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)