heureka

Divide and simplify.

t^2+8t/t^2+3t-40 ÷ t/t+5

*No brakets*

Thanks!!!

I assume ( t^2+8t ) / ( t^2+3t-40 ) ÷ t/ (t+5)  

\(\begin{array}{rcl} \dfrac{ \dfrac{( t^2+8t )} { ( t^2+3t-40 ) } } { \dfrac{ t} { (t+5) } } &=& \dfrac{ \dfrac{( t^2+8t )} { ( t+8)(t-5 ) } } { \dfrac{ t} { (t+5) } } \\\\ &=& \dfrac{( t^2+8t )} { ( t+8)(t-5 ) } \cdot \dfrac{ (t+5) }{ t} \\\\ &=& \dfrac{( t^2+8t )} { ( t+8)(t ) } \cdot \dfrac{ (t+5) }{ (t-5) } \\\\ &=& \dfrac{( t^2+8t )} { ( t^2+8t ) } \cdot \dfrac{ (t+5) }{ (t-5)} \\\\ &=& \dfrac{ (t+5) }{ (t-5)} \\\\ \end{array}\)

Divide and simplify.

3x-15/121 ÷ x-5/33

\(\begin{array}{rcl} \dfrac{ \frac{3x-15}{121} }{ \frac{x-5}{33} } &=& \dfrac{ \frac{3(x-5)}{121} }{ \frac{x-5}{33} } \\\\ &=& \frac{3(x-5)}{121} \cdot \frac{33}{(x-5)} \\\\ &=& \frac{3\cdot 33}{121} \\\\ &=& \frac{99}{121} \\\\ &=& \frac{9}{11} \end{array}\)

Find sin 2x, cos 2x, and tan 2x from the given information. tan x = −1/2 , cos x > 0

In trigonometry, the tangent half-angle formulas relate the tangent of one half of an angle to trigonometric functions of the entire angle. They are as follows:

\(\begin{array}{rcl} \boxed{~ \begin{array}{rcl} \sin{(\alpha)} &=& \dfrac{ 2\cdot \tan{\frac{\alpha}{2} } } { 1 + \tan^2{\frac{\alpha}{2} } } \\\\ \cos{(\alpha)} &=& \dfrac{ 1 - \tan^2{\frac{\alpha}{2} } } { 1 + \tan^2{\frac{\alpha}{2} } }\\\\ \tan{(\alpha)} &=& \dfrac{ 2\cdot \tan{\frac{\alpha}{2} } } { 1 - \tan^2{\frac{\alpha}{2} } } \end{array} ~} \quad \text{ or } \quad \boxed{~ \begin{array}{rcl} \sin{(2\alpha)} &=& \dfrac{ 2\cdot \tan{\alpha } } { 1 + \tan^2{\alpha } } \\\\ \cos{(2\alpha)} &=& \dfrac{ 1 - \tan^2{\alpha } } { 1 + \tan^2{\alpha } }\\\\ \tan{(2\alpha)} &=& \dfrac{ 2\cdot \tan{\alpha } } { 1 - \tan^2{\alpha } } \end{array} ~} \end{array}\)

\(\begin{array}{rcl} \sin{(2x)} &=& \dfrac{ 2\cdot \tan{x } } { 1 + \tan^2{x } } \\ \sin{(2x)} &=& \dfrac{ 2\cdot ( -\frac12 ) } { 1 + ( -\frac12 )^2 } \\ \sin{(2x)} &=& \dfrac{ -1 } { 1 + \frac14 } \\ \mathbf{\sin{(2x)}} &\mathbf{=}& \mathbf{-\frac45}\\ \\ \cos{(2x)} &=& \dfrac{ 1 - \tan^2{x } } { 1 + \tan^2{x } }\\ \cos{(2x)} &=& \dfrac{ 1 - \frac14 } { 1 + \frac14 }\\ \cos{(2x)} &=& \frac34 \cdot \frac45 \\ \mathbf{\cos{(2x)}} &\mathbf{=}& \mathbf{\frac35} \\ \\ \tan{(2x)} &=& \dfrac{ 2\cdot \tan{x } } { 1 - \tan^2{x } } \\ \tan{(2x)} &=& \dfrac{ 2\cdot ( -\frac12 ) } { 1 - ( -\frac12 )^2 }\\ \tan{(2x)} &=& \dfrac{ -1 } { 1 - \frac14 }\\ \mathbf{\tan{(2x)}} &\mathbf{=}& \mathbf{- \dfrac43}\\ \end{array}\)

Simplify by removing factors of 1.

6y^2-24/9y^2-36

\(\begin{array}{rcl} \dfrac{ 6y^2-24 } { 9y^2-36 } &=& \dfrac{ 2\cdot ( 3y^2-12 ) } { 3\cdot ( 3y^2-12 ) } \\ &=& \dfrac{ 2 } { 3 } \end{array}\)

Find the line perpendicular to 8x-2y=4 that goes through (-12,-1)

Find the line perpendicular to 8x-2y=4 that goes through (-12,-1)

\(\begin{array}{rcl} 8x-2y &=& 4 \qquad | \qquad \cdot (-1) \\ -8x+2y &=& -4 \\ 2y &=& 8x - 4 \\ y &=& \frac{8x-4}{2} \\ y &=& \frac{8x}{2} - \frac{4}{2}\\ y &=& 4x - 2 \\ \end{array}\)

\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (-12,-1) \\\\ y &=& 4x - 2 \qquad m = 4 \\ m_{\text{perpendicular}} &=& -\frac{1}{ 4 } \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-(-1)}{x-(-12)} &=& -\frac{1}{ 4 } \\ y+1 &=& -\frac{1}{ 4 } \cdot (x+12) \\ y+1 &=& -\frac{1}{ 4 } x -\frac{12}{ 4 } \\ y+1 &=& -\frac{1}{ 4 } x -3 \\ y &=& -\frac{1}{ 4 } x -3 -1 \\ y &=& -\frac{1}{ 4 } x -4 \\ \end{array}\)

If 70% of the members remembered the password but 21 members forgot, then how many members were there in all?

m = members

p = 70 %

\(\small{ \begin{array}{rcl} m &=& m\cdot p + \underbrace{(1-p)\cdot m}_{=21} \\ m &=& m\cdot p + 21\\\\ m -m\cdot p &=& 21\\\\ m(1- p) &=& 21\\\\ m &=& \frac{ 21} { 1- p} \qquad | \qquad p = 70 \% \\\\ m &=& \frac{ 21} { 1- 70 \%} \\\\ m &=& \frac{ 21} { 100 \%- 70 \%} \\\\ m &=& \frac{ 21} { 30 \%}\\\\ m &=& \frac{ 21} { \frac{30}{100} } \\\\ m &=& \frac{ 21\cdot 100} { 30 } \\\\ m &=& \frac{ 2100 } { 30 } \\\\ m &=& \frac{ 210 } { 3 } \\\\ m &=& 70 \\ \end{array} }\)

There  were 70 members in all

Wenn ich aus 196.2 m^2/s^2 die Wurzel ziehe, wird die Einheit dann berücksichtigt? Also dann zu m/s ?

\(\begin{array}{rcl} \sqrt{196.2\ \frac{m^2}{s^2} } &=& 14.0071410359 \ \frac{m}{s}\\ \end{array} \)

76/3?

\(\begin{array}{rcl} \frac{76}{3} &=& 25 \frac13\\ &=& 25.3\overline{3} \end{array} \)

Wie berechne ich die Schnittpunktkoordinaten?

In der Ebene

Wobei Punkt 19 und 31 auf einer Gerade liegen und Punkt 23 und 27 auf einer anderen Gerade.

Also wie berechne Ich jetzt den Schnittpunkt beider Geraden? :)

\(\begin{array}{rll} &\text{Gegeben sind zwei Geraden, gesucht ist ihr Schnittpunkt } S(y_s, x_s) ? \\ \end{array}\\ \begin{array}{|r|c|r|r|} \hline & Pkt.-Nr. & y-Koordinate & x-Koordinate \\ \hline \text{Gerade 1} \\ \vec{P_{23}} & 23 & 3708,24 & 1004,48 \\ \vec{P_{27}} & 27 & 3787,27 & 1086,87 \\ \vec{b}=\vec{P_{27}}-\vec{P_{23}} & & b_y=79,03 & b_x= 82,39\\ \hline \text{Gerade 2} \\ \vec{P_{19}} & 19 & 3776,30 & 1043,77 \\ \vec{P_{31}} & 31 & 3721,47 & 1069,43\\ \vec{a}=\vec{P_{31}}-\vec{P_{19}} & & a_y = -54,83 & a_x = 25,66\\ \hline \vec{c}=\vec{P_{19}}-\vec{P_{23}} & & c_y= 68,06& c_x = 39,29 \\ \hline \end{array}\)

\(\begin{array}{rcl} &\text{Wie berechne ich die Schnittpunktkoordinaten } S(y_s, x_s)? \\ \end{array}\\ \begin{array}{rcl} \vec{S} &=& \vec{P_{19}}+\lambda \vec{a} \\ \vec{c}+\lambda \vec{a} &=& \mu \vec{b} \quad | \quad \times \vec{b} \\ |\vec{c}\times \vec{b}| +\lambda |\vec{a}\times \vec{b}| &=& \mu |\vec{b}\times \vec{b}| \qquad |\vec{b}\times \vec{b}| = b^2\sin{ (0^{\circ}) } = 0\\ |\vec{c}\times \vec{b}| +\lambda |\vec{a}\times \vec{b}| &=& 0\\ \lambda &=& -\frac{ |\vec{c}\times \vec{b}| } { |\vec{a}\times \vec{b}| } \\ \vec{S} &=& \vec{P_{19}}- \frac{ |\vec{c}\times \vec{b}| } { |\vec{a}\times \vec{b}| }\cdot \vec{a} \\ \vec{S} &=& \binom{y_{19}}{x_{19}}- \frac{ |\binom{c_y}{c_x}\times \binom{b_y}{b_x}| } { |\binom{a_y}{a_x}\times \binom{b_y}{b_x}| }\cdot \binom{a_y}{a_x} \\ \vec{S} &=& \binom{3776,30}{1043,77} - \frac{ |\binom{68,06}{39,29}\times \binom{79,03}{82,39}| } { |\binom{ -54,83}{25,66}\times \binom{79,03}{82,39}| } \cdot \binom{ -54,83}{25,66} \\ \vec{S} &=& \binom{3776,30}{1043,77}- \frac{ 68,06 \cdot 82,39 - 39,29 \cdot (79,03) } { -54,83 \cdot 82,39 - 25,66 \cdot (79,03) } \cdot \binom{ -54,83}{25,66} \\ \vec{S} &=& \binom{3776,30}{1043,77}- \frac{ 2502,3747 } { -6545,3535 } \cdot \binom{ -54,83}{25,66} \\ \vec{S} &=& \binom{3776,30}{1043,77} +0,3823131478 \cdot \binom{ -54,83}{25,66} \\ \vec{S} &=& \binom{3776,30}{1043,77} + \binom{ -20,9622298935}{9,81015537248} \\ \vec{S} &=& \binom{3755,33777011}{1053,58015537}\\ \end{array}\)

Die y-Koordinate des Schnittpunktes ist 3755,33777011

Die x-Koordinate des Schnittpunktes ist 1053,58015537