10^{714}\\ z &=& 1.768076275239\ldots \cdot 10^{714}\\\\ \mathbf{5.478^{967}} &\mathbf{=}& \mathbf{1.768076275239\ldots \cdot 10^{714}} \\ \hline \end{array}\)

What is the answer to P versus NP?

... there is no answer.

The sequence x_1, x_2, x_3, . . ., has the property that x_n = x_{n - 1} + x_{n - 2}

for all n \ge 3.

If x_{11} - x_1 = 99, then determine x_6.

\(\begin{array}{|rcll|} \hline x_3 &=& x_2 + x_1 &=& 1\cdot x_2 + 1\cdot x_1 \\ x_4 &=& x_3 + x_2 &=& 2\cdot x_2 + 1\cdot x_1 \\ x_5 &=& x_4 + x_3 &=& 3\cdot x_2 + 2\cdot x_1 \\ \color{red}x_6 &\color{red}=& x_5 + x_4 &=& \color{red}5\cdot x_2 + 3\cdot x_1 \\ x_7 &=& x_6 + x_5 &=& 8\cdot x_2 + 5\cdot x_1 \\ x_8 &=& x_7 + x_6 &=& 13\cdot x_2 + 8\cdot x_1 \\ x_9 &=& x_8 + x_7 &=& 21\cdot x_2 + 13\cdot x_1 \\ x_{10} &=& x_9 + x_8 &=& 34\cdot x_2 + 21\cdot x_1 \\\\ x_{11} &=& x_{10} + x_9 &=& 55\cdot x_2 + 34\cdot x_1 \quad & | \quad x_{11} = 99 + x_1 \\ 99 + x_1 &=& 55\cdot x_2 + 34\cdot x_1 && \quad & | \quad - x_1 \\ 99 &=& 55\cdot x_2 + 33\cdot x_1 && \quad & | \quad : 11 \\ \color{red}9 &\color{red}=& \color{red}5\cdot x_2 + 3\cdot x_1 \\\\ \hline \color{red}9 &\color{red}=& \color{red}5\cdot x_2 + 3\cdot x_1 && \quad & | \quad \color{red}x_6 = 5\cdot x_2 + 3\cdot x_1\\ \mathbf{9} &\mathbf{=}& \mathbf{x_6} \\ \hline \end{array} \)