heureka

How many phone numbers can be made if the first digit must be 1,

the second digit must be a number in the range 3-5,

the third digit must be a number in the range (6-9),

and the last seven digits can be any single digit number 0-9?

 1) For the first digit, we have only one choice -- we must choose a 1.
 2) For the second digit, we have three choicees -- we must choose 3, 4, or 5.
 3) For the third digit, we have four choices -- we can choose 6, 7, 8, or 9.
 4) For the fourth digit, we have ten choices -- we can choose 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 .
 5) For the fifth digit, we have ten choices -- we can choose 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 .
 6) For the sixth digit, we have ten choices -- we can choose 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 .
 7) For the seventh digit, we have ten choices -- we can choose 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 .
 8) For the eighth digit, we have ten choices -- we can choose 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 .
 9) For the ninth digit, we have ten choices -- we can choose 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 .
10) For the tenth digit, we have ten choices -- we can choose 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 .

1 X 3 X 4 X 10 X 10 X 10 X 10 X 10 X 10 X 10 = 120 000 000  possible numbers.

How is   \(\mathbf{f_{xy}}\)    calculated   ?

\(\boxed{f(x,y) = 2x^2+3y^2+8x-24y+62 } \\ \small{ \begin{array}{|rcl|rcl|} \hline \dfrac{\partial f}{\partial x} = &f_x& = 4x + 8 \quad & \quad \dfrac{\partial f}{\partial y} = &f_y& = 6y-24 \\ \dfrac{ d\left(\dfrac{\partial f}{\partial x}\right) }{\partial x}= \dfrac{\partial^2 f}{\partial x\partial x} = &f_{xx}& = 4 ~ & ~ \dfrac{ d\left(\dfrac{\partial f}{\partial y}\right) }{\partial y}= \dfrac{\partial^2 f}{\partial y\partial y} = &f_{yy}& = 6 \\\\ \dfrac{ d\left(\dfrac{\partial f}{\partial x}\right) }{\partial y}= \dfrac{\partial^2 f}{\partial x\partial y} = &f_{xy}& = \dfrac{ d(4x+8)}{\partial y} = 0 ~ & ~ \dfrac{ d\left(\dfrac{\partial f}{\partial y}\right) }{\partial x}= \dfrac{\partial^2 f}{\partial y\partial x} = &f_{yx}& = \dfrac{d(6y-24)}{\partial x} = 0 \\\\ \hline \end{array}}\\ \boxed{f_{xy} = f_{yx}} \\ \)

What is the minimum value of the expression 2x^2+3y^2+8x-24y+62 for real x and y?

\(\begin{array}{|rclrcl|} \hline f(x,y) &=& 2x^2+3y^2+8x-24y+62 \\ f_x &=& 4x+8 \quad & \quad f_y &=& 6y-24 \\ f_x &=& 0 \quad & \quad f_y &=& 0 \\ 4x+8 &=& 0 \quad & \quad 6y-24 &=& 0 \\ x &=& \dfrac{-8}{4} \quad & \quad y &=& \dfrac{24}{6} \\ x &=& -2 \quad & \quad y &=& 4 \\ && \text{The critical Point is: } (-2,4). \\ \hline \end{array}\)

\(\begin{array}{|l|} \hline \text{Find the nature of the critical Point, we apply the second derivative test: } \\ \quad \text{If } f_{xx}\cdot f_{yy} - \left(f_{xy}\right)^2 > 0 \text{ and } f_{xx} > 0 \text{ minimum point } \\ \quad\text{If } f_{xx}\cdot f_{yy} - \left(f_{xy}\right)^2 > 0 \text{ and } f_{xx} < 0 \text{ maximum point } \\ \quad\text{If } f_{xx}\cdot f_{yy} - \left(f_{xy}\right)^2 < 0 \text{ saddle point } \\ \quad f_{xx} = 4, ~ f_{xy}=0, ~ f_{yy} = 6 \\ \quad f_{xx}\cdot f_{yy} - \left(f_{xy}\right)^2 = 4\cdot 6-0^2 = 24 > 0 \text{ and } f_{xx} = 4 > 0 \text{ minimum point } \\ \hline \end{array}\)

4. 

As shown in the diagram Angle BAC=90 degrees.

Let X be the foot of the altitude from A to line BC,

line AY be the bisector of Angle BAC, 

and line AZ be a median of Triangle ABC.

If Angle XAY=13 degrees and X is on line BY, 

then what is the measure of angle ZAC in degrees?

\(\text{Let $BZ=CZ=AZ=r$} \\ \text{Let $ZAC=x$} \\ \text{Let $ZAC=ACZ=x$}\)

\(\begin{array}{|rcll|} \hline ABC &=& 90^{\circ}-ACZ\\ &=& 90^{\circ}-x \\\\ BAX &=& 90^{\circ}-(90^{\circ}-x) \\ &=& x \\ \hline \end{array}\)

\(\text{Let $BAY=YAC$} \)

\(\begin{array}{|rcll|} \hline BAY &=& x+13^{\circ} \\ BAY=YAC &=& x+13^{\circ} \\ \mathbf{YAZ} & \mathbf{=} & \mathbf{13^{\circ} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline AYX &=& 90^{\circ} - 13^{\circ} \\ &=& 77^{\circ} \\\\ CYA &=& 180^{\circ} - AYX \\ &=& 180^{\circ} - 77^{\circ} \\ &=& 103^{\circ} \\ \hline \end{array}\)

\(\text{Let $AZY=2x$} \)

\(\begin{array}{|rcll|} \hline AZY+CYA+YAZ &=& 180^{\circ} \\ 2x+103^{\circ} +13^{\circ} &=& 180^{\circ} \\ 2x &=& 180^{\circ} -103^{\circ} -13^{\circ} \\ 2x &=& 64^{\circ} \\ \mathbf{x} &\mathbf{=}& \mathbf{32^{\circ}} \\ \hline \end{array}\)

The measure of angle ZAC in degrees is \(\mathbf{32^{\circ}}\)

Hello CPhill !

My attempt:

EXAMPLE
Triangle starts

1

1    2

1    4    6

1    6   16   22

1    8   30   68   90

1   10   48  146  304  394

1   12   70  264  714 1412 1806

A033877:
Triangular array read by rows associated with Schroeder numbers:
T(1,k) = 1;
T(n,k) = 0, if k T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k).
1,
1, 2,
1, 4, 6,
1, 6, 16, 22,
1, 8, 30, 68, 90,
1, 10, 48, 146, 304, 394,
1, 12, 70, 264, 714, 1412, 1806,
1, 14, 96, 430, 1408, 3534, 6752, 8558,
1, 16, 126, 652, 2490, 7432, 17718, 33028, 41586,
1, 18, 160, 938, 4080, 14002, 39152, 89898, 164512, 206098

Note that for the terms T(n,k) of this triangle n indicates the column and k the row.

If a,b,c are the sides of a triangle then how do I prove that

\(\large{\dfrac {ab+bc+ca}{a^2+b^2+c^2} \geq \dfrac12 }\) ?

\(\begin{array}{|lrcll|} \hline 1. & a & \leq & b+c \quad | \quad \cdot a \\ & a^2 & \leq & (b+c)a \\ & \mathbf{a^2} & \mathbf{ \leq } & \mathbf{ ab+ca } \qquad (1) \\\\ 2. & b & \leq & a+c \quad | \quad \cdot b \\ & b^2 & \leq & (a+c)b \\ & \mathbf{b^2} & \mathbf{ \leq } & \mathbf{ab+bc} \qquad (2) \\\\ 3. & c & \leq & a+b \quad | \quad \cdot c \\ & c^2 & \leq & (a+b)c \\ & \mathbf{c^2} & \mathbf{ \leq } & \mathbf{ca+bc} \qquad (3) \\ \\ \hline \\ (1)+(2)+(3): & a^2+b^2+c^2 & \leq & ab+ca+ab+bc+ ca+bc \\ & a^2+b^2+c^2 & \leq & 2\cdot (ab+bc+ca) \quad | \quad : (a^2+b^2+c^2) \\ & 1 & \leq & 2\cdot \dfrac{ab+bc+ca} {a^2+b^2+c^2} \quad | \quad :2 \\ & \dfrac12 & \leq & \dfrac{ab+bc+ca} {a^2+b^2+c^2} \\ &\mathbf{ \dfrac{ab+bc+ca} {a^2+b^2+c^2}} & \mathbf{ \geq } & \mathbf{ \dfrac12 } \\ \hline \end{array}\)

Using only the paths and the indicated directions, how many different routes are there from A to J?

\(\text{There are $\mathbf{22}$ routes from $A$ to $J$:} \\ \begin{array}{rl} 1 & \text{path of length } 3 \\ 6 & \text{paths of length } 4 \\ 10 & \text{paths of length } 5 \\ 5 & \text{paths of length } 6 \\ \end{array}\)

(the LaTeX percent doesn't work) = ?

Write \% in Latex to get % : \(\text{Percent in Latex: }~ \%\)

Wie löse ich diese Induktion bzw. wie bekomme ich den Induktionsschritt hin .

\(\bf{\text{Zeigen Sie mit vollständiger Induktion:}} \\ \displaystyle \sum\limits_{k=1}^{2n} \dfrac{ (-1)^{k-1} } { k } =\sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k }, \text{ für alle } n \in \mathbb{N} \)

\(\text{Induktionsanfang:} \\ \begin{array}{|lll|} \hline n=1 & \text{linke Seite:} & \sum\limits_{k=1}^{2} \dfrac{ (-1)^{k-1} } { k }\\ & &= \dfrac{(-1)^0}{1} +\dfrac{(-1)^1}{2} \\ & &= 1- \dfrac12 \\ & &\mathbf{= \dfrac12} \\\\ & \text{rechte Seite:} & \sum\limits_{k=1}^{1} \dfrac{ 1 } { 1+k } \\ & &= \dfrac{1}{1+1} \\ & &\mathbf{= \dfrac12} \\ \hline \end{array}\)

\(\text{Für $\mathbf{n=1}$ sind beide Seiten gleich, und die Aussage ist wahr!}\)

\(\text{Die Induktionsannahme (I.A.) lautet:} \\ \begin{array}{|rcll|} \hline \displaystyle \sum\limits_{k=1}^{2n} \dfrac{ (-1)^{k-1} } { k } &=& \displaystyle \sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } \\ \hline \end{array}\)

\(\text{Der Induktionsschluss von $\mathbf{n}$ nach $\mathbf{n+1}$:} \\ \begin{array}{|rcll|} \hline \displaystyle \sum\limits_{k=1}^{2(n+1)} \dfrac{ (-1)^{k-1} } { k } &=&\displaystyle \sum\limits_{k=1}^{n+1} \dfrac{ 1 } { (n+1)+k } \\ \hline \end{array}\)

\(\bf{\text{linke Seite:}} \\ \begin{array}{|rcll|} \hline \mathbf{ \displaystyle \sum\limits_{k=1}^{2(n+1)} \dfrac{ (-1)^{k-1} } { k } } \\ &= & \displaystyle \sum\limits_{k=1}^{2n+2} \dfrac{ (-1)^{k-1} } { k } \\\\ &= & \displaystyle \sum\limits_{k=1}^{2n} \dfrac{ (-1)^{k-1} } { k } + \overbrace{\dfrac{(-1)^{(2n+1)-1} }{2n+1}}^{\text{für }k=2n+1\text{ :}} + \overbrace{\dfrac{(-1)^{(2n+2)-1} }{2n+2}}^{\text{für }k=2n+2\text{ :}} \\\\ &= & \displaystyle \sum\limits_{k=1}^{2n} \dfrac{ (-1)^{k-1} } { k } + \overbrace{\dfrac{(-1)^{2n} }{2n+1}}^{\text{Exponent gerade :}} + \overbrace{\dfrac{(-1)^{2n+1} }{2n+2}}^{\text{Exponent ungerade :}} \\\\ &= & \displaystyle \sum\limits_{k=1}^{2n} \dfrac{ (-1)^{k-1} } { k } + \dfrac{ 1 }{2n+1} - \dfrac{ 1 }{2n+2} \\\\ &\mathbf{\overset{I.A.}{=}} & \mathbf{ \displaystyle \sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } + \dfrac{ 1 }{2n+1} - \dfrac{ 1 }{2n+2} } \\ \hline \end{array}\)

\(\small{ \bf{\text{rechte Seite Vorbereitung :}} \\ \begin{array}{|ll|} \hline \sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } &=& \dfrac{1} {n+1} +& \dfrac{1} {n+2}+ \dfrac{1} {n+3}+ \ldots +\dfrac{1} {2n} \\ \sum\limits_{k=1}^{n+1} \dfrac{ 1 } { (n+1)+k } &=& & \dfrac{1} {n+2} + \dfrac{1} {n+3}+ \ldots +\dfrac{1} {2n}+\dfrac{1} {2n+1}+\dfrac{1} {2n+2} \\ \hline \sum\limits_{k=1}^{n+1} \dfrac{ 1 } { (n+1)+k } - \sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } &=& & \dfrac{1} {2n+1}+\dfrac{1} {2n+2} - \dfrac{1} {n+1} \\ \mathbf{ \sum\limits_{k=1}^{n+1} \dfrac{ 1 } { (n+1)+k } } &=& \mathbf{ \sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } +} & \mathbf{ \dfrac{1} {2n+1}+\dfrac{1} {2n+2} - \dfrac{1} {n+1} } \\ \hline \end{array} }\)

\(\bf{\text{rechte Seite:}} \\ \begin{array}{|ll|} \hline \mathbf{ \displaystyle\sum\limits_{k=1}^{n+1} \dfrac{ 1 } { (n+1)+k } } &=& \mathbf{ \displaystyle\sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } + \dfrac{1} {2n+1}+\dfrac{1} {2n+2} - \dfrac{1} {n+1} } \\\\ &=& \displaystyle\sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } + \dfrac{1} {2n+1}+\dfrac{1} {2(n+1)} - \dfrac{1} {n+1} \\\\ &=& \displaystyle\sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } + \dfrac{1} {2n+1} - \dfrac{1} {n+1} \cdot \left(1-\dfrac12 \right) \\\\ &=& \displaystyle\sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } + \dfrac{1} {2n+1} - \dfrac{1} {n+1} \cdot \left(\dfrac12 \right) \\\\ &=& \displaystyle\sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } + \dfrac{1} {2n+1} - \dfrac{1} {2(n+1)} \\\\ &\mathbf{=}&\mathbf{ \displaystyle\sum\limits_{k=1}^{n} \dfrac{ 1 } { n+k } + \dfrac{1} {2n+1} - \dfrac{1} {2n+2} } \\ \hline \end{array}\)

A ladder 25 feet long is leaning against a wall.
The base of the ladder is sliding away from the wall at 2 feet per second.
Find the rate at which the angle between the ladder and the wall is changing
when the base of the ladder is 7 feet from the wall.

\(\begin{array}{|rcll|} \hline \sin(\phi) &=& \dfrac{x}{25\ \text{feet}} \\\\ \phi &=& \arcsin\left(\dfrac{x}{25\ \text{feet}} \right) \\\\ \dfrac{d\phi}{dx} &=& \dfrac{1} { \sqrt{ 25^2\ \text{feet}^2-x^2} } \quad | \quad \cdot dx \\\\ d\phi &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-x^2} }~dx \quad | \quad :dt \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-x^2} } \dfrac{dx}{dt} \quad | \quad \dfrac{dx}{dt}=2\dfrac{feet}{s},\quad x=7\ \text{feet} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-7^2\ \text{feet}^2} }\cdot 2\dfrac{feet}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{2}{24\ \text{feet}}\cdot \dfrac{feet}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1}{12}\cdot \dfrac{rad}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1}{12}\cdot\dfrac{180^{\circ}}{\pi\ \text{rad}} \dfrac{\text{rad}}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{ 15^{\circ}}{\pi s} \\\\ \mathbf{\dfrac{d\phi}{dt}} & \mathbf{=} & \mathbf{ \dfrac{ 4.77464829276^{\circ}}{s} } \\ \hline \end{array} \)