heureka

Find the sum of the following series by pairing two numbers to make ten(s).

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

$\begin{array}{rcrcrcrcrcrcrcrcrcrcr} \hline \hline 1 &+& 3 &+& 5 &+& 7 &+& 9 &+& 11 &+& 13 &+& 15 &+& 17 &+& 19 & \\ \hline \hline 1 & & & & & & & & & & & & & & & & &+& 19 & =& 20 \\ \hline & & 3 & & & & & & & & & & & & &+& 17 & & & =& 20 \\ \hline & & & & 5 & & & & & & & & &+& 15 & & & & & =& 20 \\ \hline & & & & & & 7 & & & & &+& 13 & & & & & & & =& 20 \\ \hline & & & & & & & & 9 &+& 11 & & & & & & & & & =& 20 \\ \hline \hline & & & & & & & & & & & & & & & & & & & =& 100\\ \hline \hline \end{array}$

( 1 + 19 ) + ( 3 + 17 ) + ( 5 + 15 ) + ( 7 + 13 ) + ( 9 + 11 ) = 20 + 20 + 20 + 20 + 20 = 100

kann mir jemand bitte erklären wie man dieses Integral löst?

2. Möglichkeit - einfacher

$\small{ \begin{array}{lrcll} & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} \\ \hline \text{Substitution : } & u &=& 1+x^2 \ dx\\ & \ du &=& 2x \ dx \qquad \ dx = \frac{\ du}{2x} \\ \hline & 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{x}{u} \cdot \dfrac{\ du}{2x} }\\ & &=& \frac72 \int \limits_{-3}^{2} { \dfrac{\ du}{u} }\\ \boxed{~ \text{Formel: } \int \limits { \dfrac{\ du}{u}} = \ln{(u)} + c ~}\\ & &=& \frac72 [~ \ln{(u)} ~]_{-3}^{2} \qquad | u = 1+x^2 \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} } & \mathbf{=} & \mathbf{ \frac72 [~ \ln{(1+x^2 )} ~]_{-3}^{2} }\\ \hline & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& \frac72 [~ \ln{(~ 1 + x^2 ~)} ~]_{-3}^{2} \\ & &=& \frac72 [~ \ln{(~ 1 + (2)^2 ~)} - \ln{(~ 1 + (-3)^2 ~)} ~] \\ & &=& \frac72 [~ \ln{(~ 1 + 4 ~)} - \ln{(~ 1 + 9 ~)} ~] \\ & &=& \frac72 [~ \ln{(~ 5 ~)} - \ln{(~ 10 ~)} ~] \\ & &=& \frac72 \ln{( \frac{5}{10} )} \\ & &=& \frac72 \ln{( \frac12 )} \\ & &=& \frac72 [ \ln{( 1 )} - \ln{( 2 )} ] \qquad \ln{( 1 )} = 0\\ & &=& \frac72 [ 0 - \ln{( 2 )} ] \\ & &=& \frac72 [ -\ln{( 2 )} ] \\ & &=& -\frac72 \ln{( 2 )} \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} }&\mathbf{=}& \mathbf{ -2.4260151320 } \\ \end{array} }$

kann mir jemand bitte erklären wie man dieses Integral löst?

$\int \limits_{-3}^{2} { \frac{7x}{1+x^2}\ dx}$

$\small{ \begin{array}{lrcll} & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} \\ \hline \text{Substitution 1: } & x &=& \sinh{(z)} \\ & \ dx &=& \cosh{(z)} \ dz\\ \cosh^2{(z)}- \sinh^2{(z)} = 1 \\ \cosh^2{(z)} = 1 + \sinh^2{(z)} \\ & 1+x^2 &=& 1 + \sinh^2{(z)} =\cosh^2{(z)} \\ \hline & 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{\sinh{(z)}}{\cosh^2{(z)}} \cdot \cosh{(z)} \ dz}\\ & &=& 7 \int \limits_{-3}^{2} { \dfrac{\sinh{(z)}}{\cosh{(z)}} \ dz}\\ \sinh{(z)} = \frac{ e^z-e^{-z} }{2} \\ \cosh{(z)} = \frac{ e^z+e^{-z} }{2} \\ & &=& 7 \int \limits_{-3}^{2} { \dfrac{e^z-e^{-z}}{e^z+e^{-z}} \ dz}\\ \hline \text{Substitution 2: } & u &=& e^z+e^{-z} \\ & \ du &=& e^z\ dz+(-1)e^{-z} \ dz\\ & \ du &=& (e^z-e^{-z}) \ dz\\ \hline = 7 \int \limits_{-3}^{2} { \dfrac{\ du}{u}}\\ = 7 [\ln{(u)}]_{-3}^{2} & | \quad u = e^z+e^{-z}\\ = 7 [\ln{(e^z+e^{-z})}]_{-3}^{2} & | \quad e^z+e^{-z} = 2\cosh{(z)}\\ = 7 [\ln{(~2\cosh{(z)}~)}]_{-3}^{2} & | \quad \cosh^2{(z)} = 1 + \sinh^2{(z)} \\ = 7 [\ln{(~2 \sqrt{1 + \sinh^2{(z)}} ~)}]_{-3}^{2} & | \quad \sinh{(z)} = x \\ = 7 [\ln{(~2 \sqrt{1 + x^2} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{(~2 \sqrt{1 + x^2} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{(~2 (1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{2}+\ln{(~2 (1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\underbrace{\ln{2}}_{\ln{2}-\ln{2}\ \text{kürzt sich raus}}+\ln{(~(1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{(~ (1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\frac12 \ln{(~ 1 + x^2 ~)}]_{-3}^{2} \\ & &=& \frac72 [ \ln{(~ 1 + x^2 ~)}]_{-3}^{2} \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} }& \mathbf{=}& \mathbf{ \frac72 [ \ln{(~ 1 + x^2 ~)}]_{-3}^{2} } \\ \hline & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& \frac72 [ \ln{(~ 1 + x^2 ~)}]_{-3}^{2} \\ & &=& \frac72 [ \ln{(~ 1 + (2)^2 ~)} - \ln{(~ 1 + (-3)^2 ~)} ] \\ & &=& \frac72 [ \ln{(~ 1 + 4 ~)} - \ln{(~ 1 + 9 ~)} ] \\ & &=& \frac72 [ \ln{(~ 5 ~)} - \ln{(~ 10 ~)} ] \\ & &=& \frac72 \ln{( \frac{5}{10} )} \\ & &=& \frac72 \ln{( \frac12 )} \\ & &=& \frac72 [ \ln{( 1 )} - \ln{( 2 )} ] \qquad \ln{( 1 )} = 0\\ & &=& \frac72 [ 0 - \ln{( 2 )} ] \\ & &=& \frac72 [ -\ln{( 2 )} ] \\ & &=& -\frac72 \ln{( 2 )} \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} }&\mathbf{=}& \mathbf{ -2.4260151320 } \\ \end{array} }$

suppose that 12 inches of wire costs 36 cents.At the same rate, how many inches of wire can be bought for 21 cents?

$\small{ \begin{array}{rcll} \dfrac{ 12\ \text{inches} } { 36\ \text{cents} } \cdot 21\ \text{cents} &=& \dfrac{12}{36}\cdot 21 \ \text{inches} \\\\ &=& \dfrac{1}{3}\cdot 21 \ \text{inches} \\\\ &=& \dfrac{21}{3} \ \text{inches} \\\\ &=& 7 \ \text{inches of wire} \\\\ \end{array} }$

A penalty in Meteor-Mania is −2 seconds. A penalty in Cosmic Calamity is −6 seconds. Yolanda had penalties totaling −10 seconds in a game of Meteor-Mania and −30 seconds in a game of Cosmic Calamity. In which game did Yolanda receive more penalties?             is it metor mania cosmic calamity or neither

$\small{ \begin{array}{rcll} \dfrac{ 1\ \text{penalty M-M} } { -2\ \text{seconds} } \cdot (-10)\ \text{seconds} &=& 5\ \text{penalty M-M} \\\\ \dfrac{ 1\ \text{penalty C-C} } { -6\ \text{seconds} }\cdot (-30)\ \text{seconds} &=& 5\ \text{penalty C-C} \end{array} }$

same penalties

Two guy wires are attached to the top of a telecommunications tower and anchored to the ground on opposite sides of the tower, as shown. The length of the guy wire is 20 m more than the height of the tower. The horizontal distance from the base of the tower to where the guy wire is anchored to the ground is one-half the height of the tower. How tall is the tower, to the nearest tenth of a metre?

$\small{ \begin{array}{rcll} \mathbf{ \text{Pythagoras:} }\\ \left( \dfrac{h}{2} \right)^2 + h^2 &=& (h+20)^2 \\ \dfrac{h^2}{4} + h^2 &=& h^2+40h+400 \\ \dfrac{h^2}{4} &=& 40h+400 \\ \dfrac{h^2}{4} -40h-400&=& 0 \qquad & | \qquad \cdot 4 \\ h^2 -160h-1600&=& 0 \\ \hline \boxed{~ \begin{array}{rcll} ax^2+bx +c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~} \\ \hline x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \qquad a = 1 \quad b = -160 \quad c = -1600 \\ h_{1,2} &=& {160 \pm \sqrt{160^2-4\cdot(-1600)} \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160^2+40 \cdot 160 } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160\cdot (160+40) } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160\cdot 200 } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{1600\cdot 20 } \over 2} \\ h_{1,2} &=& {160 \pm 40 \sqrt{20 } \over 2} \\ h_{1,2} &=& {160 \pm 40\cdot 2\cdot \sqrt{ 5 } \over 2} \\ h_{1,2} &=& {160 \pm 80\cdot \sqrt{ 5 } \over 2} \\ h_{1,2} &=& 80 \pm 40\cdot \sqrt{ 5 } \\ h &=& 80 + 40\cdot \sqrt{ 5 } \\ h &=& 40\cdot( 2 + \sqrt{ 5 } ) \\ \mathbf{ h} & \mathbf{=} & \mathbf{169.442719100} \end{array} }$

The tower is 169.4 m tall.

Write an equation of the line that is perpendicular to the line through 9,10 and 3,-2 and passes through the x-intercept of that line.

$\small{ \begin{array}{rcll} \hline \text{line } 1\quad m(\text{slope})=\ ?\\ m &=& \dfrac{y_2-y_1}{x_2-x_1} \qquad ( x_1 = 3,\ y_1=-2 ) \quad ( x_2=9,\ y_2 = 10 )\\ m &=& \dfrac{10-(-2)}{9-3} \\ m &=& \dfrac{12}{6} \\ \mathbf{m} &\mathbf{=}& \mathbf{2} \\\\ m_{\text{perpendicular}} &=& -\dfrac{1}{m} \qquad m=2\\ \mathbf{m_{\text{perpendicular}}} &\mathbf{=}& \mathbf{-\dfrac{1}{2}} \\ \hline \text{line } 1\quad x_{(intercept)}=x_i=\ ?\\ m &=& \dfrac{y-y_1}{x-x_1} \\ m &=& \dfrac{y_i-y_1}{x_i-x_1} \qquad ( x_1 = 3,\ y_1=-2 ) \quad m=2 \quad y_i = 0\\ 2 &=& \dfrac{0-(-2)}{x_i-3} \\ 2 &=& \dfrac{0-(-2)}{x_i-3} \\ 2 &=& \dfrac{2}{x_i-3} \qquad & | \qquad :2 \\ 1 &=& \dfrac{1}{x_i-3} \\ x_i-3 &=& 1 \\ x_i &=& 1+3 \\ x_i &=& 4 \\ \mathbf{\text{Point x-intercept}:\ ( x_i = 4,\ y_i=0 )}\\ \hline \text{line perpendicular:} \\ m_{\text{perpendicular}} &=& \dfrac{y-y_i}{x-x_i} \qquad ( x_i = 4,\ y_i=0 ) \quad m_{\text{perpendicular}} =-\dfrac12 \\ -\dfrac12 &=& \dfrac{y-0}{x-4} \\ -\dfrac12 &=& \dfrac{y}{x-4} \\ y &=& -\dfrac12 \cdot (x-4) \\ y &=& -\dfrac12 \cdot x - ( -\dfrac12 )(4) \\ \mathbf{y} &\mathbf{=}& \mathbf{-\dfrac12 \cdot x +2} \\ \end{array} }$

how do we solve: sqrt(x^2-10x+25)=4 ?

$x^2-10x+25 = (x-5)^2$

$\begin{array}{rcll} \sqrt{x^2-10x+25} &=& 4 \\ \sqrt{(x-5)^2} &=& 4 \\ x-5 &=& 4\\ x &=& 4+5\\ x &=& 9 \end{array}$

Wie rechnet man 1,5 mal 5 hoch -x = 2,5 mit dem logarithmus

$\small{ \begin{array}{rcll} 1,5\cdot 5^{-x} &=& 2,5\\ 1,5\cdot \dfrac{1}{5^x} &=& 2,5\\\\ \dfrac{1}{5^x} &=&\dfrac{ 2,5 } { 1,5 } \qquad & | \qquad \updownarrow\\\\ \dfrac{5^x}{1} &=&\dfrac{ 1,5 }{ 2,5 } \\\\ 5^x &=&0,6 \qquad & | \qquad \log_{10}{} \\\\ \log_{10}{ (5^x) } &=& \log_{10}{ (0,6) } \\\\ x\cdot \log_{10}{ (5) } &=& \log_{10}{ (0,6) } \\\\ x &=& \dfrac{ \log_{10}{ (0,6) } }{ \log_{10}{ (5) } } \\\\ x &=& \dfrac{ -0.2218487496 }{ 0.6989700043 } \\\\ \mathbf{x} &\mathbf{=}& \mathbf{-0.3173938055} \\\\ \end{array} }$

if I climb with my mother on a scale I weighed 103kg, and if I get on with my father, 113kg. If my father and mother together weigh 126kg, how many kilos do the three of us weigh all together?

A)168       B)169       C)170     D)171      E)172

$\small{ \begin{array}{lrcll} (1) & i + \text{ mother } &=& 103\ \text{kg} \\ (2) & i + \text{ father } &=& 113\ \text{kg} \\ (3) & \text{ father }+\text{ mother } &=& 126\ \text{kg} \\ \hline (1)+(2)+(3) & (i + \text{ mother }) + (i+ \text{ father }) + (\text{ father }+\text{ mother }) &=& 103\ \text{kg} + 113\ \text{kg} + 126\ \text{kg}\\ & 2\cdot i + 2\cdot \text{ mother } + 2\cdot \text{ father } &=& 103\ \text{kg} + 113\ \text{kg} + 126\ \text{kg}\\ & 2\cdot i + 2\cdot \text{ mother } + 2\cdot \text{ father } &=& 342\ \text{kg}\\ & 2\cdot( i + \text{ mother } + \text{ father } ) &=& 342\ \text{kg}\\ & i + \text{ mother } + \text{ father } &=& \frac{ 342 } { 2 } \ \text{kg}\\ & \mathbf{ i + \text{ mother } + \text{ father } }& \mathbf{=} & \mathbf{ 171\ \text{kg} }\\ \end{array} }$

The three of us weight all together 171 kg

The answer is D)