heureka

Wie kommt es zu dieser Lösung ? sin(x)=1/Wurzel2 x=pi/4 Bitte helft mir

Umrechnung vom Bogenmaß ins Gradmaß:

$\begin{array}{rcl} \frac{\pi}{4} \cdot \frac{360^\circ }{2\pi} &=& \frac{360^\circ}{8} = 45 ^\circ \\\\ \end{array}$

Wir erhalten also für $\frac{\pi}{4}~ \text{rad} \quad 45^\circ$

Wir suchen also $\begin{array}{rcl} \sin{ (\frac{\pi}{4}) } &=& \sin{ (45 ^\circ ) } \\ \end{array}$

Wir sehen im Schaubild, dass der $\sin{ (45 ^\circ ) }$ gleich $\frac{1}{\sqrt{2}}$ ist.

A natural number greater than 1 is said to be prime if its only factors are 1 and itself. For example, the number 7 is prime since its only factors are 1 and 7. A perfect square is an integer created by multiplying an integer by itself. The number 25 is a perfect square since it is 5 × 5 or 5 Determine the smallest perfect square that has three different prime numbers as factors.

$(2 \cdot 3\cdot 5)^2 = 2^2\cdot 3^2\cdot 5^2 = 4\cdot 9\cdot 25 =\mathbf{ 900}$

$\begin{array}{rcll} 4 \text{ people} \cdot 2.5\ h &=& \frac12\ \text{job} \quad & | \quad \cdot 2 \\ 10\text { people} \cdot x &=& \frac13\ \text{job} \quad & | \quad \cdot 3 \\ \end{array}\\ $

$2\cdot 4 \cdot 2.5 = 1 job = 3\cdot 10 \cdot x\\ x = \frac{2\cdot 4 \cdot 2.5}{3\cdot 10}\ h\\ x = \frac{20}{30} \ h\\ x = \frac23\ h \\ x = 40\ \text{minutes}$​

$\begin{array}{rcl} \begin{bmatrix} 0&-1&1\\ -5&-1&0 \end{bmatrix}\cdot \begin{bmatrix} 1&2&\\ 0&1\\ -2&-1 \end{bmatrix} &=& \begin{bmatrix} \begin{pmatrix}0\\-1\\1\end{pmatrix}\cdot \begin{pmatrix}1\\0\\-2\end{pmatrix} & \begin{pmatrix}0\\-1\\1\end{pmatrix}\cdot \begin{pmatrix}2\\1\\-1\end{pmatrix}\\ \begin{pmatrix}-5\\-1\\-0\end{pmatrix}\cdot \begin{pmatrix}1\\0\\-2\end{pmatrix} & \begin{pmatrix}-5\\-1\\-0\end{pmatrix}\cdot \begin{pmatrix}2\\1\\-1\end{pmatrix}\\ \end{bmatrix}\\\\ &=& \begin{bmatrix} 0\cdot 1 +(-1)\cdot 0 + 1\cdot(-2) & 0\cdot 2 + (-1)\cdot 1 + 1\cdot (-1)\\ (-5)\cdot 1 +(-1)\cdot 0 + 0\cdot(-2) & (-5)\cdot 2 + (-1)\cdot 1 + 0\cdot (-1)\\ \end{bmatrix}\\\\ &=& \begin{bmatrix} 0+0-2 & 0-1-1\\ -5+ 0 + 0 & -10 -1 +0\\ \end{bmatrix}\\\\ &=& \begin{bmatrix} -2 & -2\\ -5 & -11\\ \end{bmatrix} \end{array}$

Matrix H is the result.

Please give the next 4 terms of this sequence. Thanks for any help:

2, 4, 7, 10, 15, 18, 23, 26............640 (is 100th term)

Thanks heureka: brilliant work!. In my book it is stated slightly different: It is the sequence of natual or counting numbers added to prime numbers - 1. So, we have:

Counting numbers: 1, 2, 3, 4, 5, 6............

Prime numbers    + 2, 3, 5, 7, 11, 13...............

Subtract 1  from      2, 4, 7, 10, 15, 18..........etc.

each term

$\text{The formula is}\\ \boxed{~ a_n = a_{n-1} + p(n) -p(n-1) +1 \qquad n \ge 2 \qquad a_1 = 2 ~}\\ \begin{array}{rclcl} a_1 && &=& 2 \\ a_2 &=& a_1 + p(2)-p(1) + 1 = 2 + p(2)-2 + 1 &=& p(2) + 1\\ a_3 &=& a_2 + p(3)-p(2) + 1 = (~p(2) + 1~) + p(3)-p(2) + 1 &=& p(3) + 2 \\ a_4 &=& a_3 + p(4)-p(3) + 1 = (~p(3) + 2~) + p(4)-p(3) + 1 &=& p(4) + 3 \\ a_5 &=& a_4 + p(5)-p(4) + 1 = (~p(4) + 3~) + p(5)-p(4) + 1 &=& p(5) + 4\\ \cdots \\ a_n &=& a_{n-1} + p(n)-p(n-1) + 1 = (~p(n-1)+ n-2 ~)+ p(n)-p(n-1) + 1 &=& p(n) + n - 1\\ \end{array}\\ \boxed{~ a_n = n + p(n) -1 ~}$

If z = 4+i3, then |z| = ?

$\begin{array}{rcrcl} \boxed{~ z = a+ib \qquad |z| = \sqrt{a^2+b^2} ~} \end{array}$

$\begin{array}{rcrcl} z &=& 4+i\cdot 3 \\ |z| &=& \sqrt{4^2+3^2}\\ |z| &=& \sqrt{25}\\ |z| &=& 5 \end{array}$

how to solve what is the sum of the first ten terms of the geometric sequence 4,8,16,.....?

$\begin{array}{rcl} a_1 &=& 2^2 \\ a_2 &=& 2^3 \\ a_3 &=& 2^4 \\ \cdots \\ a_n &=& 2^{n+1} \end{array}$

 $\begin{array}{rcrcl} s_n &=& a_1 &+& a_2+a_3+a_4+ \cdots + a_{n-1} + a_n \\ \hline s_n &=& 2^2 &+& 2^3+2^4+2^5+ \cdots + 2^n + 2^{n+1} \\ 2\cdot s_n &=& && 2^3+2^4+2^5+ \cdots + 2^n+2^{n+1} + 2^{n+2} \\ \hline s_n - 2\cdot s_n &=& 2^2 - 2^{n+2} \\ s_n(1 - 2) &=& 2^2 - 2^{n}\cdot 2^2 \\ -s_n &=& 2^2 - 2^{n}\cdot 2^2 \\ s_n &=& 2^{n}\cdot 2^2 - 2^2\\ s_n &=& 2^2\cdot (2^{n}-1)\\\\ \mathbf{s_n} &\mathbf{=}& \mathbf{4\cdot (2^{n}-1)} \\\\ s_{10} &=& 4\cdot (2^{10}-1)\\ s_{10} &=& 4\cdot (1024-1)\\ s_{10} &=& 4\cdot 1023\\ \mathbf{s_{10}} &\mathbf{=} & \mathbf{4092} \end{array}$

The sum of the first ten terms of the geometric sequence is 4092

whats the angle for C if a = 8 b = 6 c = 7

cos-rule

$\begin{array}{rcl} c^2 &=& a^2+b^2-2ab\cos{(C)} \\ c^2+2ab\cos{(C)} &=& a^2+b^2 \\ 2ab\cos{(C)} &=& a^2+b^2-c^2 \\ \cos{(C)} &=& \frac{a^2+b^2-c^2} {2ab} \\ \cos{(C)} &=& \frac{8^2+6^2-7^2} {2\cdot 8 \cdot 6} \\ \cos{(C)} &=& \frac{64+36-49} {96} \\ \cos{(C)} &=& \frac{51} {96} = \frac{17} {32} \\ \cos{(C)} &=& 0.53125 \\ C &=& 57.9100487437^\circ \end{array}$

A karate expert breaks a stack of bricks with his bare hand. If the force applied is 520 newtons and the impact time is 0.5 × 10ȳ seconds, what is the value of impuls

I assume

$F = 520\ N$

$t = 0.5 \cdot 10^{-3} \ s$

$p = \text{impuls in}\ Ns$

$\begin{array}{rcl} p &=& F\cdot t\\ p &=& 520\ N \cdot 0.5 \cdot 10^{-3} \ s\\ p &=& 520\cdot 0.5 \cdot 10^{-3} \ Ns\\ p &=& 520\cdot 0.0005 \ Ns\\ p &=& 0.26 \ Ns\\ \end{array}$

how do you get rid of the bottom? [(x^3-2x)-(a^3-2a)]/(x-a)

$\begin{array}{rcl} \frac{ (x^3-2x)-(a^3-2a) } {x-a} &=& \frac{ x^3-2x-a^3+2a } {x-a}\\ &=& \frac{ x^3-a^3-2x+2a } {x-a}\\ &=& \frac{ x^3-a^3-2(x-a) } {x-a}\\ &=& \frac{ x^3-a^3 }{x-a} - 2\cdot( \frac{ x-a } {x-a} ) \\ &=& \frac{ x^3-a^3 }{x-a} - 2\\ \end{array}\\ \boxed{~ \text{Difference of two cubes:}\\ \begin{array}{rcl} x^3-a^3 &=& (x-a)(x^2+x\cdot a+a^2) \end{array} ~}\\ \begin{array}{rcl} \frac{ (x^3-2x)-(a^3-2a) } {x-a} &=& \frac{ x^3-a^3 }{x-a} - 2\\ &=&\frac{ (x-a)(x^2+x\cdot a+a^2) }{x-a} - 2\\ &=& x^2+x\cdot a+a^2 - 2\\ \end{array}$