heureka

quadrat : a = 26cm

Diagonale d = ?

Hallo preneukaj

$\begin{array}{rcl} d &=& \sqrt{ a^2+ a^2 } \\ d &=& \sqrt{ 2a^2 } \\ d &=& \sqrt{2}\sqrt{a^2} \qquad | \qquad \sqrt{a^2} = a\\ d &=& \sqrt{2}\cdot a \\ \hline d &=& \sqrt{2} \cdot 26\ \text{cm}\\ d &=& 1,41421356237 \cdot 26\ \text{cm}\\ d &=& 36,7695526217\ \text{cm}\\ \text{Die Diagonale ist rund } 36,77 \ \text{cm}\\ \end{array}$

Du hast richtig gerechnet!

Mutter, Vater und Sohn sind zusammen 70 Jahre alt. Der Vater ist 5 Jahre älter als die Mutter. Der Sohn ist 22 Jahre jünger als die Mutter. Wie alt sind Vater,Mutter und Sohn?

Mutter = M

Vater = V

Sohn = S

$\begin{array}{rcl} M+V+S &=&70 \\ V &=& M +5 \\ M &=& S + 22 \\ \hline V &=& M +5 \qquad M = S + 22 \\ V &=& (S+22) + 5 \\ V &=& S+27 \\ \hline M+V+S &=&70 \qquad M = S + 22 \qquad V=S+27 \\ (S + 22)+(S+27)+S &=&70 \\ 3S + 22+27 &=&70 \\ 3S + 49 &=&70 \\ 3S &=&70-49 \\ 3S &=&21 \\ S &=&\frac{21}{3} \\ \mathbf{S} &\mathbf{=}& \mathbf{7} \\ \text{Der Sohn ist } 7 \text{ Jahre alt }\\ \hline M &=& S + 22 \qquad S=7\\ M &=& 7 + 22 \\ \mathbf{M} &\mathbf{=}& \mathbf{29} \\ \text{Die Mutter ist } 29 \text{ Jahre alt }\\ \hline V &=& S+27 \qquad S=7\\ V &=& 7 + 27 \\ \mathbf{V} &\mathbf{=}& \mathbf{34}\\ \text{Der Vater ist } 34 \text{ Jahre alt }\\ \end{array}$

Here's a puzzle for when the site activity is low!

In the diagram below there are two quarter circles of radius 1 and 2 intersecting as shown.  What is the area of their intersection (i.e. the shaded region)?

$circle_1: x^2 + y^2 = 4 \qquad r_1 = 2\\ circle_2: (x-2)^2 + (y-2)^2 = 1 \qquad r_2 = 1\\$

intersections $S_1 \text{ and } S_2$:

$2 x_{s}^2 - 2a \cdot x_s + a^2 - r_1^2 = 0 \qquad a = \frac{3r_1^2-r_2^2}{2r_1} = \frac{11}{4}\\ \vec{S_1} =\dbinom{ \frac18(11-\sqrt{7}) }{ \frac18(11+\sqrt{7}) }\\ \vec{S_2} =\dbinom{ \frac18(11+\sqrt{7}) }{ \frac18(11-\sqrt{7}) }$

sectors angels (cosinus-rule) $\alpha_1 \text{ and } \alpha_2$:

$\cos{(\alpha_1)} = \frac{57}{64} \qquad \alpha_1 = 27.0481105464^{\circ}\\ \cos{(\alpha_2)} = \frac{9}{16} \qquad \alpha_2 = 55.7711336722^{\circ}\\$

Areas sectors $A_{s_1} \text{ and } A_{s_2}$:

$A_{s_1} = 4\pi \frac{27.0481105464^{\circ}}{360^{\circ}}\\ A_{s_2} = \pi \frac{55.7711336722^{\circ}}{360^{\circ}}\\$

Areas triangles $A_{t_1} \text{ and } A_{t_2}$:

$A_{t_1} = \frac{1}{2} \cdot \left| \vec{S_2}\times \vec{S_1} \right| = \frac{11}{32}\sqrt{7}\\ A_{t_2} = \frac{1}{2} \cdot \left| \left[\vec{S_1}-\binom{2}{2}\right] \times \left[\vec{S_2}-\binom{2}{2}\right] \right|= \frac{5}{32}\sqrt{7}$

$\text{The area of their intersection }\quad A = A_{s_1}-A_{t_1}+A_{s_2}-A_{t_2}$

$A=\frac{\pi}{360}(4\cdot 27.0481105464^{\circ}+55.7711336722^{\circ})-\frac{ \sqrt{7} }{2}\\ A= 1.43085212603-1.32287565553\\ \mathbf{A=0.10797647050}$

For the following geometric sequence, find a 10.

1, 1/3, 1/9, 1/27

$a_1=1 \\ a_2=\frac13 \\ a_3=\frac19 \\ a_4=\frac{1}{27} \\ \boxed{~ r = \frac{a_{n}}{a_{n-1}} ~}\\ r=\frac{ a_4 }{ a_3 }=\frac{ a_3 }{ a_2 }=\frac{ a_2 }{ a_1 }\\ r=\frac{ \frac{1}{27}}{ \frac19 }=\frac{ \frac19 }{ \frac13 }=\frac{ \frac13 }{ 1 }\\ r=\frac{9}{27}=\frac{ 3 }{9 }=\frac13\\ r=\frac{1}{3}=\frac{ 1 }{3}=\frac13\\ \mathbf{r=\frac13}$

$\boxed{~ a_n = a_1\cdot r^{n-1} ~}\\ a_1 = 1\cdot \left( \frac13 \right)^{1-1} = 1\\ a_2 = 1\cdot \left( \frac13 \right)^{2-1} = 1\cdot \frac13= \frac13\\ a_3 = 1\cdot \left( \frac13 \right)^{3-1} = 1\cdot \left( \frac13 \right)^2 = \frac19\\ a_4 = 1\cdot \left( \frac13 \right)^{4-1} = 1\cdot \left( \frac13 \right)^3 = \frac{1}{27}\\ \dots \\ a_{10} = 1\cdot \left( \frac13 \right)^{10-1} = 1\cdot \left( \frac13 \right)^9 = \frac{1}{19683}\\ \mathbf{a_{10} = \frac{1}{19683}}\\$

what is t^2+5t-84 factored

$t^2+5t-84 = (t-7)(t+12)$

1/2+2/6 = 5/6

$\begin{array}{rcl} \frac12+\frac26 \\ &=& \frac12 \cdot \frac33 +\frac26 \\ &=& \frac36 +\frac26 \\ &=& \frac56 \\ \mathbf{\frac12+\frac26} & \mathbf{=}&\mathbf{\frac56} = 0.83\overline{3} \end{array}$

Wenn ich von einem Cosinuswert den Winkelgrad berechnen will, benutze ich auf dem Taschenrechner SHIFT und dann cos-1. Hier auf dem Onlinerechner gibt es kein SHIFT. Wie kann ich also dann den Winkelgrad berechnen? Danke!

Deine Shift-Taste heißt hier: $2^{nd}$

wie second ( 2. Funktion )

What is the value of C if the F=185

$\begin{array}{rcl} \boxed{~ ^{\circ}C = \frac{ ^{\circ}F - 32 } {1.8} ~}\\ ^{\circ}C &=& \frac{ ^{\circ}F - 32 } {1.8} \qquad | \qquad ^{\circ}F = 185 \\ ^{\circ}C &=& \frac{ 185 - 32 } {1.8} \\ ^{\circ}C &=& \frac{ 153 } {1.8} \\ \mathbf{^{\circ}C }&\mathbf{=}& \mathbf{85} \\ \end{array}$

if 1 dm cubed = 1 liter, how many liters are in a vessel containing 10.5 dm cubed

$\begin{array}{rcl} 10.5\ \text{dm} \cdot \left( \frac{1\ \text{liter}}{1\ \text{dm}} \right)&=& 10.5\ \text{liter} \end{array}$

x^2-20x+99=0  x = 11 and x = 9

$\begin{array}{rcl} x^2-20x+99 &=&0 \\ \boxed{~ x = {-b \pm \sqrt{b^2-4ac} \over 2a} ~} \\ x &=& {20 \pm \sqrt{20^2-4\cdot 1 \cdot 99 } \over 2} \qquad | \qquad a = 1 \quad b = -20 \quad c = 99 \\ x &=& {20 \pm \sqrt{400-396 } \over 2} \\ x &=& {20 \pm \sqrt{4 } \over 2} \\ x &=& {20 \pm 2 \over 2} \\ x_1 &=& {20 + 2 \over 2} \\ x_1 &=& {22 \over 2} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{11} \\\\ x_2 &=& {20 - 2 \over 2} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{9}\\ \end{array}$